PS06_sol - University of Illinois Spring 2010 ECE 313 Problem Set 6 Due Wednesday March 3rd at 4 p.m Reading Ross Chapter 3 Lecture Notes 14-18

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University of Illinois Spring 2010 ECE 313: Problem Set 6 Due: Wednesday, March 3rd at 4 p.m. Reading: Ross, Chapter 3; Lecture Notes 14-18. Noncredit Exercises: DO NOT turn these in. Chapter 3: Problems 80,84 and 86; Theoretical Exercises 8,16 and 21. This Problem Set contains seven problems. 1. [Conditional Probability] Die A has four red and two white faces, whereas die B has two red and four white faces. A fair coin is flipped once. If it falls heads, the game continues by throwing die A alone. If it falls tails, die B is to be used. (a) Show that the probability of red at any throw is 1 / 2. (b) If the first throw resulted in red, what is the probability of red in the third throw? (c) If red turns up in the first n throws, what is the probability that die A is being used? Solution: Let H be the event that the fair coin lands heads. Then P ( H ) = P ( H c ) = 1 / 2. In addition, let R denote the event that the outcome of rolling a die is red. (a) The law of total probability asserts that P ( R ) = P ( H ) P ( R | H ) + P ( H c ) P ( R | H c ) = 1 / 2. (b) Let R 1 be the event that the first throw results in red, and simi- larly, let R 3 be the event that the third throw results in red. Then P ( R 3 | R 1 ) = P ( R 1 R 3 ) P ( R 1) = 1 2 · ( 2 3 ) 2 + 1 2 · ( 1 3 ) 2 1 2 = 5 9 . (c) If N denotes the event that red shows up in the first n throws, then Bayes formula gives P ( A | N ) = P ( A ) P ( A | N ) P ( N ) = (1 / 2)(2 / 3) n 1 / 2((1 / 3) n + (2 / 3) n ) = 2 n 1 + 2 n . 2. [Conditional Probability and Poisson Random Variables] Each customer who enters Laura’s clothing store will purchase a suit with probability p . If the number of customers entering the store is Poisson distributed with mean λ , what is the probability that Laura does not sell any suits? What is the probability of her selling k suits? Solution: Let X be the number of suits that Laura sells, and let N denote the number of customers who enter the store. By conditioning on N we see that P { X = 0 } = X n =0 P { X = 0 | N = n } P { N = n } = X n =0 (1 - p ) n e - λ λ n n ! = e - λp .
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Similarly, by observing that P { X = k | N = n } = ± n k ² p k (1 - p ) n - k , for n k, and equals zero otherwise, one obtains P { X = k } = e - λp ( λp ) k k ! . 3.
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This note was uploaded on 04/20/2010 for the course ECE ECE 313 taught by Professor S during the Spring '10 term at University of Illinois at Urbana–Champaign.

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PS06_sol - University of Illinois Spring 2010 ECE 313 Problem Set 6 Due Wednesday March 3rd at 4 p.m Reading Ross Chapter 3 Lecture Notes 14-18

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