131A_1_hw2-solution

# 131A_1_hw2-solution - EE 131A Probabilities Instructor:...

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EE 131A Problem Set #2 Probabilities Instructor: Vwani Roychowdhury 1. Problem 2.57. Let X denote the input and Y the output. (a) P [ Y = 0] = P [ Y = 0 | X = 0] P [ X = 0] + P [ Y = 0 | X = 1] P [ X = 1] = 0 . 5(1 - ± 1 ) + 0 . 5 ± 2 (b) P [ X = 0 | Y = 1] = P [ Y = 1 | X = 0] P [ X = 0] P [ Y = 1] = ± 1 1 - ± 2 + ± 1 P [ X = 1 | Y = 1] = 1 - P [ X = 0 | Y = 1] = 1 - ± 2 1 - ± 2 + ± 1 Input 1 is more likely given Y = 1 if P [ X = 1 | Y = 1] > P [ X = 0 | Y = 1] 1 - ± 2 > ± 1 Input 0 is more likely otherwise. 2. Problem 2.59 (a) P [ H ] = P [ H | coin 1] P [ coin 1] + P [ H | coin 2] P [ coin 2] = 0 . 5( p 1 + p 2 ) 1

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(b) Using Baye’s Rule, we have P [ coin 2 | H ] = P [ H | coin 2] P [ coin 2] P [ H ] = p 2 p 1 + p 2 3. Problem 2.61 Let’s use D to represent defective, then in this problem P [ D | A ] = 0 . 001 P [ D | B ] = 0 . 005 P [ D | C ] = 0 . 01 Since the chip is randomly selected, P [ A ] = P [ B ] = P [ C ] = 1 / 3 and we can calculate P [ D ] = P [ A ] P [ D | A ] + P [ B ] P [ D | B ] + P [ C ] P [ D | C ] = 0 . 016 / 3 . Now we need to calculate P [ A | D ], P [ B | D ],and P [ C | D ]. P [ A | D ] = P [ A D ] P [ D ] = P [ A ] P [ D | A ] P [ D ] = 1 / 16 P [ B | D ] = P [ B D ] P [ D ] = P [ B ] P [ D | B ] P [ D ] = 5 / 16 P [ C | D ] = P [ C D ] P [ D ] = P [ C ] P [ D | C ] P [ D ] = 5 / 8 4. This is a good problem to illustrate the precision and care with which one needs to ask and understand probability related questions and issues. The problem statement can be interpreted in two diﬀerent manners, leading to two diﬀerent answers. In an exam the problem statement will be made very clear (so no need to worry about grades),
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## This note was uploaded on 03/27/2012 for the course EE EE 131A taught by Professor Roychowdhary during the Spring '05 term at UCLA.

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131A_1_hw2-solution - EE 131A Probabilities Instructor:...

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