## Problem Set 4

Assigned: Oct. 20
Due: Oct. 27.

Let A, B, and C be events. You are given that

Prob(A) = 0.6.
Prob(B | A) = 0.8.
Prob(B | ~A) = 0.3. ("~A" means "not A")
Prob(C | A) = 0.9.
Prob(C | ~A) = 0.2
You are also given that B and C are conditionally independent given A; and that B and C are conditionally independent given ~A.

Compute the following values:

• a. Prob(~A) = 1 - Prob(A) = 0.4.
• b. Prob(~B | A) = 1 - Prob(B|A) = 0.2.
• c. Prob(A ^ B) = Prob(A) Prob(B|A) = 0.6*0.8 = 0.48
• d. Prob(A ^ ~B) = Prob(A) Prob(~B|A) = 0.6*0.2 = 0.12
• e. Prob(B ^ C | A) = Prob(B|A) Prob(C|A) = 0.8*0.9 = 0.72
• f. Prob(B ^ C | ~A) = Prob(B|~A) * Prob(C|~A) = 0.3*0.2 = 0.06
• g. Prob(A | B) = Prob(A) * Prob(B|A) / Prob(B)
But Prob(B) = Prob(B|A)Prob(A) + Prob(B|~A) Prob(~A) = 0.8*0.6 + 0.3*0.4 = 0.6
So Prob(A|B) = 0.6 * 0.8 / 0.6 = 0.8.
• h. Prob(A | ~B) = Prob(A) * Prob(~B|A) / Prob(~B)
But Prob(~B) = Prob(~B|A)Prob(A) + Prob(~B|~A) Prob(~A) = 0.2*0.6 + 0.7*0.4 = 0.4
So Prob(A|~B) = 0.2*0.6 / 0.4 = 0.3
• g. Prob(A | C) = Prob(A) * Prob(C|A) / Prob(C)
But Prob(C) = Prob(C|A)Prob(A) + Prob(C|~A) Prob(~A) = 0.9*0.6 + 0.2*0.4 = 0.62
So Prob(A|C) = 0.6*0.9 / 0.62 = 0.87.
• j. Prob(A | B ^ C) = Prob(B^C|A) Prob(A) / Prob(B^C).
Now Prob(B^C) = Prob(B^C^A) + Prob(B^C^~A) = Prob(B^C|A) Prob(A) + Prob(B^C|~A) Prob(~A) = 0.72*0.6 + 0.06*0.4 = 0.432 + 0.024 = 0.456
So Prob(A|B^C) = 0.72*0.6/0.456 = 0.947.