#$&* course MTH 152 Time: 2:50 PMDate: 7/16
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Given Solution: 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... %. Self-critique: How would one consider the odds to this question? ------------------------------------------------ Self-critique rating: Ok. See my question above.
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Given Solution: 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. Self-critique: Ok. ------------------------------------------------ Self-critique rating: Ok. ********************************************* Question: `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)? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: - With the above total, we use the FCP to show that there are 15 * 26 * 7 possibilities (2730) in all. - But with the constraints listed, we have {1, 3, 5, 7, 9, 11, 13, 15} = 8 odd numbers - {B, C, D, F, G, H, J, K, L, M, N, P, Q, R, S, T, V, W, X, Y, Z} = 21 consonants (or just say there are 5 vowels so 26 - 5 = 21) - {B, I, V} = 3 colors violet - Thus the number of possibilities, with constraints, comes to 8 * 21 * 3 = 540. - So the probability is 2730/540 = .19780, or 19.8%, or 20%. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: 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). Self-critique: ok. ------------------------------------------------ Self-critique rating: ok. ********************************************* Question: `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? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: - There are C(52, 5) possible 5-card hands stemming from a 52 card deck, assuming 5-card poker is being played. If it was 7-card poker, then the possible outcomes extends to C(52, 7) - If one tried to make pairs out of 13 cards, one would only reach 6 pairs. Because the order doesn’t matter, this is shown by C (4,2). - So multiplying 13 * C (4, 2) shows the maximum number of cards, multiplied by the number of ways that a pair could be made from one suit. - Now that the pair has been expressed, one must consider the rest of the hand. If a pair is chosen, this is 5 in the hand - 2 cards dealt, so there are 3 cards that are non-paired and of varying denominations. - ***One can express this by showing 52-4 = 48 for the first choice, 44 for the second choice, and 40 cards left for the third choice.*** - Assuming these three are coming from one pair, one could divide this product f 48 * 44 * 40 by 3! Resulting in 1,098,240 possible 5- card hands that would contain one pair. Now that we know how many hands contain one pair, we can extrapolate and use the number of favorable conditions (1,098, 240) over the number of un-favorable conditions C(52,5) = 2,598,960 - 1,098,240 / 2, 598,960 = .4225 - 42.3% chance of getting a hand that contains at least one pair. ******* I may have used the wrong explanation for attaining the probability here, I just saw in my notes that odds = # in favor/# opposed. I still concluded with the answer but I used the formula in the wrong manner. I amend that the probability is found by the number of possible 5-card hands / the number of possible hands derived from a 52 card deck. - The odds, then, would be used as previously mentioned. - Odds = # of events favorable/ # of events unfavorable. - My initial assumption is to use 48*44*40/3! as the numerator and the difference of 48 * 44 * 40/3! - C(52,5) = 1,500,720 - So to define, the number of events favorable is 1,098,240 and the events opposed is 1,500,720. - On this route, I’m curious as to how I can quickly reduce these numbers in order to correctly and accurately list the odds. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: 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!. STUDENT QUESTION I get confused in this step: “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.” After you get a pair, I would think you would have 44 choices for the first of the remaining three cards, 40 for the second, and 36 for the third because 52 and 48 were for the pair??? INSTRUCTOR RESPONSE: Having gotten a pair, which consists of two cards of the same denomination, you can't get another card of that denomination (if you did you would no longer have a pair, but at least three of a kind). The next card could be any card not of that denomination, and there are 48 such cards. STUDENT QUESTION: Also, I don’t understand how you simplified the answer. 13 * 6 * 48 * 44 * 40 * 5 * 4 * 3 * 2 * 1 / [ 52 * 51 * 50 * 49 * 48 * 3 * 2 * 1 ] = Why did you move 5*4*3*2*1 to the first half of the equation and 3*2*1 to the second half??? INSTRUCTOR RESPONSE: Division by a fraction is the same as multiplication by the reciprocal (for example (a / b) / (c / d) = (a / b) * (d / c) = a * d / (b * c). In this case we are dividing by [ 52 * 51 * 50 * 49 * 48 / (5*4*3*2*1) ], so we get [ 13 * 6 * 48 * 44 * 40 / 3! ] / [ 52 * 51 * 50 * 49 * 48 / (5*4*3*2*1) ] = [ 13 * 6 * 48 * 44 * 40 / (3 * 2 * 1) ] * [ (5 * 4 * 3 * 2 * 1) / (52 * 51 * 50 * 49 * 48) ]. We now multiply the numerators of the two fractions, and the denominators, to get 13 * 6 * 48 * 44 * 40 * 5 * 4 * 3 * 2 * 1 / [ 52 * 51 * 50 * 49 * 48 * 3 * 2 * 1 ] , and then continue as indicated in the given solution. Self-critique: This is a great question to flex all the brain muscles built up from the previous assignments. I had went about my answer to get the probability of being dealt one-pair, but I arrived at a similar conclusion; I’m simply more comfortable working with larger numbers in order to get a more accurate solution, being (0.4225). I normally round to four or six significant figures, as was habit from previous chemistry classes I had practiced on my own. All the formulas I have in my head, and it’s becoming second-nature to use them how and where. The only portion that I started to lose ground on was concerned with choosing the last 3 cards for the 5-card hand. For example, I eventually became able to explain why 48*44*40/3! was put to use. Since one pair is already obtained, this is a manner of mathematically saying that 52-4 = 48. I originally was in conflict with this. But I believe this relates to choosing two cards, but the next card cannot match so it is skipped and must be chosen by the following card pair. 52 - 2 = 50 -2 = 48. Still, to further concrete my understanding, paraphrase this portion of the exercise if my conclusion wasn’t spot-on. ------------------------------------------------ Self-critique rating: ********************************************* Question: `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'? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: - If a coin is tossed 5 times, this is saying 5 * 4 * 3 * 2 * 1 = 120 outcomes.
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Given Solution: 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. STUDENT QUESTION I don’t understand where you got the 32 from? Did I figure out the possiblilities of 5 tosses wrong because I came up with 80 possible outcomes. INSTRUCTOR RESPONSE On five flips there are 2 possibilities on the first flip, 2 possibilities on the second, 2 possibilities on the third, 2 possibilities on the fourth and 2 possibilities on the fifth. By the Fundamental Counting Principle there are 2 * 2 * 2 * 2 * 2 = 2^5 = 32 possible outcomes on 5 flips. You can also see this by making a 'tree' diagram of the possibilities. Self-critique: - I was on the tracks with the FCP in this question, but I apparently caught the wrong train. - Basically this is saying that, if I were to ponder how many possibilities contained 4 tails on 7 flips: “If a fair coin is tossed 7 times, how many combinations will have exactly four tails?” - Total flips = 2^7, 128. - Number of combinations containing exactly 4 tails = C(7, 4)= 35 combinations will have exactly 4 tails. - The probability of getting 4 tails out of 7 flips is thus 35/128 = 0.2734375 = 27.3% chance. ------------------------------------------------ Self-critique rating: Okay. See my comment above and my self-made problem. ********************************************* Question: `q006. What is the probability that when rolling two dice the total will be at least 4? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: - There are six sides per die, so the total number of outcomes is 6^2 = 36. - I next ponder the step to initialize how many and which combinations will produce at least 4. - So I could negate the number of combinations that produces less than 4. - (1,1) , (1,2) , (2,1) = 3 so 36 - 3 = 33 combinations that will yield at least 4. - 33/36 = 0.9166666667 = 92% chance. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ ------------------------------------------------ Self-critique Rating: ********************************************* Question: `q007. What is the probability of obtaining a 5-card hand of all red cards, when dealt from a 52-card deck? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: - Each card suit contains 13 cards. Two suits are red, two suits are black. - Thus the number of red suit cards = 13 * 2 = 26 - C(26,5) = 65,780 - Total 5-card hands = C(52, 5) - Therefore the probability of obtaining a 5-card hand consisting of only red cards is defined as - C(26,5) / C(52, 5) = 2.5% chance. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ ------------------------------------------------ Self-critique Rating: " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: #*&!