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Unformatted text preview: Problem 1. Sample space S = { all possible combinations of the two teams that can be formed } .  S  = 10 5 · 15 5 . We assume all possible team combinations are equally likely. 1. Define E = { teams chosen such that Lebron is on company A’s team AND Larry is on company B’s team } . We want to find P ( E ) and we know that P ( E ) =  E   S  . A team for company A with Lebron in it can be chosen ( 9 4 ) ways. A team for company B with Larry in it can be chosen ( 14 4 ) ways. So  E  = 9 4 · 14 4 . Therefore P ( E ) =  E   S  = ( 9 4 ) · ( 14 4 ) ( 10 5 ) · ( 15 5 ) . = 1 6 = . 167 . 2. Define F = { teams chosen such that Lebron does not play and Larry does play for company B } . We want P ( F ). A team for company A with Lebron not playing can be chosen ( 9 5 ) ways. A team for company B with Larry playing in it can be chosen ( 14 4 ) ways (as before). Then  F  = 9 5 · 14 4 ....
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This note was uploaded on 06/05/2010 for the course STAT PStat 120a taught by Professor Rohinikumar during the Spring '10 term at UCSB.
 Spring '10
 RohiniKumar

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