The question asks:
The instructor of a discrete mathematics class gave two tests. Twenty-five percent of the students received an A on the first test and 15% of the students received A’s on both tests. What percent of the students who received A’s on the first test also received A’s on the second test?
I'm still learning how to properly apply Bayes Theorem to questions like this. Bayes Theorem says:
$$ P(A \mid B) = \frac{P(B \mid A) \, P(A)}{P(B)} $$
For this problem, the probability of getting an A on the first exam (lets call that F) is .25:
$P\left(F\right)=.25$
(This would mean that complement of F, or the chance of not getting an A on the first exam, is $P\left(F^c\right)=1-.25\:=\:.75$, right? Not sure if this is pertinent to the question though).
While the probability of getting an A on both the first and second exam (let's call getting an A on the second exam S). I think this would be represented as:
$P\left(F\cap S\right)=.15$
In the end, we are attempting to compute:
$$ P(F \mid S) = \frac{P(S \mid F) \, P(F)}{P(S)} $$
...I think.
I'm struggling to figure out how to find $P\left(S\right)$ and $ P(S \mid F) $ from the above details however. Any advice for solving this?