Given this problem:
Suppose prior probabilities in a decision situation are P(S1) = 0.2, P(S2) = 0.5, and P(S3) = 0.3. With sample information I, P(I/S1) = 0.1, P(I/ S2) = 0.05 and P(I/ S3) = 0.2. Calculate the revised or posterior probabilities: P(S1/I), P(S2/I) and P(S3/I).
My question is
If P(S1) = 0.2, what is the opposite P(S1`)?
and
If P(I/S1) = 0.1, what is the opposite P(I/S1`)?
I am hesitant because we've never done this in class with 3 probabilities. We have always done it with 2. Is P(S1`) = P(S2) + P(S3)? And is P(I/S1`) = P(I/S2) + P(I/S3)?
If $S_1,\ldots,S_n$ are disjoint events, the law of total probability says $$ P(I) = \sum_{k=1}^n P(I \mid S_k)P(S_k) $$ and the bayesian theorem says $$ P(S_k \mid I) = \frac{P(I \mid S_k)P(S_k)}{P(I)} \text{.} $$ Combining the two yields $$ P(S_k \mid I) = \frac{p_k}{\sum_{k=1}^n p_k} \text{ where } p_k = P(I \mid S_k)P(S_k) \text{.} $$