Question

Suppose events H, M, and L are collectively exhaustive events. Apply Bayes’ Theorem to calculate P(H|A) with the following information: P(A|H) =0.4; P(A|M) = 0.3; P(A|L) = 0.6; P(H) = 0.5; P(M) = 0.1.


Round to Three decimals.

Answer 1

Given that
P(A|H) =0.4; P(A|M) = 0.3; P(A|L) = 0.6;
P(H) = 0.5; P(M) = 0.1.

H, M, and L are collectively exhaustive.

This means that the sum of the three probabilities is equal to 1.
P(L) + P(H) + P(M) = 1

P(L) = 1 - P(H) - P(M) = 1 - 0.5 - 0.1 = 0.4

Now we have enough information to solve for P(H|A).

P(A∩H) = P(A│H)P(H) = 0.4 x 0.5 = 0.20

P(A) = P(A│H)P(H) + P(A│M)P(M) + P(A│L)P(L)
= 0.2+ 0.3 x 0.1 + 0.6 x 0.4
= 0.2+0.03+0.24
= 0.47

P(H|A) = P(A∩H) / P(A) = 0.20/ 0.47 = 0.425

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