course Mth 152 ßÛÏ •šÕ½»Oö狇ŸÞ‰¤Ú|¢ÍµÞassignment #009
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12:14:26 Query 12.4.3 P(2 H on 3 flips)
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RESPONSE --> 1/2*1/2*1/2*3= 3/8
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12:23:21 ** 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 --> I did the C(3,2) to get 3 and mutiplied 1/2^3 to get 3/8
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12:32:16 What is the significance of .5^2 * .5 for this question?
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RESPONSE --> You would square the .5 and then mutiply it by .5 so in reality it would be .5^3
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12:33:44 ** .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 --> I'm sorry I really didn't under the question.
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12:39:45 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 --> 1/2*1/2*1/2*3 = 3/8 there would be 1 out of 2 chances that it would be heads and there are 3 tosses C(3,1) 3!/ (1! (3-1)!) = 3*2*1 / (1 * 2*1) = 3
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12:41:48 ** 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 --> I mutiplied by 3 to get 3/8 and didn't subtract because I thought there was at least 1 heads obtained
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13:42:53 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 --> 3 heads 7 flips n = 7 x = 3 p = 1/2 q = 1 - 1/2 = 1/2 7!/ 3!(7-3)! = 7*6*5*4*3*2*1/ (3*2*1)*(4*3*2*1)= (7*6*5)/(3*2*1)= 210/6 = 35 35(1/2)^3(1/2)^4 = 35(1/8)(1/16) = 35/128
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13:43:50 ** 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 --> I actually worked it out right. I pludged in the numbers into the formual and worked it like that.
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13:51:42 Query 12.4.21 P(1 success in 3 tries), success = 4 on fair die
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RESPONSE --> n = 3 p = 1/6 q = 1-p = 6/6-1/6 = 5/6 x = 1 3!/(1!(3-1)! = 3*2*1 / (1) * (2*1) = 3 3*(1/6)^1(5/6)^2 = 3*(1/6)(25/36) = 75/216 = 25/72
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13:52:29 ** 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 --> I got it right again!!!!!! I took the formula step by step and got the answer
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13:56:56 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 --> Would add the binomials
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14:00:16 ** 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 --> I'm sorry I thought it was just what probably would be used instead of working it out. I went through your answer to see exactly how you worked it.
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14:29:47 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 --> n = 8 p = 1/3 q = 1-1/3 = 2/3 x = 2 8!/(2!(8-2)!) = 8*7*6*5*4*3*2*1 / (2*1) * (6*5*4*3*2*1) = 8*7 / 2*1 = 56/2 = 28 28(1/3)^2(2/3)^6 = 28 (1/9) (64/729) = 1792/6561 =.27
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14:33:23 ** The 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**
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RESPONSE --> I got .27 instead of .448. I didn't work it the way you did. So you would find the probability for each side effect but adding and then subtracting them from 1.
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14:42:04 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 --> n = 4 p = 1/4 q = 1-p = 3/4 x = 1 4! / (1!(4-1)! = 4*3*2*1 / (1*) * (3*2*1) = 4*1 = 4 4 (1/4)^1(3/4)^3 = 4 (1/4) (27/64) = 108/256 = 54/128 = 27/64
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14:46:43 ** 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 --> I worked it wrong again.
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14:46:51 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 -->
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