course MTH 152 Resubmitting the assignments I ran for 1 - 10.
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12:20:19 `q001. Note that there are 14 questions in this assignment. List all possible 3-letter 'words' that can be formed from the set of letters { a, b, c } without repeating any of the letters. Possible 'words' include 'acb' and 'bac'; however 'aba' is not permitted here because the letter 'a' is used twice.
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RESPONSE --> abc, acb, bac, bca, cab, cba
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12:20:33 There are 2 'words' that can be formed starting with the first letter, a. They are abc and acb. There are 2 'words' that can be formed starting with the second letter, b. They are bac and bca. There are 2 'words' that can be formed starting with the third letter, c. They are cab and cba. Note that this listing is systematic in that it is alphabetical: abc, acb, bac, bca, cab, cba. It is important when listing things to be as systematic as possible, in order to avoid duplications and omissions.
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RESPONSE --> okay
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12:25:10 `q002. List all possible 3-letter 'words' that can be formed from the set of letters { a, b, c } if we allow repetition of letters. Possible 'words' include 'acb' and 'bac' as before; now 'aba' is permitted, as is 'ccc'.
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RESPONSE --> aaa, aab, aac, aba, abb, abc, aca, acb, acc, baa, bab, bac, bba, bbb, bbc, bca, bcb, bcc, caa, cab, cac, cba, cbb, cbc, cca, ccb, ccc
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12:25:34 Listing alphabetically the first possibility is aaa. There are 2 more possibilities starting with aa: aab and aac. There are 3 possibilities that start with ab: aba, abb and abc. Then there are 3 more starting with ac: aca, acb and acc. These are the only possible 3-letter 'words' from the set that with a. Thus there are a total of 9 such 'words' starting with a. There are also 9 'words' starting with b: baa, bab, bac; bba, bbb, bbc; bca, bcb and bcc, again listing in alphabetical order. There are finally 9 'words' starting with c: caa, cab, cac; cba, cbb, cbc; cca, ccb, ccc. We see that there are 9 + 9 + 9 = 27 possible 3-letter 'words'.
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RESPONSE --> okay
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12:28:11 `q003. If we form a 3-letter 'word' from the set {a, b, c}, not allowing repetitions, then how many choices do we have for the first letter chosen? How many choices do we then have for the second letter? How many choices do we therefore have for the 2-letter 'word' formed by the first two letters chosen? How many choices are then left for the third letter? How many choices does this make for the 3-letter 'word'?
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RESPONSE --> choices for first letter: 3 choices for second letter: 2 choices for 2-letter word: 6 because 3*2=6 choices for third: 1 choices for 3-letter word: 6*1= 6?
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12:28:52 There are 3 choices for the first letter. The choices are a, b and c. Recall that repetition is not permitted. So having chosen the first letter, whichever letter is chosen, there are only 2 possible choices left. The question arises whether there are now 2 + 3 = 5 or 3 * 2 = 6 possibilities for the first two letters chosen. The correct choice is 3 * 2 = 6. This is because for each of the 3 possible choices for the first letter, there are 2 possible choices for the second. This result illustrates the Fundamental Counting Principal: If we make a number of distinct choices in a sequence, the net number of possibilities is the product of the numbers of possibilities for each individual choice. By the time we get to the third letter, we have only one letter left, so there is only one possible choice. Thus the first two letters completely determine the third, and there are still only six possibilites. The Fundamental Counting Principal confirms this: the total number possibilities must be 3 * 2 * 1 = 6.
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RESPONSE --> Okay, I wasn't sure if it would have been 6, but now I know that I was right.
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12:29:40 `q004. Check your answer to the last problem by listing the possibilities for the first two letters. Does your answer to that question match your list?
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RESPONSE --> ab, ac, ba, bc, ca, cb My answer does match the answer to the last question.
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12:30:09 Listing helps clarify the situation. The first two letters could be ab, ac, ba, bc, ca or cb. Having determined the first two, the third is determined: for example if the first to letters are ba the third must be c. The possibilities for the three-letter 'words' are thus abc, acb, bac, bca, cab and cba; note that this list is obtained by simply adding the necessary letter to each of the two-letter sequences ab, ac, ba, bc, ca and cb.
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RESPONSE --> okay, I understand.
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12:31:30 `q005. If we form a 3-letter 'word' from the set {a, b, c}, allowing repetitions, then how many choices do we have for the first letter chosen? How many choices do we then have for the second letter? How many choices do we therefore have for the 2-letter 'word' formed by the first two letters chosen? How many choices are then left for the third letter? How many choices does this make for the 3-letter 'word'?
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RESPONSE --> choices for first letter: 3 choices for second letter: 3 choices for 2letter word: 3*3=9 choices for third letter: 3 choices for 3letter word: 9*3= 27
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12:31:40 As before there are 3 choices for the first letter. However this time repetition is permitted so there are also 3 choices for the second letter and 3 choices for the third. By the Fundamental Counting Principal there are therefore 3 * 3 * 3 = 27 possibilities. Note that this result agrees with result obtained earlier by listing.
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RESPONSE --> Okay, I understand.
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12:32:53 `q006. If we were to form a 3-letter 'word' from the set {a, b, c, d}, without allowing a letter to be repeated, how many choices would we have for the first letter chosen? How many choices would we then have for the second letter? How many choices would we therefore have for the 2-letter 'word' formed by the first two letters chosen? How many choices would then be left for the third letter? How many possibilities does this make for the 3-letter 'word'?
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RESPONSE --> choices for first letter: 4 choices for second letter: 3 choices for 2letter word: 4*3=12 choices for third letter: 2 choices for 3letter word: 12*2 = 24
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12:33:00 The first letter chosen could be any of the 4 letters in the set. The second choice could then be any of the 3 letters that remain. The third choice could then be any of the 2 letters that still remain. By the Fundamental Counting Principal there are thus 4 * 3 * 2 = 24 possible three-letter 'words'.
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RESPONSE --> OKay, I understand
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12:37:02 `q007. List the 4-letter 'words' you can form from the set {a, b, c, d}, without allowing repetition of letters within a word. Does your list confirm your answer to the preceding question?
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RESPONSE --> abc, abd, acb, acd, adb, adc, bac, bad, bca, bcd, bda, bdc, cab, cad, cba, cbd, cda, cdb, dab, dac, dba, dbc, dca, dcb You asked in this question for 4letter words, but I understood you wanted the 3letter words from the 4 letters because of the previous question. And, yes my answers did match up.
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12:37:59 Listing alphabetically we have abcd, abdc, acbd, acdb, adbc, adcb; bacd, badc, bcad, bcda, bdac, bdca; cabd, cadb, cbad, cbda, cdab, cdba; dabc, dacb, dbac, dbca, dcab, dcba. There are six possibilities starting with each of the four letters in the set.
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RESPONSE --> That was just confusing. Did you want 4letter words or 3letter words?
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12:40:30 `q008. Imagine three boxes, one containing a set of billiard balls numbered 1 through 15, another containing a set of letter tiles with one tile for each letter of the alphabet, and a third box containing colored rings, one for each color of the rainbow (these colors are red, orange, yellow, green, blue, indigo and violet, abbreviated ROY G BIV). If one object is chosen from each box, how many possibilities are there for the collection of objects chosen?
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RESPONSE --> choice one: 15 choice two: 26 choice three: 7 15*26*7 = 2730 possibilities
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12:40:49 There are 15 possible choices from the first box, 26 from second, and 7 from the third. The total number of possibilities is therefore 15 * 26 * 7 = 2730. It would be possible to list the possibilities: 1 a R, 1 a O, 1 a Y, ..., 1 a V 1 b R, 1 b O, ..., 1 b V, 1 c R, 1 c O, ..., 1 c V, ... , 1 z R, 1 z O, ..., 1 z V, 2 a R, 2 a O, ..., 2 a V, etc., etc. This listing would be possible, not really difficult, but impractical because it would take hours. The Fundamental Counting Principle ensures that our result is accurate.
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RESPONSE --> Okay, I understand
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12:41:59 `q009. For the three boxes of the preceding problem, how many of the possible 3-object collections contain an odd number?
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RESPONSE --> 8 odd numbers in 1st box, so 1st choice: 8 2nd choice: 26 3rd choice: 7 8*26*7 = 1456
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12:42:21 The only possible odd number will come from the ball chosen from the first box. Of the 15 balls in the first box, 8 are labeled with odd numbers. There are thus 8 possible choices from the first box which will result in the presence of an odd number. The condition that our 3-object collection include an odd number places no restriction on our second and third choices. We can still choose any of the 26 letters of the alphabet and any of the seven colors of the rainbow. The number of possible collections which include an odd number is therefore 8 * 26 * 7 = 1456. Note that this is a little more than half of the 2730 possibilities. Thus if we chose randomly from each box, we would have a little better than a 50% chance of obtaining a collection which includes an odd number.
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RESPONSE --> Okay, I understand.
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12:44:15 `q010. For the three boxes of the preceding problem, how many of the possible collections contain an odd number and a vowel?
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RESPONSE --> I'm not including Y as a vowel for this problem. 1st choice: 8odd 2nd choice: 5 vowels 3rd choice: 7 colors 8*5*7 = 280
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12:44:25 In this case we have 8 possible choices from the first box and, if we consider only a, e, i, o and u to be vowels, we have only 5 possible choices from second box. We still have 7 possible choices from the third box, but the number of acceptable 3-object collections is now only 8 * 5 * 7 = 280, just a little over 1/10 of the 2730 unrestricted possibilities.
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RESPONSE --> OKay, I understand
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12:45:48 `q011. For the three boxes of the preceding problem, how many of the possible collections contain an even number, a consonant and one of the first three colors of the rainbow?
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RESPONSE --> 1st choice: 7 even 2nd choice: 21 consonants 3rd choice: 3 colors 7*21*3 = 441
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12:46:06 There are 7 even numbers between 1 and 15, and if we count y as a constant there are 21 consonant in the alphabet. There are therefore 7 * 21 * 3 = 441 possible 3-object collections containing an even number, a consonant, and one of the first three colors of rainbow.
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RESPONSE --> Okay, I understand.
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12:51:33 `q012. For the three boxes of the preceding problem, how many of the possible collections contain an even number or a vowel?
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RESPONSE --> In this problem, the possible collections could be in 1 of 2 catagories, be even without a vowel, or be odd with a vowel, so I'm going to find the number of choices in each group and add them. even without a vowel: 7even*21consonants*7colors = 1029 choices odd with vowel: 8odd*5vowels*7colors = 280 1309 choices total
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12:52:39 There are 7 * 26 * 7 = 1274 collections which contain an even number. There are 15 * 5 * 7 = 525 collections which contain a vowel. It would seem that there must therefore be 1274 + 525 = 1799 collections which contain one or the other. However, this is not the case. Some of the 1274 collections containing an even number also contain a vowel, and are therefore included in the 525 collections containing vowels. If we add the 1274 and the 525 we are counting each of these even-number-and-vowel collections twice. We can correct for this error by determining how many of the collections in fact contain an even number AND a vowel. This number is easily found by the Fundamental Counting Principle to be 7 * 5 * 7 = 245. All of these 245 collections would be counted twice if we added 1274 to 525. If we subtract this number from the sum 1274 + 525, we will have the correct number of collections. The number of collections containing an even number or a vowel is therefore 1274 + 525 - 245 = 1555. This is an instance of the formula n(A U B) = n(A) + n(B) - n(A ^ B), where A U B is the intersection of two sets and A^B is their intersection and n(S) stands for the number of objects in the set. Here A U B is the set of all collections containing a letter or a vowel, A and B are the sets of collections containing a vowel and a consonant, respectively and A ^ B is the set of collections containing a vowel and a consonant.
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RESPONSE --> Okay, I understand why I wasn't right.
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12:54:47 `q013. For the three boxes of the preceding problem, if we choose two balls from the first box, then a tile from the second and a ring from the third, how many possible outcomes are there?
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RESPONSE --> 1st choice: 15 2nd choice: 14 3rd choice: 26 4th choice: 7 possible outcomes: 15*14*26*7= 38220
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12:56:14 There are 15 possibilities for the first ball chosen, which leaves 14 possibilities for the second. There are 26 possibilities for the tile and 7 for the ring. We thus have 15 * 14 * 26 * 7 possibilities. However the correct answer really depends on what we're going to do with the objects. This has not been specified in the problem. For example, if we are going to place the items in the order chosen, then there are indeed 15 * 14 * 26 * 7 possibilities. On the other hand, if we're just going to toss the items into a box with no regard for order, then it doesn't matter which ball was chosen first. Since the two balls in any collection could have been chosen in either of two orders, there are only half as many possibilities: we would have just 15 * 14 * 26 * 7 / 2 possible ways to choose an unordered collection.
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RESPONSE --> Okay, but was I right because the problem didn't specify this?
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12:57:19 `q014. For the three boxes of the preceding problem, if we choose only from the first box, and choose three balls, how many possible collections are there?
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RESPONSE --> 15choices*14choices*13choices = 2730 But if there is no regard as you said in the previous problem then there should only be 2730/ 3 = 910 possibilities
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12:58:03 There are 15 possibilities for the first ball chosen, 14 for the second, and 13 for the third. If the collection is going to be placed in the order chosen there are therefore 15 * 14 * 13 possible outcomes. On the other hand, if the collections are going to be just tossed into a container with no regard for order, then there are fewer possible outcomes. Whatever three objects are chosen, they could have been chosen in any of 3 * 2 * 1 = 6 possible orders (there are 3 choices for the first of the three objects that got chosen, 2 choices for the second and only 1 choice of the third). This would mean that there are only 1/6 has many possibilities. So if the order in which the objects are chosen doesn't matter, there are only 15 * 14 * 13 / 6 possibilities.
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RESPONSE --> wow, okay that's really confusing but I think I understand it enough.
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ϩrM{jwW} Student Name: assignment #002
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13:17:31 `q001. Note that there are 8 questions in this assignment. If we choose three letter tiles from a complete set, which consists of one tile for each letter, then how many three-letter 'words' are possible ?
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RESPONSE --> 1st choice: 26 2nd choice: 25 3rd choice: 24 26*25*24 = 15600
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13:18:29 There are 26 choices for the first tile chosen, 25 for the second and 24 for the third. The number of possible three-letter words with 3 distinct letters of the alphabet is therefore 26 * 25 * 24.
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RESPONSE --> Okay, I understand
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13:22:45 `q002. If we choose three letter tiles from a complete set, then how many unordered collections of three letters are possible?
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RESPONSE --> Once three letter tiles have been chosen there are 3*2*1 = 6 orders in which it can be arranged, so 26*25*24 = 15600 / 6 = 2600 possibilities
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13:23:01 If the 3-tile collections are unordered there are only 1/6 as many possibilities, since there are 3 * 2 * 1 = 6 orders in which any collection could have been chosen. Since there are 26 * 25 * 24 ways to choose the 3 tiles in order, there are thus 26 * 25 * 24 / 6 possibilities for unordered choices.
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RESPONSE --> OKay, I understand
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13:25:44 `q003. If we choose two balls from fifteen balls, numbered 1 - 15, from the first box of the preceding problem set, and do so without replacing the first ball chosen, we can get totals like 3 + 7 = 10, or 2 + 14 = 16, etc.. How many of the possible unordered outcomes give us a total of less than 29?
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RESPONSE --> All totals obtained will be less than 29, so 15*14 = 210 poss / 2 = 105 poss
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13:26:21 The smallest possible total would be 1 + 2 = 3 and the greatest possible total would be 14 + 15 = 29. We quickly see that the only way to get a total of 29 is to have chosen 14 and 15, in either order. Thus out of the 15 * 14 / 2 = 105 possible unordered combinations of two balls, only one gives us a total of at least 29. The remaining 104 possible combinations give us a total of less than 29. This problem illustrates how it is sometimes easier to analyze what doesn't happen than to analyze what does. In this case we were looking for totals less than 29, but it was easier to find the number of totals that were not less than 29. Having found that number we easily found the number we were seeking.
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RESPONSE --> Oh okay, I understand now that I should have subtracted the 14+15 total, to make 104
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13:34:02 `q004. If we place each object in all the three boxes (one containing 15 numbered balls, another 26 letter tiles, the third seven colored rings) in a small bag and add packing so that each bag looks and feels the same as every other, and if we then thoroughly mix the contents of the three boxes into a single large box before we pick out two bags at random, how many of the possible combinations will have two rings? How many of the possible combinations will have two tiles? How many of the possible combinations will have a tile and a ring? How many of the possible combinations will include a tile?
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RESPONSE --> poss with 2 rings: 7*6 = 42 poss with 2 tiles: 26*25 = 650 poss with tile & ring: 26*7 = 182 poss with tile: 26*22 = 572
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13:36:56 There are a total of 7 rings. There are thus 7 ways the first bag could have contained a ring, leaving 6 ways in which the second bag could have contained a ring. It follows that 7 * 6 / 2 = 21 of the possible combinations will contain 2 rings (note that we divide by 2 because each combination could occur in two different orders). Reasoning similarly we see that there are 26 ways the first bag could have contained a tile and 25 ways the second bag could have contained a tile, so that there are 26 * 25 / 2 = 325 possible combinations which contain 2 tiles. Since there are 26 tiles and 7 rings, there are 26 * 7 / 2 = 91 possible two-bag combinations containing a tile and a ring. There are a total of 15 + 26 + 7 = 48 bags, so the total number of possible two-bag combinations is 48 * 47 / 2. Since 15 + 7 = 22 of the bags do not contain tiles, there are 22 * 21 / 2 two-bag combinations with no tiles. The number of possible combinations which do include tiles is therefore the difference 48 * 47 / 2 - 22 * 21 / 2 between the number of no-tile combinations and the total number of possible combinations.
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RESPONSE --> I didn't divide any of the bags by 2 and that was my mistake, I understand why I should have though. I didn't understand how to find the last question, but now I know there are 48 bags and 22 don't contain tiles, so I would have subtracted the number of bags without tiles from the total number of bags.
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13:44:32 `q005. Suppose we have mixed the contents of the three boxes as described above. If we pick five bags at random, then in how many ways can we get a ball, then two tiles, then a ring, then another ball, in that order?
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RESPONSE --> 15*26*25*7*14 = 955500
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13:44:48 There are 15 bags containing balls, so there are 15 ways to get a ball on the first selection. If a ball is chosen on the first selection, there are still 26 bags containing tiles when the second selection is made. So there are 26 ways to get a tile on the second selection. At this point there are 25 tiles so there are 25 ways to get a tile on the third selection. There are still 7 rings from which to select, so that there are 7 ways the fourth choice can be a ring. Since 1 ball has been chosen already, there are 14 ways that the fifth choice can be a ball. To get the specified choices in the indicated order, then, there are 15 * 26 * 25 * 7 * 14 ways.
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RESPONSE --> Okay, I understand
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13:48:26 `q006. Suppose we have mixed the contents of the three boxes as described above. If we pick five bags at random, then in how many ways can we get two balls, two tiles and a ring in any order?
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RESPONSE --> 15*14*26*25*7 / 5*4*3*2*1 = 955500/ 120 = 7962.5
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13:49:33 There are 15 * 14 possible outcomes when 2 balls are chosen in order, and 15 * 14 / 2 possible outcomes when the order doesn't matter. There are similarly 26 * 25 / 2 possible outcomes when 2 tiles are choose without regard for order. There are 7 possible choices for the one ring. Thus we have [ 15 * 14 / 2 ] * [ 26 * 25 / 2 ] * 7 ways in which to choose 2 balls, 2 tiles and a ring. Another way to get the same result is to start with the 15 * 26 * 25 * 7 * 14 ways to choose the 2 balls, 2 tiles and one ring in a specified order, as shown in the last problem. Whichever 2 tiles are chosen, they could have been chosen in the opposite order, so if the order of tiles doesn't matter there are only half as many possible outcomes--i.e., 15 * 26 * 25 * 7 * 14 / 2 possibilities if the order of the tiles doesn't matter that the order of the balls does. If the order of the balls doesn't matter either, then we have half this many, or 15 * 26 * 25 * 7 * 14 / ( 2 * 2) ways. It should be easy to see why this expression is identical to the expression [ 15 * 14 / 2 ] * [ 26 * 25 / 2 ] * 7 obtained by the first analysis of this problem.
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RESPONSE --> Okay, I would've done the 5 factorial if all of the chosen objects were the same type. I understand why I should do it this way now.
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13:51:19 `q007. Suppose we have mixed the contents of the three boxes as described above. If we pick five bags at random, then in how many ways can we get a collection of objects that does not contain a tile?
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RESPONSE --> There are a total of 22 bags not containing a tile, so 22*21*20*19*18 / 5*4*3*2*1 = 26334
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13:51:42 Of the 48 bags, 22 do not contain a tile. If we pick five bags at random, then there are 22 * 21 * 20 * 19 * 18 ways in which the five bags could all contain something besides a tile.
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RESPONSE --> I did non specific, now I see the problem didn't specify this.
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13:52:47 `q008. Suppose the balls, tiles and rings are back in their original boxes. If we choose three balls, each time replacing the ball and thoroughly mixing the contents of the box, then two tiles, again replacing and mixing after each choice, then how many 5-character 'words' consisting of 3 numbers followed by 2 letters could be formed from the results?
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RESPONSE --> 15*15*15*26*26
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13:52:57 Since the order of the characters makes a difference when forming 'words', the order of the choices does matter in this case. We have 15 balls from which to choose, so that if we choose with replacement there are 15 possible outcomes for every choice of a ball. Similarly there are 26 possible outcomes for every choice of a tile. Since we first choose 3 balls then 2 tiles, there are 15 * 15 * 15 * 26 * 26 possible 5-character 'words'.
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RESPONSE --> I understand
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ߠLjpbŶƳJ Student Name: assignment #003
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13:59:40 `q001. Note that there are 13 questions in this assignment. As we have seen if we choose, say, 3 objects out of 10 distinct objects the number of possible results depends on whether order matters or not. For the present example if order does matter there are 10 choices for the first selection, 9 for the second and 8 for the third, giving us 10 * 9 * 8 possibilities. However if order does not matter then whatever three objects are chosen, they could have been chosen in 3 * 2 * 1 = 6 different orders. This results in only 1/6 as many possibilities, or 10 * 9 * 8 / 6 possible outcomes. We usually write this number as 10 * 9 * 8 / (3 * 2 * 1) in order to remind us that there are 10 * 9 * 8 ordered outcomes, but 3 * 2 * 1 orders in which any three objects can be chosen. If we were to choose 4 objects out of 12, how many possibile outcomes would there be if the objects were chosen in order? How many possible outcomes would there be if the order of the objects did not matter?
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RESPONSE --> in order: 12*11*10*9 no order: 12*11*10*9/(4*3*2*1)
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13:59:59 When choosing 4 objects out of 12, there are 12 choices for the first, 11 choices for the second, 10 choices for the third and 9 choices for the fourth object. If the order matters there are therefore 12 * 11 * 10 * 9 possible outcomes. If the order doesn't matter, then we have to ask in how many different orders any given collection of 4 objects could be chosen. Given any 4 objects, there are 4 choices for the first, 3 choices for the second, 2 choices for the third and 1 choice for the fourth. There are thus 4 * 3 * 2 * 1 orders in which a given set of 4 objects could be chosen. We therefore have 12 * 11 * 10 * 9 / ( 4 * 3 * 2 * 1) possible outcomes when order doesn't matter.
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RESPONSE --> okay, I understand
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14:00:39 `q002. If order does not matter, then how many ways are there to choose 5 members of a team from 23 potential players?
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RESPONSE --> 23*22*21*20*19/ (5*4*3*2*1)
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14:00:52 If order did matter then there would be 23 * 22 * 21 * 20 * 19 ways choose the five members. However order does not matter, so we must divide this number by the 5 * 4 * 3 * 2 * 1 ways in which any given set of five individuals can be chosen. We therefore have 23 * 22 * 21 * 20 * 19 / ( 5 * 4 * 3 * 2 * 1) possible 5-member teams.
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RESPONSE --> Okay, I understand.
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14:01:43 `q003. In how many ways can we line up 5 different books on a shelf?
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RESPONSE --> 5*4*3*2*1 = 120
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14:01:55 It should be clear that there are 5 * 4 * 3 * 2 * 1 ways, since there are 5 choices for the first book, 4 for the second, etc.. If we multiply these numbers out we get 5 * 4 * 3 * 2 * 1 = 120. It might be a little bit surprising that there should be 120 ways to order only 5 objects.
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RESPONSE --> Okay I understand
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14:04:10 `q004. The expression 5 * 4 * 3 * 2 * 1 is often written as 5 ! , read 'five factorial'. More generally if n stands for any number, then n ! stands for the number of ways in which n distinct objects could be lined up. Find 6 ! , 7 ! and 10 ! .
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RESPONSE --> 6! = 6*5*4*3*2*1 = 720 7! = 7*6*5*4*3*2*1 = 5040 10! = 10*9*8*7*6*5*4*3*2*1 = 3628800
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14:05:26 6 ! = 6 * 5 * 4 * 3 * 2 * 1 = 720. 7 ! = 7 * 6 * 5 * 4 * 3 * 2 * 1 = 5040. 10 ! = 3,628,800. These numbers grow at an astonishing rate. The last result here shows is that there are over 3 million ways to arrange 10 people in a line. The rapid growth of these results like in part explain the use of the ! symbol to designate factorials.
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RESPONSE --> Okay, I understand
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14:07:13 `q005. What do we get if we simplify the expression (10 ! / 6 !) ?
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RESPONSE --> If we simplified this we would get 10*9*8*7 because the 6! would be taken out
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14:07:36 10 ! / 6 ! = 10 * 9 * 8 * 7 * 6 * 5 * 4 * 3 * 2 * 1 / ( 6 * 5 * 4 * 3 * 2 * 1). We can simplify this by rewriting it as 10 * 9 * 8 * 7 * (6 * 5 * 4 * 3 * 2 * 1) / ( 6 * 5 * 4 * 3 * 2 * 1) = 10 * 9 * 8 * 7. We see that the 6 * 5 * 4 * 3 * 2 * 1 in the numerator matches the same expression in the denominator, so when divided these expressions give us 1 and we end up with just 10 * 9 * 8 * 7 * 1 = 10 * 9 * 8 * 7. Note that this is just the number of ways in which 4 objects can be chosen, in order, from a collection of 10 objects.
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RESPONSE --> Okay, I understand
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14:08:31 `q006. We saw above that there are 23 * 22 * 21 * 20 * 19 ways to choose 5 individuals, in order, from 23 potential members. How could we express this number as a quotient of two factorials?
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RESPONSE --> 23!/ (23-5)!
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14:08:48 If we divide 23 ! by 18 ! , the numbers from 18 down to 1 will occur in both the numerator and denominator and when we divide we will be left with just the numbers from 23 down to 19. Thus 23 * 22 * 21 * 20 * 19 = 23 ! / 18 !.
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RESPONSE --> I had the same thing not simplified. I understand this
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14:09:17 `q007. How could we express the number of ways to rank 20 individuals, in order, from among 100 candidates?
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RESPONSE --> 100!/80!
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14:09:35 There are 100 choices for the first candidate, 99 for the second, 98 for the third, etc.. For the 20th candidate there are 81 choices. You should convince yourself of this if you didn't see it originally. Our product is therefore 100 * 99 * 98 * ... * 81, which can be expressed as 100 ! / 80 !. We see that the denominator must be (100 - 20)! . For this example 100 - 20 represents the difference between the number of individuals available and the number selected.
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RESPONSE --> I understand
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14:10:19 `q008. How could we express the number of ways to rank r individuals from a collection of n candidates?
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RESPONSE --> n!/ (n-r)!
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14:10:34 By analogy with the preceding example, we should divide n ! by ( n - r ) !. The number is therefore n ! / ( n - r ) !.
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RESPONSE --> I understand
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14:12:52 `q009. The expression n ! / ( n - r ) ! denotes the number of ways in which r objects can be chosen, in order, from among n objects. When we choose objects in order we say that we are 'permuting' the objects. The expression n ! / ( n - r ) ! is therefore said to the the number of permutations of r objects chosen from n possible objects. We use the notation P ( n , r ) to denote this number. Thus P(n, r) = n ! / ( n - r ) ! . Find P ( 8, 3) and explain what this number means.
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RESPONSE --> P (8, 3) = 8!/ 5! = 336 This is number of ways in which 8 objects can be taken 3 ways
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14:12:52 `q009. The expression n ! / ( n - r ) ! denotes the number of ways in which r objects can be chosen, in order, from among n objects. When we choose objects in order we say that we are 'permuting' the objects. The expression n ! / ( n - r ) ! is therefore said to the the number of permutations of r objects chosen from n possible objects. We use the notation P ( n , r ) to denote this number. Thus P(n, r) = n ! / ( n - r ) ! . Find P ( 8, 3) and explain what this number means.
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RESPONSE -->
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14:13:20 P(n, r) = n! / ( n - r) !. To calculate P(8, 3) we let n = 8 and r = 3. We get P(8, 3) = 8 ! / ( 8 - 3) ! = 8 ! / 5 ! = 8 * 7 * 6 * 5 * 4 * 3 * 2 * 1 / ( 5 * 4 * 3 * 2 * 1) = 8 * 7 * 6. This number represents the number of ways 3 objects can be chosen, in order, from 8 objects.
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RESPONSE --> I understand
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14:14:43 `q010. In how many ways can an unordered collection of 3 objects be chosen from 8 candidates?
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RESPONSE --> 8!/ 3! 5! = 56
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14:14:52 There are 8 * 7 * 6 ways to choose 3 objects from 8, in order, and 3! ways to order any unordered collection of 3 objects, so there are 8 * 7 * 6 / ( 3 * 2 * 1 ) possible unordered collections. This number is easily enough calculated. Since 3 goes into 6 twice and 2 goes into 8 four times, we see that 8 * 7 * 6 / ( 3 * 2 * 1) = 4 * 7 * 2 = 56. There are 56 different unordered collections of 3 objects chosen from 8.
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RESPONSE --> Okay, I understand
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14:16:12 `q010. How could the result of the preceding problem be expressed purely in terms of factorials?
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RESPONSE --> 8!/ 3!5!
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14:16:28 The product 8 * 7 * 6 is just 8 ! / 5 !, and the expression 8 * 7 * 6 / ( 3 * 2 * 1) can therefore be expressed as 8 ! / ( 5 ! * 3 !).
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RESPONSE --> Okay, I understand.
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14:16:58 `q011. In terms of factorials, how would we express the number of possible unordered collections of 5 objects chosen from 16?
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RESPONSE --> 16!/ (5! *11!)
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14:17:12 There are 16 ! / ( 16 - 5) ! Possible ordered sets of 5 objects chosen from the 16. There are 5 ! ways to order any unordered collection of 5 objects. There are thus 16 ! / [ ( 16 - 5 ) ! * 5 ! ] possible unordered collections of 5 objects from the 16.
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RESPONSE --> Okay, I understand
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14:17:56 `q012. In terms of factorials, how would we express the number of possible unordered collections of r objects chosen from n objects?
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RESPONSE --> n!/ [r! * (n-r)!]
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14:18:06 There are P(n, r) = n ! / ( n - r ) ! possible ordered collections of r objects. There are r ! ways to order any unordered collection of r objects. There are thus P ( n, r ) / r! = n ! / [ r ! * ( n - r) ! ] possible unordered collections of r objects chosen from n objects.
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RESPONSE --> OKay, I understand.
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14:18:44 `q013. When we choose objects without regard to order, we say that we are forming combinations as opposed to permutations, which occur when order matters. The expression we obtained in the preceding problem gives us a formula for combinations: C ( n , r ) = P ( n, r) / r! = n ! / [ r ! ( n - r) ! ] gives us to number of possible combinations, or unordered collections, of r objects chosen from a set of n objects.
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RESPONSE --> Okay, I understand this completely.
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}wKzf Student Name: assignment #004
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14:37:31 `q001. Note that there are 9 questions in this assignment. In how many ways can we get a total of 9 when rolling two fair dice?
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RESPONSE --> 1st dice: 1, 2, 3, 4, 5, 6 2nd dice: 1, 2, 3, 4, 5, 6 9 = (3+6), (4+5), (5+4), (6+3) There would be 4 ways to get a total of 9 when rolling 2 fair dice
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14:37:46 There are two dice. Call one the 'first die' and the other the 'second die'. It is possible for the first die to come up 3 and the second to come up 6. It is possible for the first die to come up 4 and the second to come up 5. It is possible for the first die to come up 5 and the second to come up 4. It is possible for the first die to come up 6 and the second to come up 3. These are the only possible ways to get a total of 9. Thus there are 4 ways. We can represent these 4 ways as ordered pairs: (3,6), (4, 5), (5, 4), (6, 3).
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RESPONSE --> OKay, I understand
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14:38:18 `q002. In how many ways can we choose a committee of three people from a set of five people?
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RESPONSE --> 5*4*3 = 60 ways
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14:38:57 A committee when first chosen is understood to consist of equal individuals. The committee is therefore unordered, and we see that in choosing a committee of three people from a set of five people we are forming a combination of 3 people from among 5 candidates. The number of such combinations is C ( 5, 3) = 5 ! / [ 3 ! ( 5 - 3) ! ] = 5 ! / [ 3 ! * 2 ! ] = 5 * 4 * 3 * 2 * 1 / [ ( 3 * 2 * 1 ) * ( 2 * 1) ] = 5 * 4 / ( 2 * 1) = 5 * 2 = 10. There are 10 possible 3-member committees within a group of 5 individuals.
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RESPONSE --> Oh okay, I did the P(5,3) I understand why I was wrong. Because, in a committee order doesn't matter.
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14:40:05 `q003. In how many ways can we choose a president, a secretary and a treasurer from a group of 10 people?
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RESPONSE --> P(10, 3) = 10!/ 7! = 86400
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14:40:40 This choice is ordered. The order of our choice determines who becomes president, who becomes secretary and who becomes treasurer. Since we are choosing three people from 10, and order matters, we are looking for the number of permutations of three objects chosen from 10. This number is P(10, 3) = 10! / (10-3)! = 10! / 7! = 10 * 9 * 8 = 720.
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RESPONSE --> I must not have cleared my calculator, because that's the path I was on and miscalculated.
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14:41:15 `q004. In how many ways can we arrange six people in a line?
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RESPONSE --> 6*5*4*3*2*1 = 720
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14:41:23 There are 6 ! = 720 possible orders in which to arrange six people.
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RESPONSE --> Okay, I understand
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14:42:03 `q005. In how many ways can we rearrange the letters in the word 'formed'?
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RESPONSE --> There are 6 letters, so there are 6! = 720 ways to rearrange the letters.
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14:42:15 There are six distinct letters in the word 'formed'. Thus we can rearrange the letters in 6 ! = 720 different ways.
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RESPONSE --> Okay, I understand
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14:42:57 `q006. In how many ways can we rearrange the letters in the word 'activities'?
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RESPONSE --> We can rearrange the letters in 10! = 3,628,800 ways
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14:43:50 There are 10 letters in the word 'activities', but some of them are repeated. There are two t's and three i's. If we think of these 10 letters as being placed on letter tiles, there are 10! ways to rearrange the tiles. However, not all of these 10 ! ways lead to different words. For any spelling the three 'i' tiles can be arranged in 3 ! = 6 different ways, all of which spelled same word. And for any spelling the two 't' tiles can be arranged in 2 ! = 2 different ways. We must thus divide the 10! by 3! and by 2!, leading to the conclusion that there are 10 ! / ( 3 ! * 2 !) different spellings of the rearranged tiles.
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RESPONSE --> OH my gosh, you tricked me. That's not fair. I understand
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14:44:41 `q007. In how many ways can we line up four people, chosen from a group of 10, for a photograph?
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RESPONSE --> Order won't matter, so C(10, 4) = 10! / (4! * 6!)
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14:45:03 We are arranging four people chosen from 10, in order. The number of possible arrangements is therefore P ( 10, 4) = 10! / ( 10-4)! = 10 ! / 6 ! = 10 * 9 * 8 * 7.
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RESPONSE --> Okay, I was wrong. I thought order wouldn't matter
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14:49:48 `q008. In how many ways can we get a total greater than 3 when rolling two fair dice?
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RESPONSE --> 1 dice: 1, 2, 3, 4, 5, 6 2 dice: 1, 2, 3, 4, 5, 6 >3 = (1+3), (1+4), (1+5), (1+6), (2+2), (2+3), (2+4), (2+5), (2+6), (3+1), (3+2), (3+3), (3+4), (3+5), (3+6), (4+1), (4+2), (4+3), (4+4), (4+5), (4+6), (5+1), (5+2), (5+3), (5+4), (5+5), (5+6), (6+1), (6+2), (6+3), (6+4), (6+5), (6+6) 33 ways
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14:50:08 It would not be difficult to determine the number of ways to get totals of 4, 5, 6, etc. However it is easier to see that there is only one way to get a total of 2, which is to get 1 on both dice; and that there are 2 ways to get a total of 3 (we can get 1 on the first die and 2 on the second, or vice versa). So there are 3 ways to get 3 or less. Since there are 6 possible outcomes for the first die and 6 possible outcomes for the second, there are 6 * 6 = 36 possible outcomes for the two dice. Of these 36 we just saw that 3 give a total of 3 or less, so there must be 36 - 3 = 33 ways to get more than 3.
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RESPONSE --> Okay, I understand how you did it, I was just more comfortable listing.
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14:52:16 `q009. A committee consists of 5 men and 7 women. In how many ways can a subcommittee of 4 be chosen if the number of men and women must be equal?
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RESPONSE --> P(5, 2) = 20 P(7, 2) = 42 20*42 = 840 ways
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14:52:54 If the numbers of women and men are equal, then there must be 2 of each. Recall that a committee is regarded as unordered. If order doesn't matter there are C(5, 2) ways to choose 2 men out of 5, and C(7, 2) ways to choose 2 women out of 7. We have to choose 2 men AND we have to choose 2 women, so the Fundamental Counting Principal tells us that there are C(5, 2) * C(7, 2) = 10 * 21 = 210 possible subcommittees.
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RESPONSE --> I did the same type with permutation, but the order didn't matter so I should have used the combination.
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\ㅴDKpw} Student Name: assignment #005
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09:39:26 `q001. Note that there are 10 questions in this assignment. List the possible outcomes if a fair coin is flipped 2 times.
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RESPONSE --> ht, hh, tt, th
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09:39:42 There are 2 coins. Call one of them the first and the other the second coin. We can get Heads on the first and Heads on the second, which we will designate HH. Or we can get Heads on the first and Tails on the second, which we will designate HT. The other possibilities can be designated TH and TT. Thus there are 4 possible outcomes: HH, HT, TH and TT.
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RESPONSE --> OKay
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09:40:56 `q002. List the possible outcomes if a fair coin is flipped 3 times.
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RESPONSE --> HHH, HHT, HTH, HTT, THH, THT, TTH, TTT
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09:41:24 The possible results for the first 2 flips are HH, HT, TH and TT. We can obtain all possible results for 3 flips by appending either H or T to this list. We start out by writing the list twice: HH, HT, TH, TT HH, HT, TH, TT We then append H to each outcome in the first row, and T to each outcome in the second. We obtain HHH, HHT, HTH, HTT THH, THT, TTH, TTT Note that this process shows clearly why the number of possibilities doubles when the number of coins increases by one. With two coins we had 4 possible outcomes and with three coins we had 8 outcomes, twice as many as with two coins.
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RESPONSE --> I understand
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09:43:28 `q003. List the possible outcomes if a fair coin is flipped 4 times.
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RESPONSE --> HHHH, HHHT, HHTH, HHTT, HTHH, HTHT, HTTH, HTTT, THHH, THHT, THTH, THTT, TTHH, TTHT, TTTH, TTTT
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09:43:59 We can follow the same strategy as in the preceding problem. We first list twice all the possibilities for 3 coins: HHH, HHT, HTH, HTT, THH, THT, TTH, TTT HHH, HHT, HTH, HTT, THH, THT, TTH, TTT Then we append H to the front of one list and T to the front of the other: HHHH, HHHT, HHTH, HHTT, HTHH, HTHT, HTTH, HTTT THHH, THHT, THTH, THTT, TTHH, TTHT, TTTH, TTTT Again we see why the number of possibilities doubles when the number of coins increases by one. With three coins we had 8 possible outcomes and with four coins we had 16 outcomes, twice as many as with two coins.
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RESPONSE --> Okay, I understand
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09:47:36 `q004. If a fair coin is flipped 4 times, how many of the outcomes contain exactly two 'heads'?
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RESPONSE --> HHTT, HTHT, THHT, THTH, TTHH (5)
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09:48:38 The two 'heads' can occur in positions 1 and 2 (HHTT), 1 and 3 (HTHT), 1 and 4 (HTTH), 2 and 3 (THHT), 2 and 4 (THTH), or 3 and 4 (TTHH). These six possibilities can be expressed by the sets {1, 2}, {1, 3}, {1, 4}, {2, 3}, {2, 4}, {3, 4}. Thus the possibilities are represented by sets of two numbers chosen from the set {1, 2, 3, 4}. When choosing 2 numbers from a set of four, there are 4 * 3 / 2 possible combinations. Since in this case it doesn't matter in which order the two positions are picked, this will be the number of possible outcomes with exactly two 'heads'. The number of possibilities is thus C(4, 2) = 6.
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RESPONSE --> I missed one when I was looking through my list, but I see it now. Okay.
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09:59:34 `q005. If a fair coin is flipped 7 times, how many of the outcomes contain exactly three 'heads'?
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RESPONSE --> 35 possibilities for 3 head positions
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10:00:15 The possible positions for the three 'heads' can be numbered 1 through 7. We have to choose three positions out of these seven possibilities, and the order in which our choices occur is not important. This is equivalent to choosing three numbers from the set {1, 2, 3, 4, 5, 6, 7} without regard for order. This can be done in C(7,3) = 7 * 6 * 5 / 3! = 35 ways. There are thus 35 ways to obtain 3 'heads' on 7 flips.
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RESPONSE --> I did this problem but using the C(7,3). I took me awhile to figure out just how to do it, but that's how I got my answer.
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10:02:13 `q006. If we flip a fair coin 6 times, in how many ways can we get no 'heads'? In how many ways can we get exactly one 'head'? In how many ways can we get exactly two 'heads'?
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RESPONSE --> C(6,0) = 1 way to get no heads C(6,1) = 6 ways to get one head C(6,2) = 15 ways to get two heads
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10:02:32 In how many ways can we get exactly three 'heads'?
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RESPONSE --> C(6,3) = 20 ways to get three heads
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10:03:09 In how many ways can we get exactly four 'heads'? In how many ways can we get exactly five 'heads'?
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RESPONSE --> C(6,4) = 15 ways to get four heads C(6,5) = 6 ways to get five heads
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10:03:41 In how many ways can we get exactly six 'heads'?
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RESPONSE --> C(6,6) = 1 way to get six heads
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10:05:12 In how many ways can we get exactly seven 'heads'?
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RESPONSE --> With six coins you can't get 7 heads, in formula: C(6,7) = 0
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10:05:36 The number of ways to get no 'heads' is C(6,0) = 1. The number of ways to get exactly one 'head' is C(6,1) = 6. The number of ways to get exactly two 'heads' is C(6,2) = 15. The number of ways to get exactly three 'heads' is C(6,3) = 20. The number of ways to get exactly four 'heads' is C(6,4) = 15. The number of ways to get exactly five 'heads' is C(6,5) = 6. The number of ways to get exactly six 'heads' is C(6,6) = 1. These numbers form the n = 6 row of Pascal's Triangle: 1 6 15 20 15 6 1 See your text for a description of Pascal's Triangle. Note also that these numbers add up to 64, which is 2^6, the number of possible outcomes when a coin is flipped 6 times.
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RESPONSE --> I understand completely
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10:14:50 `q007. List all the subsets of the set {a, b}. Then do the same for the set {a,b,c}. Then do the same for the set {a,b,c,d}.
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RESPONSE --> a, b, ab a, b, c, ab, ac, bc, abc a, b, c, d, ab, ac, ad, bc, bd, cd, abc, abd, acd, bcd, abcd
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10:16:13 The set {a, b} has four subsets: the empty set { }, {a}, {b} and {a, b}. These four sets are also subsets of {a, b, c}, and if we add the element c to each of these four sets we get four different subsets of {a, b, c}. The subsets are therefore {}{ }, {a}, {b}, {a, b}, {c}, {a, c}, {b, c} and {a, b, c}. We see that the number of subsets doubles when the number of elements in the set increases by one. This seems similar to the way the number of possible outcomes when flipping coins doubles when we add a coin. The connection is as follows: To form a subset we can go through the elements of the set one at a time, and for each element we can either choose to include it or not. This could be done by flipping a coin once for each element of the set, and including the element if the coin shows 'heads'. Two different sequences of 'heads' and 'tails' would result in two different subsets, and every subset would correspond to exactly one sequence of 'heads' and 'tails'. Thus the number of possible subsets is identical to the number of outcomes from the coin flips.
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RESPONSE --> If I had even thought about the empty set I would have gotten it right. I understand what I should have done. The empty set was missing on all lists of subsets.
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10:18:39 `q008. How many subsets would there be of the set {a, b, c, d, e, f, g, h}?
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RESPONSE --> 4 elements had a number of subsets equal to 16, so 5 elements would have 32 subsets, 6 elements would have 64 subsets, 7 elements would have 128 subsets, and 8 elements would have 256 subsets.
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10:19:56 There are 2 possible subsets of the set {a}, the subsets being { } and {a, b}. The number doubles with each additional element. It follows that for a set of 2 elements there are 2 * 2 subsets (double the 2 subsets of a one-element set), double this or 2 * 2 * 2 subsets of a set with 3 elements, double this or 2 * 2 * 2 * 2 subsets of a set with 4 elements, etc.. There are thus 2 * 2 * 2 * 2 * 2 * 2 * 2 * 2 = 256 subsets of the given set, which has 8 elements. This number is also written as 2^8. }{More generally there are 2^n subsets of any set with n elements.
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RESPONSE --> I understand.
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10:25:35 `q009. How many 4-element subsets would there be of the set {a, b, c, d, e, f, g, h}?
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RESPONSE --> 4 elements would have one 4 element subset so it would double up to 8 with 5 elements, 16 with 6 elements, 32 with 7 elements and 64 with 8 elements
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10:26:20 To form a 4-elements subset of the given 8-elements set, we have to choose 4 elements from the 8. Since the order of elements in a set does not matter, order will not matter in our choice. The number of ways to choose 4 elements from a set of 8, without regard for order, is C(8, 4) = 8 * 7 * 6 * 5 / ( 4 * 3 * 2 * 1) = 70.
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RESPONSE --> I was stuck on the whole doubling up. But, this is much more accurate and I understand why I should do it this way, plus it's easier.
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10:27:56 `q010. How many subsets of the set {a,b,c,d} contain 4 elements? How many subsets of the set {a,b,c,d} contain 3 elements? How many subsets of the set {a,b,c,d} contain 2 elements? How many subsets of the set {a,b,c,d} contain 1 elements? How many subsets of the set {a,b,c,d} contain no elements?
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RESPONSE --> C(4,4) = 1 C(4,3) = 4 C(4,2) = 6 C(4,1) = 4 C(4,0) = 1
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10:28:05 The number of 4-element subsets is C(4,4) = 1. The number of 3-element subsets is C(4,3) = 4. The number of 2-element subsets is C(4,2) = 6. The number of 1-element subsets is C(4,1) = 4. The number of 0-element subsets is C(4,0) = 1. We note that these numbers form the n = 4 row 1 4 6 4 1 of Pascal's Triangle, and that they add up to 2^4 = 16, the number of possible subsets of a 4-element set.
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RESPONSE --> I understand.
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wrC_S⩀Q Student Name: assignment #006
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10:49:43 `q001. Note that there are 8 questions in this assignment. Using a standard deck of cards, in how many ways is it possible to get a hand containing exactly two 5's?
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RESPONSE --> C(4, 2) * C(48,3) = 1,098,240
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10:50:15 In order to get a hand containing exactly two 5's we must select, without regard for order, two of the four 5's, then we must select the remaining 3 cards from the 48 cards that are not 5's. There are C(4,2) ways to select two 5's from the four 5's in the deck. There are C(48,3) ways to select the 3 remaining cards from the 48 cards which are not 5's. We must do both, so by the Fundamental Counting Principle there are C(4,2) * C(48, 3) ways to obtain exactly two 5's.
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RESPONSE --> I'm not sure if I had the multiplication right, but I had the understanding of how to do it.
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10:51:23 `q002. Using a standard deck of cards, in how many ways is it possible to get a hand containing exactly two 5's and exactly two 9's?
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RESPONSE --> C(4,2) * C(4,2) * C(44,1) = 1584
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10:51:35 There are C( 4, 2) ways to select exactly two 5's and C(4, 2) ways to select exactly two 9's. There are 44 remaining cards which are neither 9 nor 5. The total number of possible ways is therefore C(4, 2) * C(4, 2) * 44.
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RESPONSE --> I understand.
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10:52:30 `q003. Using a standard deck of cards, in how many ways is it possible to get a 'full house' consisting of two 5's and three 9's?
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RESPONSE --> C(4, 2) * C(4, 3) = 24
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10:52:43 There are C(4, 2) ways to get two 5's and C(4, 3) ways to get three 9's. It follows that the number of ways to get the specified 'full house' with two 5's and three 9's is C(4,2) * C(4,3).
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RESPONSE --> I understand completely.
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10:53:38 `q004. Using a standard deck of cards, in how many ways is it possible to get a 'full house' consisting of two 5's and three identical face cards?
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RESPONSE --> The same, because there are 3 sets of 4 face cards. C(4,2) * C(4,3)
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10:53:48 There are C(4,2) ways to get two 5's and C(4,3) ways to get three of any given face card. There are 3 possible face cards, so the number of ways to get a 'full house' consisting of two 5's and three identical face cards is 3 * C(4,2) * C(4,3).
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RESPONSE --> I understand.
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10:54:25 `q005. Using a standard deck of cards, in how many ways is it possible to get a 'full house' consisting of two of one denomination and three of another?
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RESPONSE --> C(4,2) *C(4,3)
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10:55:43 For any two denominations, e.g., a pair of 5's and three 9's, there are C(4,2) * C(4,3) different full houses. There are 13 possible choices for the pair, which leaves 12 possible choices for the three-of-a-kind, which to by the Fundamental Counting Principle makes 13 * 12 possible choices for the two denominations. Note that order does matter when choosing the denominations because one has to be the pair and the other the three-of-a-kind. Again by the Fundamental Counting Principle we conclude that there are 13 * 12 * C(4,2) * C(4,3) possible full houses.
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RESPONSE --> I understand what I should have done completely. Becuase there is more than two denominations to choose from.
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10:58:43 `q006. Using a standard deck of cards, in how many ways is it possible to get a 'flush' consisting of five cards all of the same suit?
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RESPONSE --> 4 different suits 4 * C(13, 5)
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10:58:51 There are 13 cards in any given suit, so there are C(13,5) hands consisting of all 5 cards in that suit. There are 4 suits, so there are 4 * C(13,5) possible flushes.
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RESPONSE --> I understand.
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10:59:52 `q007. Using a standard deck of cards, in how many ways is it possible to get a 'straight' consisting of exactly one each of the denominations 5, 6, 7, 8 and 9?
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RESPONSE --> Because there are 4 of each of the cards, 4 * 4 * 4 * 4 * 4 or 4^5
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10:59:59 There are four 5's, four 6's, four 7's, four 8's and four 9's. So there are 4 * 4 * 4 * 4 * 4 = 4^5 possible straights from 5 to 9.
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RESPONSE --> I understand.
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11:02:25 `q008. Using a standard deck of cards, in how many ways is it possible to get a 'straight' consisting of five cards of consecutive denominations, assuming that the 'ace' can be either 'high' or 'low'?
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RESPONSE --> 4^5 * number of consecutive denominations 4^5 * 10
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11:03:01 There are 10 possible denominations for the 'low' card of a straight (any card from ace through 10 can serve as the 'low' card of a straight). There are 4^5 possible straights for each possible low card. It follows that there are 10 * 4^5 possible straights.
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RESPONSE --> I understand this completely. I understood this whole section fairly well.
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dXyvՠ Student Name: assignment #007
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11:37:08 `q001. Note that there are 5 questions in this assignment. Suppose we toss two dice. How many possible outcomes are there for the numbers on the two dice? How many of these outcomes given a total greater than 4? What therefore is the probability that the total on a toss of two dice is greater than 4?
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RESPONSE --> 6 poss for 1st, 6 poss for 2nd, so poss outcomes = 6*6= 36 Sets that give a total greater than 4: 30 30/36 = 5/6
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11:37:20 There are 6 possible outcomes on the first die and 6 on the second. The number of possible outcomes is therefore 6 * 6 = 36 (e.g., (1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6), (2, 1), (2, 2), etc.). The six outcomes (1, 1), (1, 2), (1, 3), (2, 1), (2, 2), (3, 1) are all the possible outcomes which are 4 or less. The remaining 36 - 6 = 30 outcomes are all greater than 4. It follows that the probability of obtaining a result greater than 4 is 30 / 36 = 5/6 or .833... or 83.33... %.
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RESPONSE --> I understand.
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11:38:30 `q002. What are the odds that the total on a toss of two dice will be greater than 4?
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RESPONSE --> 30 possibilities of being greater than 4 and 36 total poss, so 30 to 6 odds
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11:38:53 As seen in the previous question, there are 30 possible outcomes or the total is greater than 4 and 6 outcomes where the outcome is less than or equal to 4. The odds in favor of any event are expressed as odds = number in favor to number opposed. {}In this case the odds of a result greater than 6 are 30 to 6, which reduces to 5 to 1. These odds can also be expressed as 30 : 6 or 5 : 1.
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RESPONSE --> Okay, I understand how I can simplify this now.
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11:43:36 `q003. Suppose we have three boxes, one containing balls numbered 1-15, another tiles labeled a-z, and another one ring for each of the seven colors of the rainbow. How many possibilities are there for the collection of items we obtain if we choose one item from each box? How many of these possibilities contain an odd number, a consonant and a 'blue-type' color (blue, indigo or violet)? If we choose one item from each box, what is the probability that our collection will contain an odd number, a consonant and a 'blue-type' color (blue, indigo or violet)?
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RESPONSE --> total poss: 15*26&7 = 2730 odd,con,btype poss: 8*21*3 = 504 504/2730 = .18 or 18%
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11:43:49 There are 15 * 26 * 7 possibilities for the collection obtained by choosing one item from each box. There are 8 odd numbers, 21 consonants if we include 'y' and three 'blue-type' colors. So there are 8 * 21 * 3 possible combinations consisting of an odd number, a consonant and a 'blue-type' color. The desired probability is therefore ( 8 * 21 * 3) / ( 15 * 26 * 7).
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RESPONSE --> I understand completely.
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11:50:39 `q004. How many possible 5-card hands can be dealt from a 52-card deck? How many of these hands contain exactly one pair? What therefore is the probability that a hand dealt from a well-shuffled deck will contain exactly one pair? What are the odds in favor of such a deal resulting in a hand with exactly one pair?
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RESPONSE --> poss 5card hands: C(52,5) hands w/one pair: C(48,3) *C(4,2) [C(48,3)*C(4,2)] / C(52,5) = .42 or 42% 1098240 to 1500720 or
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11:51:40 There are C(52, 5) possible hands. There are C(4,2) ways to get a pair of any given denomination and 13 denominations, and there are then 48 choices for the first of the remaining three cards, 44 for the second and 40 for the third. Any given combination of the three remaining cards can be chosen in any of 3! ways so there are 48 * 44 * 40 possible choices of these 3 cards. Thus there are 13 * C(4, 2) * ( 48 * 44 * 40 / 3!) hands containing of exactly one pair. The probability of exactly one pair is therefore [ 13 * C(4,2) * (48 * 44 * 40) / 3! ] / C(52,5). This expression is easily enough written out and reduced [ 13 * 6 * 48 * 44 * 40 / 3! ] / [ 52 * 51 * 50 * 49 * 48 / (5*4*3*2*1) ] = 13 * 6 * 48 * 44 * 40 * 5 * 4 * 3 * 2 * 1 / [ 52 * 51 * 50 * 49 * 48 * 3 * 2 * 1 ] = 6 * 44 * 4 * 5 * 4 / [ 4 * 51 * 5 * 49 ] = 6 * 44 * 4 / [ 51 * 49 ] = (24 * 44) / (51 * 49) = (8 * 44) / ( 17 * 49) = .42 approx. Further explanation: This builds on the ideas of permutations and combinations developed in previous assignments. To get a hand you have to 'choose' 5 of the 52 cards, and order doesn't matter. There are C(52, 5) ways of doing this To get a pair of 5's, for example, you have to choose 2 of the four 5's in the deck. There are C(4, 2) ways to do this. There are 13 denominations (2's, 3's, 4's, ..., Queens, Kings). The pair could be from anyone of these denominations so there are 13 * C(4,2) ways to get a pair. After choosing the pair, you can't choose another card of that denomination or you would no longer have a pair. That leaves only 48 cards from which to choose the third. You already have a pair so the next card can't match the denomination of the third, so you have only 44 cards from which to choose the fourth. Similar reasoning shows that there are only 40 cards from which to choose the fifth card. These last three cards could have been chosen in any of 3! orders. So the number of ways of choosing the last three cards is 48*44*40/3!. So by the fundamental counting principle, since we have to choose a pair and then choose three other cards not matching the denomination of the pair or of one another, the number of possible ways to accomplish this is 13 * C(4,2) * 48 * 44 * 40 / 3!.
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RESPONSE --> I understand what I did wrong, I didn't incorporate that there were 13 poss for a pair
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11:55:12 `q005. If a fair coin is tossed five times, how many possible outcomes are there? How many of these outcomes will have exactly 3 'heads'? What therefore is the probability that on 5 tosses of a fair coin we will obtain exactly 3 'heads'?
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RESPONSE --> There are 32 poss. Of these 12 have 3 heads. 12/32 = .375 = 37.5% 0r 38%
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11:56:32 On 5 flips there are C(5,3) = 10 possible outcomes with exactly 3 'heads'. There are 2^5 = 32 possible outcomes altogether. The probability of 3 'heads' on 5 flips is therefore 10 / 32 = 5/16 = .3125.
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RESPONSE --> I always automatically try to count up the number of heads instead of doing the logical and easier thing by using C(,). I need to learn to use it more and my results might come out correct more often.
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QGᶶ_rrĜ Student Name: assignment #008
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12:20:18 `q001. Note that there are 7 questions in this assignment. Suppose that a card is dealt from a well-shuffled deck, and that you can tell by the reflection in your opponent's reading glasses that the card is a red face card. However you can't tell any more than that. What is the probability that the card is the Jack or the Queen of Diamonds?
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RESPONSE --> 26 red cards, only 1 Jack of Diamonds, only 1 Queen of Diamonds, so probability would be 2/26 or 1/13 or .08 = 8%
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12:21:41 In this case your knowledge that the card is a red face card limits the possibilities to six: The Jack of Hearts or Diamonds, the Queen of Hearts or Diamonds, or the King of Hearts or Diamonds. The probability that the card is one of the two specified cards is therefore 2 / 6 = 1/3. Note that without any limits on the possibilities, the probability that the card is the Jack or Queen of Diamonds is only 2 / 52 = 1 / 26. Note also that the probability that a card is a red face card is 6 / 52 = 3/26. If we divide the first probability by the second we get 1/26 / ( 3/26) = 1/26 * 26/3 = 1/3. Thus the probability that a card is the Jack or Queen of Diamonds, given that it is a red face card, is equal to the probability that it is the Jack or Queen of Diamonds (and a face card), divided by the probability that it is a red face card. This statement has the form 'The probability of B, given A, is equal to the probability of A ^ B divided by the probability of A'. This statement is abbreviated to the form P(B | A) = P(A ^ B) / P(A). This is the formula for Conditional Probability. In this problem the outcome was Jack or Queen of Diamonds, and the condition was that we have a red face card.
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RESPONSE --> OH okay, when you said red face card, I thought you just meant that it was a red card.
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12:22:42 `q002. Suppose that a face card is the first card dealt from a full deck of well-shuffled cards. What is the probability that the next card dealt (without replacement) will also be a face card?
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RESPONSE --> There would be 15 face cards left in the deck, so the probability would be 15/51 = .29 = 29%
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12:24:27 We know that after the first card is dealt there are 11 face cards left out of the original 12, and 51 cards left in the deck. The probability is therefore obviously 11/51. We can also analyze this situation as a conditional probability. B stands for 'a face card is dealt on the second card' while A stands for 'a face card is dealt on the first card'. So the event A ^ B stands for 'a face card is dealt on the first card and on the second', with probability 12/52 * 11/51. A stands for 'a face card is dealt on the first card', with probability 12 / 52. So P(B | A) stands for 'a face card is deal on the second card given that a face card is dealt on the first'. By the formula we have P(B | A) = P ( A ^ B ) / P(A) = [ 12 / 52 * 11 / 51 ] / [ 12 / 52 ] = 11 / 51, which of course we already knew from direct analysis.
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RESPONSE --> I need to just calm down. I'm getting so cofused at myself. On that one I was counting aces as face cards not even thinking about it. But I know why this is how it should be worked.
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12:30:39 `q003. Given that the first clip of a coin is Heads, what is the probability that a five-flip sequence will result in exactly four Heads?
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RESPONSE --> 4 poss, total sequences with heads being first flip: 16 4/16 = 1/4
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12:31:05 If we were to list the 2^5 = 32 possible outcomes for five flips, we would find that 16 of them have 'heads' on the first flip, and that of these 16 there are 4 outcomes with exactly four 'heads'. The probability therefore looks like 4 / 16 = 1/4, which is correct. To verify this by the formula P( B | A ) = P( A ^ B) / P(A), we let B stand for the desired event of exactly four 'heads' and A for the 'given' event of 'heads' on the first flip. On five flips, P(A) = 16 / 32 = 1/2 (probability of 'heads' on the first flip), which P(B ^ A) = 4 / 32 (four of the 32 possible outcomes have 'heads' on the first flip and exactly four 'heads'). The formula therefore gives us P( B | A ) = P( A ^ B) / P(A) = (4/32) / (2/1) = (4 / 32) * (2 / 1) = 4 / 16 = 1/4.
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RESPONSE --> Okay, I understand.
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12:37:32 `q004. Given that the first of two dice comes up even, what is the probability that the total on the two dice will be greater than 9? How does this compare with the unconditional probability that the total of two fair dice will be greater than 9?
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RESPONSE --> 3/9 = 1/3 prob total unconditional prob is 6/36 = 1/5 It is more probable to get a total greater than 9 when two fair dice are rolled and come out even.
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12:38:37 We can list the sample space of dice possibilities for which the first number is even. The sample space is { (2,1), (2,2), (2,3), (2,4), (2,5), (2,6), (4,1), (4,2), (4,3), (4,4), (4,5), (4,6), (6,1), (6,2), (6,3), (6,4), (6,5), (6,6) }. We note that there are 18 elements in the sample space. We then find the corresponding totals, which are 3, 4, 5, 6, 7, 8 5, 6, 7, 8, 9, 10 7, 8, 9, 10, 11, 12. Of these 18 totals, 4 are greater than 9. Thus the probability that the total of two dice will be greater than 9, given that the first is even, is 4/18 = 2/9. To verify this by the formula P( B | A ) = P( A ^ B) / P(A), we let B stand for the set of all dice pairs which give a total greater than 9, and A for the set of all dice pairs where the first die shows an even number. We have seen that A = { (2,1), (2,2), (2,3), (2,4), (2,5), (2,6), (4,1), (4,2), (4,3), (4,4), (4,5), (4,6), (6,1), (6,2), (6,3), (6,4), (6,5), (6,6) }. Listing the elements in B we find that B = { (4, 6), (6, 4), (5, 5), (6, 5), (5, 6), (6, 6) }. There are 6 elements in this set. A ^ B consists of the set of elements common to both A and B, or { (4, 6), (6, 4), (6,5), (6, 6) }. Since there are 4 elements in A ^ B, 18 elements in A, and 36 elements in the sample space for two dice, it follows that P(A) = 18 / 36 = 1/2 and P(A ^ B) = 4 / 36 = 1/9. Therefore the probability we are looking for, P(B | A), is given by P(B | A) = P(A ^ B) / P(A) = (1/9) / (1/2) = (1/9) * (2/1) = 2/9. This is in agreement with the previous result obtained by listing.
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RESPONSE --> I must not be reading anything right. I thought you meant both of the dice when rolled came out even, not just one of them.
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12:39:52 `q005. A spinner has numbers 2, 3, 4, 5 and 6. Given that the first number is odd, what is the probability that the sum of the results on two consecutive spins is even?
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RESPONSE --> The second number would have to be odd, so the probability would be 2/5
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12:40:21 The set of possibilities for which the first number is odd is { (3, 2), (3, 3), (3, 4), (3, 5), (3, 6), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6) }. There are therefore 10 possibilities. Of these 4 add up to an even total, so the probability that the total is even, given that the first number is odd, is Probability of B given A = 4/10 = 2/5. To verify this by the formula P( B | A ) = P( A ^ B) / P(A), we let B stand for the set of all pairs that add up to an even number and A for the set of all pairs for which the first number is even. The sample space for two spins has 5 * 5 = 25 elements. Of these, only the four outcomes (3, 3), (3, 5), (5, 3) and (5, 5) for which both spinners land on odd numbers are in the set A ^ B. Thus P(A | B) = 4/25. The set A consists of the 10 pairs listed earlier. So P(A) = 10/25 = 2/5. Thus P(B | A) = P(A ^ B) / P(A) = (4/25) / (2/5) = (4/25) * (5/2) = 2/5 in agreement with our previous result.
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RESPONSE --> I understand.
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12:42:17 `q006. What is the probability that two consecutive cards dealt (without replacement) from a full deck will both be Hearts?
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RESPONSE --> 13/52 * 12/52 = .06 or 6%
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12:45:45 When the first card is dealt there are 13 Hearts in a deck of 52. When the second card is dealt there are 12 Hearts left in the remaining 51 cards. The order in which the cards are dealt does not matter, and there are two possible orders for any 2-card 'hand'. The number of ways to get 2 Hearts is therefore C(13,2) = 13 * 12 / 2 = 156 / 2 = 78. The number of possible 2-card 'hands' is C(52, 2) = 52 * 51 / 2 = 1326. The probability of obtaining two Hearts is therefore 78 / 1326, which can be reduced or expressed as a decimal.
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RESPONSE --> I got the right answer although I didn't actually calculate right. I should have done 13/52 * 12/51 because it was without replacement, which would have given me the same answer.
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12:46:58 `q007. What is the probability that two consecutive cards dealt (without replacement) from a full deck will both be of the same suit?
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RESPONSE --> 4 suits, so 4 * (13/52 *12/51) = .24 or 24%
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12:47:54 A quick common-sense solution tells us that sense the first card can be anything, then since of the 51 remaining cards, there remain remain 12 cards that match the suit of the first the probability must be 12/51 = 4/17.
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RESPONSE --> I did mine a little different, but had the same answer. I understand this too.
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ԔqҧE Student Name: assignment #009
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Ǯӣɴw Student Name: assignment #009
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12:59:43 `q001. Note that there are 5 questions in this assignment. What is the probability that on two rolls of a fair die, we obtain exactly two 3's?
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RESPONSE --> 1/36
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13:00:02 The probability of obtaining a 3 on a single role is 1/6 (one of the six possible outcomes is a 3). Since the two rolls are independent, it follows that if two dice are rolled the probability of obtaining two 3's is 1/6 * 1/6 = 1/36.
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RESPONSE --> I understand how we got this conclusion.
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13:02:32 `q002. What is the probability that on three rolls of a fair die, we obtain exactly two 5's?
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RESPONSE --> prob of getting a 5 is 1/6 prob not getting 5 is 5/6 1/6 * 1/6 * 5/6 = 5/216
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13:04:44 On three rolls of a fair die, the two 5's can occur on the first and second, the first and third or the second and third rolls. That is, of the three available positions the two positions in which the 5's occur can occur in C(3,2) = 3 different ways. Since the probability of a 5 on any roll is 1/6 and the probability of not getting a 5 on a roll is 5/6. Any one of the three ways of getting two 5's and one non-5 is therefore (1/6) * (1/6) * (5/6 ) = 5/216. Since each of the three ways to get the desired outcome occurs with probability 5/216, it follows that Probability of exactly two 3's on three rolls = 3 * 5/216 = 15/216 = 5/72.
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RESPONSE --> I understand.
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13:07:27 `q003. What is the probability that on six rolls of a fair die, we obtain exactly two 5's?
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RESPONSE --> 1/6 * 1/6 * 5/6 * 5/6 *5/6 *5/6 = 625/ 46656 = .01 or 1%
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13:08:04 In order to get exactly two 5's on six rolls of the fair die, we must get two 5's and four results that are not 5. The probability of getting a 5 on any roll is 1/6, and the probability of getting a result other than 5 is 5/6. Therefore given any two positions out of the six the probability of obtaining 5's in two of the positions and non-5's in the remaining four positions is by the Fundamental Counting Principle Probability of 5's in exactly two of the six positions = (1/6) * (1/6) * (5/6) * (5/6) * (5/6) * (5/6) = (1/6)^2 * ( 5/6)^4. There are C(6,2) ways in which the positions of the two 5's can be selected from the six available positions. Thus we have Probability of exactly two 5's on six flips = C(5,2) * (1/6)^2 * (5/6)^4.
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RESPONSE --> I understand I think. No, I do.
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13:09:31 `q004. If we let p stand for the probability of getting a 5 on a roll of a die and q for the probability of not getting a 5 on a roll, then how would we expressed a probability of getting exactly r 5's on n rolls?
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RESPONSE --> We use the binomial probabilty formula P(x) = C(n,x)p^x q^n-x
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13:09:58 By analogy with the preceding problem, we see that to get r 5's on n rolls we must get 5 the total of r times and non-5 a total of (n-r) times. Since probability of getting a 5 is p, the probability of getting 5 a total of r times is represented by p^r. Since the probability of getting a non-5 is q, then the probability of getting a non-5 a total of (n-r) times is represented by q^(n-r). There are C(n, r) ways to place fives in r of n positions, so the probability of getting 5 fives and n non-fives is C(n, r) * p^r * q^(n-r).
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RESPONSE --> Same thing, different terms or variables used.
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13:11:50 `q005. Explain why, if p is the probability of getting a 5 on a single roll of a die, it follows that the probability q of not getting a 5 is q = 1-p. How would we therefore express the formula C(n, r) * p^r * q^(n-r) only in terms of p?
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RESPONSE --> Because probability is a fraction, so when this fraction of p is subtracted from 1 it gives the probability of not getting a 5 on a single roll, q. I'm not sure how to express the formula only in terms of p.
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13:12:17 If we roll a single die, we either get 5 or we don't. The two events are mutually exclusive -- they can both happen on the same roll. They also cover all possibilities. The sum of the probabilities is therefore 1. So we conclude that p + q = 1, and from this it follows immediately that q = 1 - p. Substituting 1 - p for q in the expression C(n, r) * p^r * q^(n-r) we obtain Probability of r fives on n rolls = C(n, r) * p^r * (1-p) ^ (n-5).
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RESPONSE --> Oh okay, you wanted substitution used. I understand.
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醮șTȥK Student Name: assignment #010
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13:45:32 `q001. Note that there are 9 questions in this assignment. In a certain lottery the probability of winning $100 is .005, the probability of winning $1000 is .0002 and the probability of winning $10,000 is .00001. Otherwise you win nothing. What is the probability of winning nothing?
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RESPONSE --> If you subtract these three probabilities from 100%, then the probability of winning nothing would be .99479 or 99.5%
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13:45:43 The probability of winning something is the sum .005 + .0002 + .00001 =.00521. The events of winning something and winning nothing are mutually exclusive, and they comprise all possible outcomes. It follows that the probability of winning something added to the probability winning nothing must give us 1, and that therefore Probability of winning nothing = 1 - .00521 = .99479.
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RESPONSE --> I understand.
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13:49:26 `q002. In the same lottery , where the probability of winning $100 is .005, the probability of winning $1000 is .0002 and the probability of winning $10,000 is .00001, if you bought a million tickets how many would you expect to win the $100 prize? How many would you expect to win the $1000 prize? How many would you expect to win the $10,000 prize?
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RESPONSE --> .5% of winning $100. 1,000,000 (.005) = 5,000 for $100 1,000,000 (.0002) = 200 for $1000 1,000,000 (.00001) = 10 for &10,000
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13:49:34 The probability of winning the $100 prize is.005, so out of a million tries we would expect to win the $100 a total of .005 * 1,000,000 = 5,000 times. Similarly we would expect to win the $1000 prize a total of .0002 * 1,000,000 = 200 times. The expected number of times we would win the $10,000 prize would be .00001 * 1,000,000 = 10.
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RESPONSE --> I understand.
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13:51:20 `q003. In the lottery of the preceding problem, if you were given a million tickets how much total money would you expect to win?
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RESPONSE --> $100 (5000) = $500,000 $1,000 (200) = $200,000 $10,000 (10) = $100,000 $800,000 total.
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13:51:27 As seen in preceding problem, you would expect to win $100 a total of 5,000 times for a total of $500,000, you would expect to win the $1000 prize 200 times for a total of $200,000, and you expect to win the $10,000 prize 10 times for total of $100,000. The expected winnings from a million tickets would therefore be the total $800,000 of these winnings.
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RESPONSE --> I understand.
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13:55:23 `q004. In the lottery of the preceding problem, if you bought a million tickets for half a million dollars would you most likely come out ahead?
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RESPONSE --> According to the probability, you would come out with $800,000 subtracting the $500,000 spent. You would according to probability come out with $300,000.
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13:56:14 You would expect on the average to win $800,000, and your probability of winning at least $500,000 would seem to be high. You would have a very good expectation of coming out ahead.
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RESPONSE --> Okay that was worded kind of weird. But I understand.
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14:01:14 `q005. In the lottery of the preceding problem, how much would you expect to win, per ticket, if you bought a million tickets? Would the answer change if you bought 10 million tickets?
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RESPONSE --> Using $300,000 as the winning and the 1,000,000 tickets. There would be an average of $.30 per ticket winning. This average would not change with buying more tickets the winnings would increase as well.
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14:01:44 Your expected winnings would be $800,000 on a million tickets, which would average out to $800,000/1,000,000 = $.80, or 80 cents. If you bought 10 million tickets you expect to win 10 times as much, or $8,000,000 for an average of $8,000,000 / 10,000,000 = $.80, or 80 cents. The expected average wouldn't change. However you might feel more confident that your average winnings would be pretty close to 80 cents if you have 10 million chances that if you had 1 million chances.
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RESPONSE --> Okay, I thought you wanted the the money not lost to be the only thing counted.
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14:04:50 `q006. If we multiply $100 by the probability of winning $100, $1000 by the probability of winning $1000, and $10,000 by the probability of winning $10,000, then add all these results, what is the sum? How does this result compare with the results obtained on previous problems, and why?
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RESPONSE --> We get .8 to be the result. This is the average winnings we already obtained earlier.
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14:05:04 We get $100 * .005 + $1,000 * .0002 + $10,000 * .00001 = $.50 + $.20 + $.10 = $.80. This is the same as the average per ticket we calculated for a million tickets, or for 10 million tickets. This seems to indicate that a .005 chance of winning $100 is worth 50 cents, a .0002 chance of winning $1,000 is worth 20 cents, and a .00001 chance of winning $10,000 is worth 10 cents.
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RESPONSE --> I understand completely.
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14:07:51 `q007. The following list of random digits has 10 rows and 10 columns: 3 8 4 7 2 3 0 8 3 9 1 8 3 7 3 2 9 1 0 3 4 3 3 0 2 1 4 9 8 2 4 3 4 9 9 2 0 1 3 9 8 3 4 1 3 0 5 3 9 7 2 4 7 4 5 3 7 2 1 8 3 6 9 0 2 5 9 5 2 3 4 5 8 5 8 8 2 9 8 5 9 3 4 6 7 4 5 8 4 9 4 1 5 7 9 2 9 3 1 2. Starting in the second column and working down the column, if we let even numbers stand for 'heads' and odd numbers for 'tails', then how many 'heads' and how many 'tails' would we end up with in the first eight flips? Answer the second question but starting in the fifth row and working across the row. Answer once more but starting in the first row, with the second number, and moving diagonally one space down and one to the right for each new number.
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RESPONSE --> 4 heads and 6 tails in first eight flips. 3 heads and 7 tails in fifth row 4 heads and 6 tails diagonally
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14:08:20 Using the second column, the first eight flips would be represented by the numbers in the second column, which are 8, 8, 3, 3, 3, 4, 6, and 5. According to the given rule this correspond to HHTTTHHT, total of four 'heads' and four 'tails'. Using the fifth row we have the numbers 8 3 4 1 3 0 5 3, which according to the even-odd rule would give us HTHTTHTT, or 3 'heads' and 5 'tails'. Using the diagonal scheme we get 8, 3, 0, 9, 0, 7, 5, 8 for HTHTHTTH, a total of four 'heads' and four 'tails'.
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RESPONSE --> Oh, I thought I was looking for 10 numbers
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14:13:10 `q008. Using once more the table 3 8 4 7 2 3 0 8 3 9 1 8 3 7 3 2 9 1 0 3 4 3 3 0 2 1 4 9 8 2 4 3 4 9 9 2 0 1 3 9 8 3 4 1 3 0 5 3 9 7 2 4 7 4 5 3 7 2 1 8 3 6 9 0 2 5 9 5 2 3 4 5 8 5 8 8 2 9 8 5 9 3 4 6 7 4 5 8 4 9 4 1 5 7 9 2 9 3 1 2 let the each of numbers 1, 2, 3, 4, 5, 6 stand for rolling that number on a die-e.g., if we encounter 3 in our table we let it stand for rolling a 3. If any other number is encountered it is ignored and we move to the next. Starting in the fourth column and working down, then moving to the fifth column, etc., what are the numbers of the first 20 dice rolls we simulate? If we pair the first and the second rolls, what is the total? If we pair the third and fourth rolls, what is the total? If we continue in this way what are the 10 totals we obtain?
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RESPONSE --> 1 4 5 6 2 3 2 3 5 2 3 2 1 2 3 5 4 2 4 5 5, 11, 5, 5, 7, 5, 3, 8, 6, 9
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14:14:25 The numbers we get in the fourth column are 7, 7, 0, 9, 1, 4, 0, 5, 6, 7, then in the fifth column we get 2, 3, 2, 9, 3, 5, 2, 8, 7, 9 and in the sixth column we get 3, 2, 1, 2, 0, 3, 5, 8, 4, 2. We hope to get 20 numbers between 1 and 6 from this list of 30 numbers, but we can be sure that this will be the case. If it is, we will add some numbers from the seventh column. Omitting any number on our current list not between 1 and 6 we get 1, 4, 5, 6 from the fourth column, then from the fifth column we get 2, 3, 2, 3, 5, 2 and from the sixth column we get 3, 2, 1, 2, 3, 5, 4, 2. This gives us only 18 numbers between 1 and 6, and we need 20. So we go to the seventh column, which starts with 0, 9, 4, 0, 5. The first number we encounter between 1 6 is 4. The next is 5. This completes our list. Our simulation therefore gives us the list 1, 4, 5, 6, 2, 3, 2, 3, 5, 2, 3, 2, 1, 2, 3, 5, 4, 2, 4, 5. This list represents a simulated experiment in which we row of a fair die 20 times. The first and second rolls were 1 and 4, which add up to 5.{} The second and third rolls were 5 and 6, which add up to 11. The remaining rolls give us 2 + 3 = 5, 2 + 3 = 5, 5 + 2 = 7, 3 + 2 = 5, 1 + 2 = 3, 3 + 5 = 8, 4 + 2 = 6, and 4 + 5 = 10. The totals we obtain our therefore 5, 11, 5, 5, 7, 5, 3, 8, 6, and 10.
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RESPONSE --> Okay, except your 4+5=10
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14:16:14 `q009. According to the results of the preceding question, what proportion of the totals were 5, 6, or 7? How do these proportions compare to the expected proportions?
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RESPONSE --> 4 totals of 5, 1 total of 6, and 1 total of 7. 4/10 , 1/10 , 1/10 these make up together 1/2 of the totals.
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14:17:09 We obtain four 5's, one 6 and one 7. Thus 6 of our 10 results were 5, 6 or 7. We saw earlier that of the 36 possible outcomes of rolling two dice, four give us a total of 5, while five give us a total of 6 and six give the total of 7. If we add these numbers we see that 15 of the 36 possible outcomes in the sample space are 5, 6 or 7 for probability 15/36. Our simulation results in 6/10, a higher proportion than the probabilities would lead us to expect. However since the simulation resulted from random numbers it is certainly possible that this will happen, just as it is possible that if we rolled two dice 10 times 7 of the outcomes would be in this range.
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RESPONSE --> I don't know why I said that 6/10 was 1/2. That was my mistake. I understand all of this.
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