hw3solns - STA 4321/5325 Spring 2010 Solutions to Homework...

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Unformatted text preview: STA 4321/5325 - Spring 2010 Solutions to Homework 3 2.77 (a) There are 6 possible outcomes when rolling the first dice, for each outcome, rolling the second dice can result in 6 possible outcomes, thus there are 6 × 6 = 36 outcomes in the sample space S . (b) There are 6 outcomes that gives the sum equals 7. Those outcomes are (1,6), (6,1), (2,5), (5,2), (3,4), and (4,3). The probability of interest is number of outcomes if sum is 7 total number of equally likely outcomes = 6 36 = 1 6 2.80 We know that P ( T ) = 4 12 , P ( D ) = 6 12 The event that neither a three-point spets nor a defensive standouts is selected can be denoted by ¯ T ∩ ¯ D , the probability of this event is P ( ¯ T ∩ ¯ D ) = 5 12 (a) The event that the selected player is a three-point spet and a defensive standout can be denoted by T ∩ D , the probability of interest is P ( T ∩ D ) = P ( T ) + P ( D )- P ( T ∪ D ) = 4 12 + 6 12- parenleftBig 1- P ( T ∪ D ) parenrightBig = 4 12 + 6 12- ( 1-...
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This note was uploaded on 06/05/2011 for the course STA 4321 taught by Professor Staff during the Spring '08 term at University of Florida.

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hw3solns - STA 4321/5325 Spring 2010 Solutions to Homework...

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