IE111_F09_HW05_soln

IE111_F09_HW05_soln - IE 111 Fall 2009 Homework #5,...

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IE 111 Fall 2009 Homework #5, solutions Question 1. To check on the security screeners at an airport, we put 4 fake bombs into a set of 52 baggage pieces. Suppose that the screeners actually just pick 6 pieces randomly (without replacement) to inspect. Let X be the number of our fake bombs that they find. a) According to which well known distribution is X distributed? X~ Hypergeometric (M=52, r=4, N=6) b) Find the Probability Mass Function f(x). Remember to specify an appropriate domain for x. P(X=0) = 48 C 6 * 4 C 0 / 52 C 6 = 0.6028, P(X=1) = 48 C 5 * 4 C 1 / 52 C 6 = 0.3364, P(X=2) = 48 C 4 * 4 C 2 / 52 C 6 = 0.0573, P(X=3) = 48 C 3 * 4 C 3 / 52 C 6 = 0.0034, P(X=4) = 48 C 2 * 4 C 4 / 52 C 6 = 0.000055, or in functional form: P X = x = ۴۸ ۶ x  ۴ x ۵۲ ۶ , x = ۰,۱,۲,۳,۴. c) Find E(X) E[X]= 0* P(X=0) + 1* P(X=1)+ 2* P(X=2)+ 3*P(X=3)+ 4* P(X=4) = 0.461538462. Also notice that if we approximate this hypergeometric distribution with the binomial distribution with parameters N=6, p=r/M=4/52, then the approximate expected value is Np=24/52=0.4615. Actually, though the binomial distribution is only an approximation, the expected value is exact. Question 2. An urn has 6 red marbles and 14 green marbles. Careful on this one, the distribution to use keeps changing! a) If I pick 3 marbles with replacement , and let X= the number of red marbles picked, what is the P(X=1)? X~ Binomial (N=3, p=6/20) P(X=1) = 3 C 1 p 1 (1-p) 2 =3* (6/20) 1 *(1-6/20) 2 = 0.441 b) If I keep picking marbles with replacement until I get my first red marble, what is the probability I picked exactly 7 times? X~ Geometric (p=6/20) P(X=7) = (1-p) 6 p =(1-6/20) 6 *(6/20) = 0.0353 c) If I pick 3 marbles without replacement and let X= the number of red marbles picked, what is the P(X=1)? X~ Hypergeometric (M=20, N=3, r=6) P(X=1)= 6 C 1 * 14 C 2 / 20 C 3 = 0.4789 d) If I pick 3 marbles without replacement and let X= the number of red marbles picked, what is the expected value of X? 1
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E[X]= N*(r/M) = 3*6/20 =9/10 e) Suppose I have picked 8 marbles with replacement and have not yet gotten a red marble. How many more picks do I expect will be required before my first red marble? E[X] = 1/p for geometric distribution. From the lack of memory property of geometric distribution, no matter that I did not find red in first 8 trials, expected number of picks until I find red after that time is 1/(6/20), which is 20/6. f) Suppose I have picked 9 marbles without replacement and have not yet gotten a red marble. How many more picks (without replacement) do I expect will be required before my first red marble? This is a tricky problem! We can't use any of our favorite distributions; you have to think through it from first principles. Here's a hint to get you started. Let X = number of remaining picks. Compute P(X=1), P(X=2), P(X=3), etc. (it will help to think of the Geometric distribution, but it's not quite the same) and then compute the mean value from there. After 9 picks, there are 6 red and 5 green marbles left. So
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This note was uploaded on 02/21/2010 for the course IE 111 taught by Professor Storer during the Spring '07 term at Lehigh University .

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IE111_F09_HW05_soln - IE 111 Fall 2009 Homework #5,...

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