course Mth 152 Àë²Á…|ê‹¡^¬fƒX©¢aÛ‚•íùD £wܯassignment #009
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14:25:36 Query 12.4.3 P(2 H on 3 flips)
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RESPONSE --> 3/8
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14:26:18 ** On three flips you can get HHH, HHT, HTH, HTT, THH, THT, TTH, TTT. Of there 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. If you have n flips, there are C(n,r) ways to get r Heads. This number appears in the n+1 row, as the r+1 entry, of Pascal's triangle. **
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RESPONSE --> ok
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14:28:26 What is the significance of .5^2 * .5 for this question?
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RESPONSE --> There are three coins tossed and each time there is a .5 chance that it will be heads. There are three .5's multiplied by each other, representing the number of times that the coin is being flipped.
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14:28:35 ** .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. **
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RESPONSE --> ok
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14:31:18 Query 12.4.6 P(>= 1 H on 3 flips) Give the requested probability and explain how you obtained your result.
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RESPONSE --> HHH, HHT, HTH, HTT, THH, THT, TTH, TTT Only one of these possibilities contains zero heads, so the answer is: 7/8
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14:31:33 ** 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. **
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RESPONSE --> oi
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14:32:40 Query 12.4.15 P(3 H on 7 flips) Give the requested probability and explain how you obtained your result.
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RESPONSE --> C(7,3)*(1/2)^3*(1/2)^4 = 35/128
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14:32:46 ** 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. **
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RESPONSE --> ok
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14:33:37 Query 12.4.21 P(1 success in 3 tries), success = 4 on fair die
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RESPONSE --> 25/72
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14:33:42 ** 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 = 4 * 1/6 * 25 / 36 = 100 / 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. **
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RESPONSE --> ok
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14:34:43 Query 12.4.33 P(exactly 7 correct answers), 3-choice mult choice, 10 quest. What is the desired probability?
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RESPONSE --> C(10,7)*(1/3)^7*(2/3)^3 = .016
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14:35:04 ** 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 . **
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RESPONSE --> ok
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14:35:24 Query 12.4.39 P(more than 2 side effect on 8 patients), prob of side effect .3 for each
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RESPONSE --> .448
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14:35:38 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?
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RESPONSE --> ok
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14:36:23 ** The 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. **
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RESPONSE --> ok
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14:38:55 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?
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RESPONSE --> C(10,8)*(1/2)^8*(1/2)^2 = .044
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14:39:03 ** 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. **
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RESPONSE --> ok
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14:39:10 Query Add comments on any surprises or insights you experienced as a result of this assignment.
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RESPONSE --> Alright
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