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HWK1_sols

# HWK1_sols - IEOR 4701 1 Ross 4.28 Assignment 1 Proposed...

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IEOR 4701 Assignment 1 Proposed Solutions Summer 2011 1. Ross, 4.28 A sample of 3 items is selected at random from a box containing 20 items of which 4 are defective. Find the expected number of defective items in the sample. Solution. Let X denote the number of defective items in sample. If we assume that the sampling was done with replacement, then X Bin(3 , 4 / 20) and the expected number of defective items in the sample is 3 · 1 / 5 = 3 / 5 . Otherwise, then X HyperGeo( N = 20 , n = 3 , m = 4) , and E { X } = 3 · 4 20 = 3 / 5 . 2. Ross, 4.41 A man claims to have extrasensory perception. As a test, a fair coin is flipped 10 times, and the man is asked to predict the outcome in advance. He gets 7 out of 10 correct. What is the probability that he would have done at least this well if he had no ESP? Solution. Let X = number of right guesses. Then X Bin(10 , 1 / 2) . P { have done as well } = P { X 7 } = 1 - P { X 6 } = 1 - 0 . 8281 = 0 . 1719 . 3. Ross, Theoretical 4.28. Let X be a negative binomial random variable with parameters r and p , and let Y be a binomial random variable with parameters n and p . Show that P { X > n } = P { Y < r } by means of probabilistic interpretation of these random variables. Do not attempt to give any analytical proof of the preceding. Solution. We give a description of each of these events: { X > n } = { at least n + 1 trials needed to obtain r successes } = { not more than n trials needed to have r successes } C = { in a sequence of n trials there are at least r successes } C { Y < r } = { less than r successes happen in n trials } = { at least r successes happen in n trials } C 4. At 8:59 AM, A, B, C are waiting (in that order) outside a bank that will open at 9 AM. The bank has two tellers; each of the requirements that A, B, and C are bringing to the bank are i.i.d. exponentially distributed random variables with mean 1. What is the probability that A is the last one to leave the bank? Solution. Let T A , T B , and T C the times that A, B, and C take to be served, respectively. We are interested in computing P { T A > T B + T C } . Intuitively, using the memory-less property of the exponential distribution and the fact that the three random variables are i.i.d., we would Page 1 of 7

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IEOR 4701 Assignment 1 Proposed Solutions Summer 2011 expect this probability to be P { T A > T B } · P { T A > T C } = 1 2 · 1 2 = 1 4 . We verify this intuition below: P { T A > T B + T C } = P { T A > T B + T C | T A > T B } P { T A > T B } + P { T A > T B + T C | T A T B } P { T A T B } = P { T A > T B + T C | T A > T B } P { T A > T B } + 0 = = 1 2 · 1 2 = 1 4 . P { T A > T C } = P { T A > T B } = 1 / 2 is due to the fact that T A , T B , T C are i.i.d. You can also verify it formally: P { T A > T B } = Z 0 P { T A > t } · f T B ( t ) dt = Z 0 e - t · e - t dt = Z 0 e - 2 t dt = 1 2 . You were expected to know how to do the above. For those who are curious, we present yet another way of thinking about this. We use the fact that min(Exp( λ 1 ) , Exp( λ 2 )) Exp( λ 1 + λ 2 ), and that P { Exp( λ 1 ) = min(Exp( λ 1 ) , Exp( λ 2 )) } = λ 1 λ 1 + λ 2 . Using this and the
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