HW6-soln - IE 111 Fall 2007 Homework #6 Solutions Due...

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IE 111 Fall 2007 Homework #6 Solutions Due Friday October 19 at the beginning of class 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~ Hyper geometric ( N=52,K=4), E[X]= np = 6*(4/52) = 0.462 , V[X] = np*(1-p)*(N-n)/(N-1) = 6*(4/52) *48/52*46/51 = 0.384 b) Find the Probability Mass Function f(x). Remember to specify an appropriate range for x. P X (x=0) = 48 C 6 * 4 C 0 / 52 C 6 = 0.6028, P X (x=1) = 48 C 5 * 4 C 1 / 52 C 6 = 0.3364, P X (x=2) = 48 C 4 * 4 C 2 / 52 C 6 = 0.0573, P X (x=3) = 48 C 3 * 4 C 3 / 52 C 6 = 0.0034 , P X (x=4) = 48 C 2 * 4 C 4 / 52 C 6 = 0.000055 c) Find E(X) E[X] = 0* P X (x=0) + 1* P X (x=1)+ 2* P X (x=2)+ 3* P X (x=3)+ 4* P X (x=4) = 0.4615 d) Find V(X) V[X] = 0* P X (x=0) + 1* P X (x=1)+ 4* P X (x=2)+ 9* P X (x=3)+ 16* P X (x=4) – E[X] 2 = 0.384 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)?
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X~ Hyper geometric ( N=20,K=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? E[X]= n*p = 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 homework help was uploaded on 02/26/2008 for the course IE 111 taught by Professor Storer during the Spring '07 term at Lehigh University .

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HW6-soln - IE 111 Fall 2007 Homework #6 Solutions Due...

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