exercise25answers - Exercise 25 Answers Philosophy 102...

Info iconThis preview shows page 1. Sign up to view the full content.

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

Unformatted text preview: Exercise 25 Answers Philosophy 102 (Maher) Spring 2009 1. p(D|T ) = = = 1/203 ≈ 0.005. Paul almost certainly doesn’t have the disease. 2. p(B|S) = p(S|B)p(B) p(S|B)p(B)+P (S|∼B)p(∼B) (1/4)(1/5) (1/4)(1/5)+(1/36)(4/5) T |D)p(D) p(T |D)p(D)+p(T |∼D)p(∼D) (0.99)(0.0001) (0.99)(0.0001)+(0.02)(0.9999) = = 9/13 ≈ 0.69. 3. p(B|W ) = = = 12/29 ≈ 0.41. In surveys, most people say 0.8. 4. (a) p(U3 |B) = p(W |B)p(B) p(W |B)p(B)+p(W |∼B)p(∼B) (0.8)(0.15) (0.8)(0.15)+(0.2)(0.85) = = 10/11 ≈ 0.91. Getting a black ball is strong evidence for U3 . (b) p(U1 |∼B) = p(B|U3 )p(U3 ) p(B|U1 )p(U1 )+p(B|U2 )p(U2 )+p(B|U3 )p(U3 ) (1/10)(1/3) (0)(1/3)+(1/100)(1/3)+(1/10)(1/3) = = 100/289 ≈ 0.35. Getting a white ball is only slight evidence for U1 . 5. (a) p(F |H1 ) = p(H1 |F )p(F ) p(H1 |F )p(F )+p(H1 |∼F )p(∼F ) (1/2)(2/3) (1/2)(2/3)+(1)(1/3) p(H1 .H2 |F )p(F ) p(H1 .H2 |F )p(F )+p(H1 .H2 |∼F )p(∼F ) (1/4)(2/3) (1/4)(2/3)+(1)(1/3) p(∼B|U1 )p(U1 ) p(∼B|U1 )p(U1 )+p(∼B|U2 )p(U2 )+p(∼B|U3 )p(U3 ) (1)(1/3) (1)(1/3)+(99/100)(1/3)+(9/10)(1/3) = = 1/2. (b) p(F |H1 .H2 ) = = = 1/3. The evidence for ∼F is now stronger. 6. p(D|T ) = p(T |D)p(D) p(T |D)p(D)+P (T |∼D)p(∼D) (0.99)(1/700) (0.99)(1/700)+(0.05)(699/700) = = 33/1198 ≈ 0.03. Belinda probably doesn’t have the disease. ...
View Full Document

This note was uploaded on 04/13/2009 for the course PHIL 102 taught by Professor Weinberg during the Spring '08 term at University of Illinois at Urbana–Champaign.

Ask a homework question - tutors are online