Question

Problem 4-23 (Algorithmic) Suppose that you are given a decision situation with three possible states of nature: S1, S2, and S3. The prior probabilities are P(S1) = 0.20, P(S2) = 0.57, and P(S3) = 0.23. With sample information I, P(I | S1) = 0.12, P(I | S2) = 0.04, and P(I | S3) = 0.20. Compute the revised or posterior probabilities: P(S1 | I), P(S2 | I), and P(S3 | I). If required, round your answers to four decimal places.

Answer 1

Answer with explanation:

Given : The three possible states of nature: S1, S2, and S3. The prior probabilities are P(S1) = 0.20  P(S2) = 0.57, and P(S3) = 0.23.

With sample information I, P(I | S1) = 0.12, P(I | S2) = 0.04, and P(I | S3) = 0.20.

Using the law of total probability , we have

$P(I)=P(S1)\times P(I | S1)+P(S2)\times P(I | S2)+ P(S3)\times P(I | S3)\\\\ =0.20\times 0.12+0.57\times 0.04+ 0.23\times0.20 =0.0928$

Using Bayes theorem , we have

$P(S1 | I)=\dfrac{P(I | S2)P(S1)}{P(I)}\\\\=\dfrac{0.12\times 0.20}{0.0928}\\\\\approx0.2586$

$P(S2 | I)=\dfrac{P(I | S2)P(S2)}{P(I)}\\\\=\dfrac{0.04\times0.57}{0.0928}\approx0.2457$

$P(S1 | I)=\dfrac{P(I | S3)P(S3)}{P(I)}\\\\=\dfrac{0.20\times 0.23}{0.0928}\\\\=0.4957$