#$&* course Mth 152 11/10/2012 5:46 009. ``q Query 9
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Given Solution: On three flips you can get HHH, HHT, HTH, HTT, THH, THT, TTH, TTT. Of these 8 possibilities, only 3 of them have two Heads. Thus the probability is 3 / 8. You can get this result without listing. There are 2 possibilities for each flip which gives you 2*2*2 = 2^3 = 8 possible outcomes. To get 2 `heads' you must get `heads' in exactly 2 of the 3 positions. There are C(3, 2) = 3 possible choices of the 3 positions so the probability is C(3,2) / 2^3 = 3/8. More generally, if you have n flips, there are C(n,r) ways to get r Heads. The value of C(n, r) appears in the n+1 row, as the r+1 entry, of Pascal's triangle. STUDENT COMMENT: I solved this question using the given solution for binomials, it might have been more work, but I’m guessing it’s ok? INSTRUCTOR RESPONSE: Your solution is fine. Make sure you also understand the given solution, which is a reasoning process as opposed to a formula. From your previous work I'm confident you do. Knowing how the formula represents the reasoning process, you can then use the formula with confidence. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating: OK ********************************************* Question: What is the significance of .5^2 * .5 for this question? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: .5^2 is the probability of getting the same flip times in a row, and .5 is the probability of getting heads or tails. So .5^2 * .5 can be the probability of getting HHT, and the probability is the same for any other combination of two heads and a tail. There are C(3, 2) = 3 possibly orders of two heads and a tail, so 3 * .5^2 * .5 = 3 * .125 = 375 = 3/8. confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: .5^2 is the probability of getting Heads twice in a row. .5 is the probability of a Tails. .5^2 * .5 is therefore the probability of getting HHT. Since the probabilities are independent, you have the same probability of getting two Heads and a Tail in some different order. Since there are C(3,2) possible orders for 2 Heads on 3 coins, the probability of getting 2 Heads and one Tail is C(3,2) * .5^2 * .5 = 3 * .125 = .375, the same as the 3/8 we obtained by listing. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating: OK ********************************************* Question: `q Query 12.4.6 P(>= 1 H on 3 flips) Give the requested probability and explain how you obtained your result. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: There is one possibility that doesn’t give us one head or more and that is TTT. That leaves 7 ways to get >= 1 head so there are 7/8 probability of getting >= 1 head. confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: Probability of getting no heads on three flips is P(TTT) = .5 * .5 * .5 = .125, or 1/8, obtained by multiplying the probability of getting a tails for each of 3 independent flips. Subtracting this from 1 gives .875, or 7/8. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating: OK &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: Query 12.4.15 P(3 H on 7 flips) Give the requested probability and explain how you obtained your result. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: There are C(7, 3) = 7 * 6 * 5/3 * 2 * 1 = 35 possible orders for 3 heads out of 7 flips. There are (1/2)^3 * (1/2)^4 = (1/2)^7 = 1/128 probability of getting 3 heads out of 7 flips in any order, so there are 35 * 1/128 = 35/128 probability of getting 3 heads out of 7 flips. confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: There are C(7,3) = 7 * 6 * 5 / 3! = 35 ways to choose three of the 7 `positions' for Heads on 7 flips. So there are C(7,3) = 7 * 6 * 5 / 3! = 35 ways to get three heads on 7 flips. The probability of any of these ways is (1/2)^3 * (1/2)^4 = 1 / 2^7 = 1 / 128. The probability of 3 Heads on 7 flips is therefore 35 * 1/128 = 35 / 128. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating: OK `q Query 12.4.21 P(1 success in 3 tries), success = 4 on fair die YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: There is a 1/6 probability of being successful and a 5/6 probability of not being successful. So there is (1/6) * (5/6)^2 probability of being successful once in 3 tries in any order. There are C(3, 1) = 3 possibly orders, so there are 3 * (1/6) * (5/6)^2 = 3* (1/6) * (25/36) = 75/ 216 = 25/72 probability of being successful once in 3 tries. confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: To get 1 success on 3 tries you have to get 1 success and 2 failures. On any flip the probability of success is 1/6 and the probability of failure is 5/6. For any ordered sequence with 1 success and 2 failures the probability is 1/6 * (5/6)^2. Since there are C(3,1) = 3 possible orders in which exactly 1 success can be obtained, the probability is C(3,1) * 1/6 * (5/6)^2 = 3 * 1/6 * 25 / 36 = 75 / 216 = 25 / 72. This matches the binomial probability formula C(n, r) * p^r * q^(n-r), with prob of success p = 1/6, prob of failure q = 1 - 1/6 = 5/6, n = 3 trials and r = 1 success. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating: OK `q Query 12.4.33 P(exactly 7 correct answers), 3-choice mult choice, 10 quest. What is the desired probability? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: There is a 1/3 probability of getting the correct answer, and a 2/3 probability of getting an incorrect answer. So there is a (1/3)^7 * (2/3)^3 probability of getting 7 correct answers out of 10 question in any order. There are C(10, 7) = 120 possibly orders, so there are 120 * (1/3)^7 * (2/3)^3 = 0.0163 probability of getting 7 correct answers out of 10 question. confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: The probability of a correct answer from a random choice on any single question is 1/3. For any sequence of 7 correct answers and 3 incorrect the probability is (1/3)^7 * (2/3)^3. There are C(10,7) possible positions for 7 correct answers among 10 questions. So the probability is C(10,7) * (1/3)^7 * (2/3)^3 = 320/19683 = 0.0163 approx. This matches the binomial probability formula C(n, r) * p^r * q^(n-r), with prob of success p = 1/3, prob of failure q = 1 - 1/3 = 2/3, n = 10 trials and r = 7 success. ANOTHER SOLUTION: There are C(10,7) ways to distribute the 7 correct answers among the 10 questions. The probability of any single outcome with 7 successes and 3 failures is the product of (1/3)^7, representing 7 successes, and (2/3)^3, representing 3 failures. The probability of exactly seven correct questions is therefore prob = C(10,7) * (2/3)^3 * (1/3)^7 . ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating: OK `q Query 12.4.39 P(more than 2 side effect on 8 patients), prob of side effect .3 for each YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: The probability of 2 side effects is C(8, 2) * .3^2 * .7^6 = .296, the probability of 1 side effect is C(8,1) * .7^7 * .3^1 = 0.198, the probability of 0 side effects is C(8,0) * .7^8 = .058. .296 + .198 + .058 = .552, 1 - .552 = .448 probability of 2 or more side effects on 8 patients. confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `aThe probability of 0 side effects is C(8,0) * .7^8. The probability of 1 side effect is C(8,1) * .7^7 * .3^1. The probability of 2 side effects is C(8,2) * .7^6 * .3^2. The sum of these two probabilities is the probability that two or fewer patients will have side effects. We subtract this probability from 1 to get the probability that more than 2 will experience side effects. The result is approximately .448. DER** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating: OK `q Query 12.4.48 P(4 th child is 1 st daughter) What is the probability that the fourth child is the first daughter and how did you obtain your result? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: There is .5 probability for having a son or daughter, so for the fourth child to be a daughter the first three would have to be sons. So there is .5^3 * .5 = .625 = 1/16 chance of the 4 child being a daughter. confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `aThe fourth child will be the first daughter if the sequence is SSSD, S standing for son and D for daughter. The probability of S on any birth is .5, and the probability of G is .5. The probability of SSSD is .5^3 * .5 = .0625 or 1/16. ** `q Query 12.4.54 10-step rnd walk, 1 dim; P(6 South) What is the probability of ending up 6 blocks South of the starting point and how did you obtain it? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: We must got 8 steps south and 2 steps north to get to 6 steps south. There are C(10, 8) = 45 orders that we can do this in. There are (1/2)^8 * (1/2)^2 = (1/2)^10 = 1/1024 probability of getting 6 south in any order, so there is 45 * 1/1024 = 45/1024 probability of getting 6 south in 10 coin flips. confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: To end up 6 blocks South requires 8 steps South and 2 steps North. Thus exactly 8 of the 10 steps must be South, and there are C(10,8) ways for this to happen. The probability of any given combination of 8 South and 2 North is (1/2)^8 * (1/2)^2 = 1 / 2^10 = 1 / 1024. The probability of ending up 6 blocks South is therefore prob = C(10,8) * (1/2)^8 * (1/2)^2 = 45 * (1/2)^10 = 45 / 1024, or about .044. ** STUDENT QUESTION: Do we find the 8 steps South and 2 steps North by trial and error? INSTUCTOR RESPONSE: You have to take 10 steps, each north or south. If n stands for the number of steps north and s for the number of steps south, then n and s add up to 10, while the net number of steps south is s - n. We could even up the system of equations n + s = 10 s - n = 6. and solve for n and s. However if you think about the situation, the answer is fairly obvious, so I didn't complicate the solution with the details of this reasoning. Instead I just made the assertion 'To end up 6 blocks South requires 8 steps South and 2 steps North.' &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating: OK Query Add comments on any surprises or insights you experienced as a result of this assignment. " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: Query Add comments on any surprises or insights you experienced as a result of this assignment. " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: #*&!