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14:10:25 `q001. Note that there are 5 questions in this assignment. What is the probability that on two rolls of a fair die, we obtain exactly two 3's?
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RESPONSE --> C(6,2) x (1/6)^2 x (5/6)^4 = 3125/15552 confidence assessment: 2
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14:11:01 The probability of obtaining a 3 on a single role is 1/6 (one of the six possible outcomes is a 3). Since the two rolls are independent, it follows that if two dice are rolled the probability of obtaining two 3's is 1/6 * 1/6 = 1/36.
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RESPONSE --> ok self critique assessment: 3
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14:13:26 `q002. What is the probability that on three rolls of a fair die, we obtain exactly two 5's?
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RESPONSE --> C(3,2) x (1/6)^2 x (5/6)^1 = 5/72 confidence assessment: 2
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14:13:48 On three rolls of a fair die, the two 5's can occur on the first and second, the first and third or the second and third rolls. That is, of the three available positions the two positions in which the 5's occur can occur in C(3,2) = 3 different ways. Since the probability of a 5 on any roll is 1/6 and the probability of not getting a 5 on a roll is 5/6. Any one of the three ways of getting two 5's and one non-5 is therefore (1/6) * (1/6) * (5/6 ) = 5/216. Since each of the three ways to get the desired outcome occurs with probability 5/216, it follows that Probability of exactly two 3's on three rolls = 3 * 5/216 = 15/216 = 5/72.
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RESPONSE --> ok self critique assessment: 3
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14:14:56 `q003. What is the probability that on six rolls of a fair die, we obtain exactly two 5's?
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RESPONSE --> C(6,2) x (1/6)^2 x (5/6)^4 = 3125/15552 confidence assessment: 2
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14:15:38 In order to get exactly two 5's on six rolls of the fair die, we must get two 5's and four results that are not 5. The probability of getting a 5 on any roll is 1/6, and the probability of getting a result other than 5 is 5/6. Therefore given any two positions out of the six the probability of obtaining 5's in two of the positions and non-5's in the remaining four positions is by the Fundamental Counting Principle Probability of 5's in exactly two of the six positions = (1/6) * (1/6) * (5/6) * (5/6) * (5/6) * (5/6) = (1/6)^2 * ( 5/6)^4. There are C(6,2) ways in which the positions of the two 5's can be selected from the six available positions. Thus we have Probability of exactly two 5's on six flips = C(5,2) * (1/6)^2 * (5/6)^4.
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RESPONSE --> ok self critique assessment: 3
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14:19:33 `q004. If we let p stand for the probability of getting a 5 on a roll of a die and q for the probability of not getting a 5 on a roll, then how would we expressed a probability of getting exactly r 5's on n rolls?
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RESPONSE --> C(n,r) x (p)^r x (q)^(n - r) confidence assessment: 2
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14:19:59 By analogy with the preceding problem, we see that to get r 5's on n rolls we must get 5 the total of r times and non-5 a total of (n-r) times. Since probability of getting a 5 is p, the probability of getting 5 a total of r times is represented by p^r. Since the probability of getting a non-5 is q, then the probability of getting a non-5 a total of (n-r) times is represented by q^(n-r). There are C(n, r) ways to place fives in r of n positions, so the probability of getting 5 fives and n non-fives is C(n, r) * p^r * q^(n-r).
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RESPONSE --> ok self critique assessment: 3
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14:23:17 `q005. Explain why, if p is the probability of getting a 5 on a single roll of a die, it follows that the probability q of not getting a 5 is q = 1-p. How would we therefore express the formula C(n, r) * p^r * q^(n-r) only in terms of p?
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RESPONSE --> C(n,r)*p^r*(1-p)^(n-r) confidence assessment: 1
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14:23:56 If we roll a single die, we either get 5 or we don't. The two events are mutually exclusive -- they can both happen on the same roll. They also cover all possibilities. The sum of the probabilities is therefore 1. So we conclude that p + q = 1, and from this it follows immediately that q = 1 - p. Substituting 1 - p for q in the expression C(n, r) * p^r * q^(n-r) we obtain Probability of r fives on n rolls = C(n, r) * p^r * (1-p) ^ (n-5).
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RESPONSE --> ok self critique assessment: 3
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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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