Can $P(A \cap B)$ be computed as $P(A)*P(B|A)$ and vice versa in any example?

Question

Following Bayes example is taken from here

A math teacher gave her class two tests. 25% of the class passed both tests and 42% of the class passed the first test. What percent of those who passed the first test also passed the second test?

The Bayes formula for this example is:

$P(S|F)$ = $\frac{P(S)*P(F|S)}{P(F)}$

Using the following formula:

$P(S \cap F)$ = $P(S)*P(F|S)$

the above formula can be rewritten to this:

$P(S|F)$ = $\frac{P(S \cap F)}{P(F)}$

after assigning proper numbers this can be computed and this is the result:

$P(S|F)$ = $\frac{0.25}{0.42} = 0.60$

My question is if is possible to compute also $P(S)$ and $P(F|S)$ (as a compounds of $P(S \cap F)$) when $P(S|F)$ is known already?

Answer 1

If you know only P(S|F) and P(F), you can't definetely compute P(S) and P(F|S). Lets use your example: we have P(S|F)=0.6 and P(F)=0.42. Here we have Solution1: P(S) = 0.5, P(F|S) = 0.5 fits. But Solution2: P(S) = 0.25, P(F|S) = 1 fits as well. So the only thing you can tell is that P(S∩F) = 0.25 but you can't definetely compute P(S) and P(F|S) from this inputs.