Homework3

# Homework3 - HOMEWORK PROBLEMS#3 SOLUTIONS TO EVEN PROBLEMS...

This preview shows pages 1–2. Sign up to view the full content.

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

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: HOMEWORK PROBLEMS #3 SOLUTIONS TO EVEN PROBLEMS 114. Let P i = the detector shows a positive reading (lie) for person i , T i = person i is telling the truth. Then we have: P ( P i | T i ) = . 10 ,P ( P i | T i ) = . 95. (a) By independence we have that P (positive reading for both suspects) = P ( P 1 ∩ P 2 | T 1 ∩ T 2 ) = P ( P 1 | T 1 ) P ( P 2 | T 2 ) = . 10 × . 95 = . 095 (b) P ( P 1 ∩ P 2 | T 1 ∩ T 2 ) = P ( P 1 | T 1 ) P ( P 2 | T 2 ) = (1- . 10) × . 95 = . 855 (c) P ( P 1 ∩ P 2 | T 1 ∩ T 2 ) = P ( P 1 | T 1 ) P ( P 2 | T 2 ) = . 10 × (1- . 95) = . 005 (d) P ( P 1 ∪ P 2 | T 1 ∩ T 2 ) = P ( P 1 | T 1 ) + P ( P 2 | T 2 )- P ( P 1 ∩ P 2 | T 1 ∩ T 2 ) = . 10 + . 95- . 095 = . 955 116. Let L i = line i fails. Then P (does not completely fail) = 1- P (fails) = 1- P ( L 1 ∩ L 2 ∩ L 3 ) = 1- P ( L 1 ) P ( L 2 ) P ( L 3 ) = 1- . 01 3 = . 999999 120. (a) Number of ways to test all 6 refrigerators: P (6 , 6) = 6!. Number of ways to test all 6 refrigerators with a defective one in the first three and a defective one in the fourth position:...
View Full Document

## This note was uploaded on 02/23/2011 for the course MATH 444 taught by Professor Any during the Fall '10 term at Roosevelt.

### Page1 / 2

Homework3 - HOMEWORK PROBLEMS#3 SOLUTIONS TO EVEN PROBLEMS...

This preview shows document pages 1 - 2. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online