20105ee131A_1_ps2_2010fall_Sol

20105ee131A_1_ps2_2010fall_Sol - EE 131A Probability...

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EE 131A Problem Set 2 Probability Wednesday, October 6, 2010 Instructor: Lara Dolecek Due: Wednesday, October 13, 2010 Reading: Chapter 2 of Probability, Statistics, and Random Processes by A. Leon-Garcia 100 points total 1. Multiple Choice Test. Problem 2.44, page 86 of ALG Solution : There are 3 10 possible answers to the questions. Assuming each paper selects answer at random, then P { two papers have the same answers } = 3 10 3 10 × 1 3 10 = 1 3 10 = 1 . 69 × 10 - 5 2. Restless toddler. Problem 2.49, page 87 of ALG Solution : There are 3! possible permutations of three volumes, then P { the books are in the correct order } = 1 3! = 1 6 3. Testing With Replacement. Problem 2.56, page 87 of ALG Solution : (a) With replacement, the number of total possible outcomes is given by 50 5 . The problem is to pick k bad items and 5 - k good items, then the number of outcomes obtaining k defective items is ( 5 k ) 10 k 40 5 - k . Therefore, P [ X = k ] = ( 5 k ) 10 k 40 5 - k / 50 5 = ( 5 k ) ( 10 50 ) k ( 40 50 ) 5 - k binomial probabilities (b) Without replacement, the number of total possible outcomes is given by ( 50 5 ) . The
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