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)* P(I | S1)+P(S2)* P(I | S2)+ P(S3)* P(I | S3)\\\\ =0.20* 0.12+0.57* 0.04+ 0.23*0.20 =0.0928`

Using Bayes theorem , we have

`P(S1 | I)=(P(I | S2)P(S1))/(P(I))\\\\=(0.12* 0.20)/(0.0928)\\\\\approx0.2586`

`P(S2 | I)=(P(I | S2)P(S2))/(P(I))\\\\=(0.04*0.57)/(0.0928)\approx0.2457`

`P(S1 | I)=(P(I | S3)P(S3))/(P(I))\\\\=(0.20* 0.23)/(0.0928)\\\\=0.4957`