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?