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

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?

• PS: Can I somewhere find the documentation or cheat sheet for stackexchange equation editor? – Wakan Tanka Jun 3 '15 at 11:46
• Don't know about stats.SE but there is a very useful MathJax basic tutorial and quick reference over on meta.math.SE. – Dilip Sarwate Jun 3 '15 at 12:58