The prior probabilities for events A1 and A2 are P(A1) = .40 and P(A2) = .60. It

Question

The prior probabilities for events A1 and A2 are P(A1) = .40 and P(A2) = .60. It is also known that P(A1 A2) = 0. Suppose P(B | A1) = .20 and P(B | A2) = .06.

The prior probabilities for events A1 and A2 are P(A1) = .40 and P(A2) = .60. It is also known that P(A1 and A2) = 0. Suppose P(B | A1) = .20 and P(B | A2) = .06.

Are events A1 and A2 mutually exclusive?

Compute P(A1 and B) (to 4 decimals).

Compute P(A2 and B) (to 4 decimals).

Compute P(B) (to 4 decimals).

Apply Bayes' theorem to compute P(A1 | B) (to 4 decimals).

Also apply Bayes' theorem to compute P(A2 | B) (to 4 decimals).

Explanation / Answer

P(A1) = .40; P(A2) = .60; P(B | A1) = .20; P(B | A2) = .06 . Also,P(A1 and A2) = 0 (given)

a) Since we know P(A1 and A2) = 0, the events are mutually exclusive

b) P(A1 and B)=P(B/A1)*P(A1)= 0.2*0.4 =0.08

c) P(A2 and B)=P(B/A2)*P(A2)=0.06*0.6=0.036

d) P(B)=P(B/A1)*P(A1)+P(B/A2)*P(A2)=0.2*0.4+0.06*0.60=0.116

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