sol2 - Problem 81/82 A = { (3 , 1) , (3 , 2) , (3 , 3) , (3...

Info iconThis preview shows pages 1–6. Sign up to view the full content.

View Full Document Right Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

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 ....
View Full Document

Page1 / 17

sol2 - Problem 81/82 A = { (3 , 1) , (3 , 2) , (3 , 3) , (3...

This preview shows document pages 1 - 6. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online