course ø¤»pî£ú÷æÔÆ»ú‚¸I òÙªËÙÄ€w¿assignment #009
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19:14:48 Query 12.4.3 P(2 H on 3 flips)
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RESPONSE --> P(3)= 3C3*(1/2)^3(1/2)^0= 1*1/8=1/8
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19:18:05 ** 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 gave the answer for #4 (3 heads) however, my number 3 is: P(2)=3C2*(1/2)^2*(1/2)= 3*(1/4)*(1/2)=3/8 Which gives the same solution as shown here
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19:20:06 What is the significance of .5^2 * .5 for this question?
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RESPONSE --> it is p(possibilites) .5 raised to 2 successes times .5 (probibility of failures)
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19:21:09 ** .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 understand where these numbers come from
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19:23:56 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 --> P(1,2,or3)=P(1)+P(2)+P(3) 3/8+3/8+1/8=7/8 I worked out the prob of obtaining each 1, 2, and 3 and then added the results togather
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19:24:36 ** 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 understand that it would have been quicker to use this method
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19:26:32 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 --> P(3)= 7C3*(1/2)^3*(1/3)^4= 35*1/8*1/16 =35/128
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19:26:56 ** 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 understand how I abtained the correct solution
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19:28:20 Query 12.4.21 P(1 success in 3 tries), success = 4 on fair die
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RESPONSE --> P(1)=3C1*(1/6)*(1/2)^2=3*1/6*1/36=1/72
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19:30:38 ** 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 understand that I should have used 5/6^2 instead of 1/6 because that is the prob of a failure
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19:39:56 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 --> P(7)=C(10,7)*(1/3)^7*(2/3)^3
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19:40:52 ** 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 understand how to get the solution.
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19:57:16 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 --> P(1,2)=P(1)+P(2) P(1)=C(8,1)*(.3)*(.7)^7= 8*.30*.0823=.197 P(2)=C(8,1)*(.3)^2*(.7)^6=28*.09*.117=.294 .197+.294=.491 1-.491=.509(prob of getting more than 2 with side effects)
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19:57:48 ** 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 left out the prob of 0, but I get the idea
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20:00:43 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 -->
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20:01:40 ** 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 --> My 48 is about walking 1 block north for heads and 1 block south for tails sorry
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20:09:42 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 --> ??? P(6)=C(10,6)*(1/2)^6*(1/2)^4
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20:11:20 ** 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 --> Oh ok I understand now that you go 8 blocks south
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20:11:29 Query Add comments on any surprises or insights you experienced as a result of this assignment.
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RESPONSE -->
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