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exercise25answers - Exercise 25 Answers Philosophy...

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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. ...
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