Chapt 1c

# Chapt 1c - 1.7 Conditional Probability 21 = ( . 1 )( . 4 )...

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Unformatted text preview: 1.7 Conditional Probability 21 = ( . 1 )( . 4 ) ( . 9 )( . 6 ) + ( . 1 )( . 4 ) = . 069 (c) The probability that x 1 was transmitted, given that y 1 was received, is given by P ( x 1 | y 1 ) = P ( x 1 y 1 ) P ( y 1 ) = P ( y 1 | x 1 ) P ( x 1 ) P ( y 1 | x 1 ) P ( x 1 ) + P ( y 1 | x 2 ) P ( x 2 ) = ( . 9 )( . 6 ) ( . 9 )( . 6 ) + ( . 1 )( . 4 ) = . 931 (d) The probability that x 2 was transmitted, given that y 2 was received, is given by P ( x 2 | y 2 ) = P ( x 2 y 2 ) P ( y 2 ) = P ( y 2 | x 2 ) P ( x 2 ) P ( y 2 | x 1 ) P ( x 1 ) + P ( y 2 | x 2 ) P ( x 2 ) = ( . 9 )( . 4 ) ( . 1 )( . 6 ) + ( . 9 )( . 4 ) = . 857 (e) The unconditional probability of error is given by P e = P ( x 1 ) P 12 + P ( x 2 ) P 21 = ( . 6 )( . 1 ) + ( . 4 )( . 1 ) = . 1 trianglesolid Example 1.8 The quarterback for a certain football team has a good game with probability 0.6 and a bad game with probability 0.4. When he has a good game, he throws at least one interception with a probability of 0.2; and when he has a bad game, he throws at least one interception with a probability of 0.5. Given that he threw at least one interception in a particular game, what is the probability that he had a good game? Solution Let G denote the event that the quarterback has a good game and B the event that he had a bad game. Similarly, let I denote the event that he throws 22 Chapter 1 Basic Probability Concepts at least one interception. Then we have that P ( G ) = . 6 P ( B ) = . 4 P ( I | G ) = . 2 P ( I | B ) = . 5 P ( G | I ) = P ( G I ) P ( I ) According to the Bayes formula, the last equation becomes P ( G | I ) = P ( G I ) P ( I ) = P ( I | G ) P ( G ) P ( I | G ) P ( G ) + P ( I | B ) P ( B ) = ( . 2 )( . 6 ) ( . 2 )( . 6 ) + ( . 5 )( . 4 ) = . 12 . 32 = 3 / 8 = . 375 trianglesolid Example 1.9 Two events A and B are such that P [ A B ] = . 15, P [ A B ] = . 65, and P [ A | B ] = . 5. Find P [ B | A ] . Solution P [ A B ] = P [ A ]+ P [ B ] P [ A B ] . 65 = P [ A ]+ P [ B ] . 15. This means that P [ A ] + P [ B ] = . 65 + . 15 = . 80. Also, P [ A B ] = P [ B ] P [ A | B ] . This then means that P [ B ] = P [ A B ] P [ A | B ] = . 15 . 50 = . 30 Thus, P [ A ] = . 80 . 30 = . 50. Since P [ A B ] = P [ A ] P [ B | A ] , we have that P [ B | A ] = P [ A B ] P [ A ] = . 15 . 50 = . 30 trianglesolid Example 1.10 A student went to the post office to mail a package to his parents. He gave the postal attendant a bill he believed was \$20. However, the postal at- tendant gave him change based on her belief that she received a \$10 bill from the student. The student started to dispute the change. Both the student and 1.7 Conditional Probability 23 the postal attendant are honest but may make mistakes. If the postal attendants drawer contains 30 \$20 bills and 20 \$10 bills, and she correctly identifies bills 90% of the time, what is the probability that the students claim is valid?...
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## This note was uploaded on 01/05/2010 for the course STAT 350 taught by Professor Carlton during the Fall '07 term at Cal Poly.

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Chapt 1c - 1.7 Conditional Probability 21 = ( . 1 )( . 4 )...

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