midterm_solutions

midterm_solutions - EE 464 Midterm Exam – G. Caire –...

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Unformatted text preview: EE 464 Midterm Exam – G. Caire – 10/4/2006 1 USC EE 464 – Probability Theory for Engineers – Fall 2006 Midterm Exam: Solutions 10/4/2006 1. In each packet of Corn Flakes a plastic figurine of a Star Wars character can be found. There are four possible characters, Luke, Leila, Darth Vader and Obi-Wan Kenobi, appearing independently and with equal probability. Find the probability that Luke, Leila and Darth Vader are found in a bulk purchase of five packets. Solution: Define A i = { object i is found } . Then, without loss of generality, the probability that Luke, Leila, and Darth Vader are found can be written as P ( A 1 ,A 2 ,A 3 ). To determine this probability, write: P ( A 1 ,A 2 ,A 3 ) = 1- P parenleftBigg 3 uniondisplay i =1 A c i parenrightBigg where, since all A i are equiprobable, P parenleftBigg 3 uniondisplay i =1 A c i parenrightBigg = 3 P ( A c 1 )- 3 P ( A c 1 ,A c 2 ) + P ( A c 1 ,A c 2 ,A c 3 ) = 3(1- 1 4 ) 5- 3(1- 1 2 ) 5 + (1- 3 4 ) 5 = 317 512 Hence, P ( A 1 ,A 2 ,A 3 ) = 1- P parenleftBig uniontext 3 i =1 A c i parenrightBig = 1- 317 512 = 195 512 ≈ . 3809. 2. There are 2 bins, B 1 ,B 2 . Bin B 1 contains 3 red balls and 7 green balls, and bin B 2 contains 6 red balls and 4 green balls. You pick at random one bin with probability 1 / 2, and remove two balls from it. (a) find the probability that the second ball is green. (b) find the probability that the second ball is green given that the first is green. Solution: Define the events: (a) b i : Box B i is chosen: i = 1 , 2. (b) g i : i th ball chosen is green: i = 1 , 2. (c) r i : i th ball chosen is red: i = 1 , 2. (a) The probability that the second ball is green is given by P ( g 2 ), given by: EE 464 Midterm Exam – G. Caire – 10/4/2006 2 P ( g 2 ) = P ( g 2 | b 1 ,g 1 ) · P ( g 1 | b 1 ) · P ( b 1 ) + P ( g 2 | b 2 ,g 1 ) · P ( g 1 | b 2 ) · P ( b 2 ) + P ( g 2 | b 1 ,r 1 ) · P ( r 1 | b 1 ) · P ( b 1 ) + P ( g 2 | b 2 ,r 1 ) · P ( r 1 | b 2 ) · P ( b 2 ) = ( 6 9 )( 7 10 )( 1 2 ) +( 3 9 )( 4 10 )( 1 2 ) +( 7 9 )( 3 10 )( 1 2 ) +( 4 9 )( 6 10 )( 1 2 ) = 11 20 (b) The probability that the second ball is green given the first ball is green is given by P ( g 2 | g 1 ) = P ( g 2 ,g 1 ) P ( g 1 ) , which is expanded as: P ( g 2 | g 1 ) = P ( g 2 ,g 1 ) P ( g 1 ) = P ( g 2 ,g 1 | b 1 ) P ( b 1 ) + P ( g 2 ,g 1 | b 2 ) P ( b 2 ) P ( g 1 | b 1 ) P ( b 1 ) + P ( g 1 | b 2 ) P ( b 2 ) = P ( g 2 | g 1 ,b 1 ) P ( g 1 | b 1 ) P ( b 1 ) + P ( g 2 | g 1 ,b 2 ) P ( g 1 | b 2 ) P ( b 2 ) P ( g 1 | b 1 ) P ( b 1 ) + P ( g...
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This note was uploaded on 05/03/2008 for the course EE 441 taught by Professor Neely during the Fall '08 term at USC.

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midterm_solutions - EE 464 Midterm Exam – G. Caire –...

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