Solution4 - SOLUTIONS FOR HOMEWORK 4 1 Let E be the event that at least one die lands on 6 and let F denote the event that the two dice land on

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1. Let E be the event that at least one die lands on 6, and let F denote the event that the two dice land on different numbers. Then E = { (6 , j ) : j = 1 , 2 , 3 , 4 , 5 , 6 } ∪ { ( i, 6) : i = 1 , 2 , 3 , 4 , 5 , 6 } , F = { ( i, j ) : i 6 = j, i, j = 1 , 2 , 3 , 4 , 5 , 6 } , and EF = { (6 , i ) , ( i, 6) : i = 1 , 2 , 3 , 4 , 5 } . It is easy to see that F has 30 outcomes and EF has 10 outcomes. Therefore P ( E | F ) = 10 30 = 1 3 . 6. Let W denote the event that exactly 3 white balls are drawn, and let E denote the evemnt that the first and third balls drawn are white. (a) We first consider the non-replacement case. The number of outcomes in W is ± 4 3 ² × 8 × 7 × 6 × 4. The number of outcomes in EW is 8 × 7 × × ± 2 1 ² × 6 × 4. Therefore the desired conditional probability is P ( E | W ) = 8 × 7 × × ± 2 1 ² × 6 × 4 ± 4 3 ² × 8 × 7 × 6 × 4 = ± 2 1 ² ± 4 3 ² = 1 2 . (b) We next consider the replacement case. The number of outcomes in W is ± 4 3 ² × 8 × 8 × 8 × 4 and the number of outcomes in EW is 8 × 8 × ± 2 1 ² × 8 × 4. Thus the desired probability is P ( E | W ) = 8 × 8 × ± 2 1 ² × 8 × 4 ± 4 3 ² × 8 × 8 × 8 × 4 = 1 2 . 11. (a) The number of outcomes in A s is 51 and the number of outcomes in BA s is 3. Thus P ( B | A s ) = 3 51 = 1 17 . (b) The number of outcomes in A is ± 4 2 ² + ( 4 ) × ± 48 a ² = 6 + 192 = 198, and the number of outcomes in BA is ± 4 2 ² = 6. Thus P ( B | A ) = 6 198 = 1 33 . 14. Let
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This homework help was uploaded on 04/13/2008 for the course M 362k taught by Professor Berg during the Spring '08 term at University of Texas at Austin.

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Solution4 - SOLUTIONS FOR HOMEWORK 4 1 Let E be the event that at least one die lands on 6 and let F denote the event that the two dice land on

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