#$&* course Mth 152 November 10 - 7:19pm 007. *********************************************
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Given Solution: There are 3 possible odd outcomes and four outcomes less that 5 which would add up to 7 outcomes, except that 2 of the outcomes < 5 are alrealdy odd and won't be counted. Thus the number of outcomes which are odd or less that 5 is 3 + 4 - 2 = 5 (this expresses the rule that n(A U B) = n(A) + n(B) - n(A ^ B), where U and ^ stand for union and intersection, respectively ). Thus the probability is 5/6. In terms of the specific sample space: The sample space for the experiment is {1, 2, 3, 4, 5, 6}. Success corresponds to events in the subset {1, 2, 3, 4, 5}. There are 6 elements in the sample space, 5 in the subset consisting of successful outcomes. Thus the probability is 5/6. ** Self-critique: Your solution was a bit confusing for me but I understand. ------------------------------------------------ Self-critique Rating: 2 ********************************************* question: Query 12.2.15 drawing neither heart nor 7 from full deck YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Getting my head together: 52 cards. 5 cards in a deck. 13 demonations. 4 suits. In the suit of hearts there are 13 cards. 52-13=39. There are 39 cards that are not hearts. Next we look at 7’s. We know there is four suits so that means there are only 3 other 7’s that aren’t hearts. That takes the cards down to 36. The probability is 36/52. confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `aThe sample space consists of the 52 cards in a full deck. There are 39 cards that aren't hearts, four 7's but only three aren't hearts so there are 36 cards that aren't hearts or seven. The probability is therefore 36/52 = 9/13. The odds in favor of the event are 16 to 36 (number favorable to number unfavorable), which in reduced form is 4 to 9. ** Self-critique: could have simplified further. OKAY ------------------------------------------------ Self-critique Rating: OKAY ********************************************* question: 12.2.24 prob of black flush or two pairs YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: What is a black flush? If it’s okay with you I’m going to have to look that one up. So literally a black color flush, okay. Since there are 13 denominations and 5 cards (a hand), it’d be C (13,5) which equals 1,287. The black flush… well half the cards in a deck are black right? I don’t understand what a black flush is. But since there are four suits, we’ll say that makes 2. So since we have 1,287 possibilities we’ll times that by 2. 1,287 times 2 = 2574. Is that right? Now we go to, two pairs. All the ways of getting two pairs equal up to 123,552 (page 727) The probability of the two would just be the two added together against the total, which is also listed on page 727, which is 2,598,960. Therefore we have the probability of 126,126 / 2,598,960. confidence rating #$&*: 2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: There are C(13,5) = 1287 ways to get a flush in a given suit--gotta choose the 5 cards from the 13 cards in that suit. There are two black suits so there are 2 * 1287 = 2574 possible black flushes. As the text tells you there are 123,552 ways to get two pairs. You can incidentally get this as 13 * C(4, 2) * 12 * C(4, 2) * C(44, 1) / 2 (2 of the 4 cards in any of the 13 denominations, then 2 of the 4 cards in any of the remaining 12 denominations, divide by 2 because the two denominations could occur in any order, then 1 of the 44 remaining cards not in either of the two denominations. There is no way that a hand can be both a black flush and two pairs, so there is no overlap to worry about (i.e., n(A and B) = 0 so n(A or B) = n(A) + n(B) - n(A and B) = n(A) + n(B) ). Thus there are 123,552 + 2574 = 126,126 ways to get one or the other. The probability is therefore 126,126 / 2,598,960 = .0485, approx. ** Self-critique: OKAY ------------------------------------------------ Self-critique Rating: OKAY ********************************************* question: 12.2.33 - x is sum of 2-digit numbers from {1, 2, ..., 5}; prob dist for random vbl x from book: “let x denote the sum of two distinct numbers selected randomly from the set { 1,2,3,4,5 } Construct the probability distribution for the random variable x.” - 743 YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: If two numbers are selected randomly from the set they will equal x. It says two distinct numbers but also says randomly so that doesn’t suggest any particular order on allowing repetition or not, right? But in the case that they do have an order and cannot be repeated there would only be 9 combinations. 1 (invalid, isn’t two numbers) 1+1 (invalid, doesn’t work because you can’t repeat) 1+2 = 3 (valid) 1+3 = 4 (valid) 1+4, 2+3 = 5 (valid) 1+5, 2+4 = 6 (valid) 2+5, 1+6 = 7 (valid) 3+5 = 8 (valid - remembering 7,8,9,10 were not in set) 4+5 = 9 (highest we can reach without repeating or using numbers outside set.) To check: 1 2 3 4 5 1 3 4 5 6 2 5 6 7 3 7 8 4 9 Looking at the numbers and there probability we get: X P (x) 3 .1 4 .1 5 .2 6 .2 7 .2 8 .1 9 .1 confidence rating #$&*: 2.5 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `aIf 2 different numbers are chosen from the set (1, 2, 3, 4, 5} then the sum 3 can appear only as 1+2. 4 can appear only as 1+3, assuming numbers can't be repeated (so, for example, 2+2 is not allowed). 5 can occur as 1+4 or as 2+3. 6 can occur as 1+5 or as 2+3. 7 can occur as 2+5 or as 1+6. 8 can occur only as 3+5. 9 can occur only as 4+5. Of the 10 possible combinations, the sums 3, 4, 8 and 9 can occur only once each, so each has probability .1. The sums 5, 6 and 7 can occur 2 times each, so each has probability .2. The possible sums are as indicated in the table below. 1 2 3 4 5 1 3 4 5 6 2 5 6 7 3 7 8 4 9 This assumes selection without replacement. There are C(5, 2) = 10 possible outcomes, as can be verified by counting the outcomes in the table. 3, 4, 8 and 9 appear once each as outcomes, so each has probability 1/10. 5, 6 and 7 appear twice each as outcomes, so each has probability 2/10. x p(x) 3 .1 4 .1 5 .2 6 .2 7 .2 8 .1 9 .1 ** Self-critique: OKAY - In your solution for getting 5 you have written, “6 can occur as 1+5 or as 2+3.” I think you mean 2+4, right?