Question

Probabilities with possible states of nature: s1, s2, and s3. Suppose that you are given a decision situation with three possible state of nature: s1, s2, and s3.The prior probabilities are P(s1) =.1, P(s2) = .6, and P(s3) = .3.With sample information I, P(I|s1) =.15, P(I|s2) = .2, and P(I|s3) = .1.Compute P(s1|I)Compute P(s2|I)Compute P(s3|I)

Answer 1

1.
`P(s_1|I)=(1)/(11)`

2.
`P(s_2|I)=(8)/(11)`

3.
`P(s_3|I)=(2)/(11)`

Explanation:

Given information:

`P(s_1)=0.1, P(s_2)=0.6, P(s_3)=0.3`

`P(I|s_1)=0.15,P(I|s_2)=0.2,P(I|s_3)=0.1`

(1)

We need to find the value of P(s₁|I).

`P(s_1|I)=(P(I|s_1)P(s_1))/(P(I|s_1)P(s_1)+P(I|s_2)P(s_2)+P(I|s_3)P(s_3))`

`P(s_1|I)=((0.15)(0.1))/((0.15)(0.1)+(0.2)(0.6)+(0.1)(0.3))`

`P(s_1|I)=(0.015)/(0.015+0.12+0.03)`

`P(s_1|I)=(0.015)/(0.165)`

`P(s_1|I)=(1)/(11)`

Therefore the value of P(s₁|I) is
`(1)/(11)`.

(2)

We need to find the value of P(s₂|I).

`P(s_2|I)=(P(I|s_2)P(s_2))/(P(I|s_1)P(s_1)+P(I|s_2)P(s_2)+P(I|s_3)P(s_3))`

`P(s_2|I)=((0.2)(0.6))/((0.15)(0.1)+(0.2)(0.6)+(0.1)(0.3))`

`P(s_2|I)=(0.12)/(0.015+0.12+0.03)`

`P(s_2|I)=(0.12)/(0.165)`

`P(s_2|I)=(8)/(11)`

Therefore the value of P(s₂|I) is
`(8)/(11)`.

(3)

We need to find the value of P(s₃|I).

`P(s_3|I)=(P(I|s_3)P(s_3))/(P(I|s_1)P(s_1)+P(I|s_2)P(s_2)+P(I|s_3)P(s_3))`

`P(s_3|I)=((0.1)(0.3))/((0.15)(0.1)+(0.2)(0.6)+(0.1)(0.3))`

`P(s_3|I)=(0.03)/(0.015+0.12+0.03)`

`P(s_3|I)=(0.03)/(0.165)`

`P(s_3|I)=(2)/(11)`

Therefore the value of P(s₃|I) is
`(2)/(11)`.