course Mth 152 U]ݾYԹYՎ}Wxa{assignment #008
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11:18:00 `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 --> n- number of trials n=2 p- probability of success p=1/2 q- 1-p = 1-1/2 q=1/2 x- number of success = x=2 P(2) = C(2,2) (1/2)^2 (1/2)^0 2!/ (2!(2-2)!) = 2!/(2!*0!) = 2*1/0 = 0
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11:19:19 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 --> I tried to work it through the binomial probability formula. self critique assessment: 2
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11:21:15 `q002. What is the probability that on three rolls of a fair die, we obtain exactly two 5's?
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RESPONSE --> 1/6*1/6*5/6 = 5/216 of the three rolls we would get 1 with a 5 and another with a five and the other would be something other than a 5 confidence assessment: 2
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11:23:26 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 --> I got to 5/216 and did not mutiple by 3 to get 15/216 self critique assessment: 2
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11:27:33 `q003. What is the probability that on six rolls of a fair die, we obtain exactly two 5's?
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RESPONSE --> 1/6*1/6*5/6*5/6*5/6*5/6 = 625/46656 *6 = 3750/46656 = 1375/23328 = .06 confidence assessment: 1
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11:28:27 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 --> That is how I worked it. self critique assessment: 2
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11:33:37 `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 --> P(r) = C(n,r)(p)^r(1-r)^(n-r)
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11:35:20 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 --> I tried to put the letters in the binomial formula self critique assessment: 2
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11:40:39 `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 --> Because the probability of not getting the give would be the number subtracted from the probability of getting the five, the not probability would be what is left after the probability p= 5/5 - 1/5 confidence assessment: 1
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11:41:52 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 --> I said it would be what was left of probable to get not probable self critique assessment: 2
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