Question

Consider a decision situation with four possible states of nature: s₁, s₂, s₃, and s₄. The prior probabilities are P(s₁) = 0.35, P(s₂) = 0.15, P(s₃) = 0.20, P(s₄) = 0.30. The conditional probabilities are P(C|s₁) = 0.2, P(C|s₂) = 0.09, P(C|s₃) = 0.15, and P(C|s₄) = 0.20. Find the revised (posterior) probabilities P(s₁|C), P(s₂|C), P(s₃|C), and P(s₄|C).

Answer 1

We are given the prior probabilities here as:
P(s1) = 0.35,
P(s2) = 0.15,
P(s3) = 0.2,
P(s4) = 0.3

Also the conditional probabilities here are given as:
P(c | s1) = 0.2,
P(c | s2) = 0.09,
P(c | s3) = 0.15,
P(c | s4) = 0.2

Using law of total probability, we have here:
P(c) = P(c | s1)P(s1) + P(c | s2)P(s2) + P(c | s3)P(s3) + P(c | s4)P(s4)

P(c) = 0.2*0.35 + 0.09*0.15 + 0.15*0.2 + 0.2*0.3

P(c) = 0.07 + 0.0135 + 0.03 + 0.06 = 0.1735

Using Bayes theorem now, we get the posterior probabilities here as:
P(s1 | c) = P(c | s1)P(s1) / P(c) = 0.07 / 0.1735 = 0.4035
P(s2 | c) = 0.0135 / 0.1735 = 0.0778
P(s3 | c) = 0.03 / 0.1735 = 0.1729
P(s4 | c) = 0.06 / 0.1735 = 0.3458

The required probabilities are put in bold above.