Consider a decision making problem with three states of nature: S1, S2, and S3, for which P(S1) = .1 and P(S2) = .3.Suppose also that there are two possible sample indicators, I1 and I2, and the following conditional probabilities hold: P(I1|S1) = .2, P(I1|S2) = .4, and P(I1|S3) = .6.The P(S2|I2) is:
A).18
B).24
C).36
D).50

Question

Quizplus · Accepted Answer

We don't have the conditional probability of I2 given S2, so we have to use Bayes' Theorem to calculate P(S2|I2):

P(I2|S2) = 0.6 (since 1 - 0.4 = 0.6)
P(I2|S1) = ? (we don't have this information)
P(I2|S3) = 0.3 (since 1 - 0.7 = 0.3)

We can calculate P(I2) by the law of total probability:

P(I2) = P(I2|S1)P(S1) + P(I2|S2)P(S2) + P(I2|S3)P(S3)
P(I2) = (1 - 0.2) × 0.1 + (1 - 0.4) × 0.3 + (1 - 0.7) × 0.6
P(I2) = 0.48

Now we can use Bayes' Theorem:

P(S2|I2) = P(I2|S2)P(S2) / P(I2)
P(S2|I2) = 0.4 × 0.3 / 0.48
P(S2|I2) = 0.3

So the answer is (C) 0.36.

Consider a Decision Making Problem with Three States of Nature