Tutorial Exercise Suppose that you are given a decision situation with three possible states of nature: S,, 52, and s;. The prior probabilities are P(s, ) = 0.1, P(52) = 0.5, and P(5;) = 0.4. With sample information I, P(IIs, ) = 0.1, P(I|s2) = 0.05, and P(I|s?) = 0.2. Compute the revised or posterior probabilities: P(s, II), P(52 |!), and P(s; |1). Step 1 Posterior probabilities are conditional probabilities based on the outcome of the sample information. These can be computed by developing a table using the following process. 1. Enter the states of nature in the first column, the prior probabilities for the states of nature, P(IIs,), in the second column and the conditional probabilities in the third column. 2. In column 4 compute the joint probabilities by multiplying the prior probability values in column 2 by the corresponding conditional probabilities in column 3. 3. Sum the joint probabilities in column 4 to obtain the probability of the sample information I, P(I). 4. In column 5, divide each joint probability in column 4 by P(!) to obtain the posterior probabilities, P(s, |1). The prior probabilities are given to be P(s, ) = 0.1, P(s2) = 0.5, and P(s?) = 0.4. The conditional probabilities given each state of nature are P(IIs, ) = 0.1, P(IIs,) = 0.05, and P(IIs;) = 0.2. Use the given prior and conditional probabilities to compute the joint probabilities. States of Nature Prior Probabilities P(5;) Conditional Probabilities P(IIs;) Joint Probabilities P(In s;) P(Ins, ) = P(5, )P(IIS, ) 51 P(s, ) = 0.1 P(I|s ) = 0.1 = 0.1(0.1) = 52 P(S,) = 0.5 P(I|s,) = 0.05 P(53) = 0.4 P(I|S?) = 0.2 The sum of the Joint Probabilities column gives P(I) =