course Mth 152 öÐÊé‘®Øȯoï±êðû肱¢¯üˆÓhÚassignment #009
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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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