# sol2 - Problem 81/82 A =(3 1(3 2(3 3(3 4(3 5(3 6,P A = 1 6...

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Unformatted text preview: Problem 81/82 A = { (3 , 1) , (3 , 2) , (3 , 3) , (3 , 4) , (3 , 5) , (3 , 6) } ,P ( A ) = 1 6 . B = { (1 , 4) , (2 , 4) , (3 , 4) , (4 , 4) , (5 , 4) , (6 , 4) } ,P ( B ) = 1 6 . C = { (1 , 6) , (2 , 5) , (3 , 4) , (4 , 3) , (5 , 2) , (6 , 1) } ,P ( C ) = 1 6 . P ( AB ) = P ( { (3 , 4) } ) = 1 36 . P ( AC ) = P ( { (3 , 4) } ) = 1 36 . P ( BC ) = P ( { (3 , 4) } ) = 1 36 . P ( ABC ) = P ( { (3 , 4) } ) = 1 36 . P ( AB ) = 1 36 = 1 6 1 6 = P ( A ) P ( B ) . P ( AC ) = 1 36 = 1 6 1 6 = P ( A ) P ( C ) . P ( BC ) = 1 36 = 1 6 1 6 = P ( B ) P ( C ) . P ( ABC ) = 1 36 6 = 1 6 1 6 1 6 = P ( A ) P ( B ) P ( C ) . Hence, the three events are pairwise independent, however, they are not (mutually) inde- pendent. 1 Problem 83/104 Let R denote the need of rework. We have the following setup: P ( A 1 ) = 0 . 50 ,P ( A 2 ) = 0 . 30 ,P ( A 3 ) = 0 . 20 , P ( R | A 1 ) = 0 . 05 ,P ( R | A 2 ) = 0 . 08 ,P ( A 3 | J ) = 0 , 069 . By the law of total probabilities: P ( R ) = P ( R | A 1 ) P ( A 1 ) + P ( R | A 2 ) P ( A 2 ) + P ( R | A 3 ) P ( A 3 ) = 0 . 069 . By Bayes rule: P ( A 1 | R ) = P ( R | A 1 ) P ( A 1 ) P ( R ) = 0 . 362 . P ( A 2 | R ) = P ( R | A 2 ) P ( A 2 ) P ( R ) = 0 . 348 . P ( A 3 | R ) = P ( R | A 3 ) P ( A 3 ) P ( R ) = 0 . 290 . 2 Problem 99/24 (a) F ( x ) is a step function and thus X is a discrete rv. Furthermore, the “jumps” of F ( x ) occur at all the possible values of F ( x ), and the “size” of each jump is the probability of the possible value. Therefore the set of such values is { 1 , 3 , 4 , 5 , 12 } , and p X (1) = F (1)- 0 = 0 . 30 . p X (3) = F (3)- F (1) = 0 . 10 . p X (4) = F (4)- F (3) = 0 . 05 . p X (6) = F (6)- F (4) = 0 . 15 . p X (12) = F (12)- F (6) = 0 . 40 . (b) P (3 ≤ X ≤ 6) = P ( X ≤ 6)- P ( X < 3) = F (6)- F (1) = 0 . 30 . P (4 ≤ X ) = 1- P ( X < 4) = 1- F (3) = 0 . 60 . 3 Problem 107/32 (a) E ( X ) = (13 . 5)(0 . 2) + (15 . 9)(0 . 5) + (19 . 1)(0 . 3) = 16 . 38 . E ( X 2 ) = (13 . 5) 2 (0 . 2) + (15 . 9) 2 (0 . 5) + (19 . 1) 2 (0 . 3) = 272 . 298 . V ( X ) = E ( X 2 )- [ E ( X )] 2 = 272 . 298- 16 . 38 2 = 3 . 9936 . (b) E (25 X- 8 . 5) = 25 E ( X )- 8 . 5 = 401 . (c) V (25 X- 8 . 5) = 25 2 V ( X ) = 2496 . (d) E ( X- . 01 X 2 ) = E ( X )- . 01 E ( X 2 ) = 13 . 657 . 4 Problem 107/36 Let pr be the (constant) premium that the company charges. Considering the deductible, the profit of the company is then computed as pr- ( X- 500) + , where ( X- 500) + = ( X- 500 , if X > 500 , , if X ≤ 500 ....
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sol2 - Problem 81/82 A =(3 1(3 2(3 3(3 4(3 5(3 6,P A = 1 6...

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