EE465-USC-Fall08-Assignment1_Solution

EE465-USC-Fall08-Assignment1_Solution - EE 465 Homework 1...

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EE 465 : Homework 1 Solutions 1. n [ i =1 E i = E 1 [ ± E 1 E 2 ² [ ± E 1 E 2 E 3 ² [ ... [ ± E 1 E 2 E n 1 E n ² Since the terms on the right hand side are mutually exclusive, we have P [ i E i ! = P ( E 1 )+ P ± E 1 E 2 ² + + P ± E 1 E 2 E n 1 E n ² Since P ( AB ) P ( A ) A,B ,wehave P ( [ i E i ) P ( E 1 P ( E 2 + P ( E n ) P n [ i =1 E i ! n X i =1 P ( E i ) (Induction can also be used to prove the bound) 2. Using the theorem of total probability, P ( win )= 12 X i =2 P ( win | throw i ) P ( throw i ) P ( win | throw i 1 ,i =7 , 11 0 =2 , 3 , 12 P ( ibefore 7) =4 , 5 , 6 , 8 , 9 , 10 Using the result from 1.12 (also discussed in the discussion session), for any two mutually exclu- sive events E and F , P { event E occurs before F } = P ( E ) P ( E P ( F ) Now since i and 7 are mutually exclusive events for any i ± P ( 7) = P ( i ) P ( i P (7) We have P (2) = P (12) = 1 36 ,P (3) = P (11) = 2 36 (4) = P (10) = 3 36 (5) = P (9) = 4 36 (6) = P (8) = 5 36 (7) = 6 36 and P ( win )=(1)[ P (7) + P (11)] + (0) [ P (2) + P (3) + P (12)] + 10 X i =4 ,i ± =7 ± P ( i ) P ( i P (7) ² P ( i ) Substituting the P ( i ) values in the above equation, we get P ( win )=0 . 49
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3. Let C = event that a person is blind. Using Bayes’ Formula,
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This note was uploaded on 08/05/2009 for the course EE 465 taught by Professor Chow during the Fall '04 term at USC.

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EE465-USC-Fall08-Assignment1_Solution - EE 465 Homework 1...

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