Conditional Probability Concept Question The organizers of a cycling competition know that about 8% of the racers use steroids. They decided to employ a test that will help them identify steroid-users. The following is known about the test: When a person uses steroids, the person will test positive 96% of the time; on the other hand, when a person does not use steroids, the person will test positive only 9% of the time. The test seems reasonable enough to the organizers. The one last thing they want to find out is this: Suppose a cyclist does test positive, what is the probability that the cyclist is really a steroid-user.
S be the event that a randomly selected cyclist is a steroid-user and P be the event that a randomly selected cyclist tests positive.
***My questions is Can someone please translate and explain P(P|S) and P(S|P) ?
 A: We are given that $P(S)=0.08$ (hence $P(\neg S)=0.92$), $P(P|S)=0.96$ and $P(P|\neg S)=0.09$.
What we want to knwo is $P(S|P)$.
Note that $P(S\cap P)=P(S|P)\cdot P(P)$ as well as $P(S\cap P)=P(P|S)\cdot P(S)$, therefore
$$ P(S|P) = \frac{P(P|S)\cdot P(S)}{P(P)}.$$
Thus we first need $P(P)$, which we get from $P(P)=P(P|S)P(S)+P(P|\neg S)P(\neg S)$. At last now all is reduced to given values.
A: $\Pr(P|S)$ is the probability that the person tests positive, given that she uses steroids. We are told explicitly that this is $0.96$. 
$\Pr(S|P)$ is the probability she is a steroid user, given that she tests positive. That is what we are asked to find.  Informally, if we confine attention to the people who test positive, $\Pr(S|P)$ measures the proportion of them that really are steroid users. 
Since the problem says that the proportion of steroid users is not high (sure!), many of the positives will be false positives.  Thus I would expect that $\Pr(S|P)$ will not be very high: the test is not as good as it looks on first sight.
For computing, there are two ways I would suggest, the first very informal and probably not acceptable to your grader, and the second more formal.
$(1)$: Imagine $1000$ cyclists. About how many of them will test positive? About how many of these will be steroid users?  Divide the second number by the first, since $\Pr(S|P)$ asks us to confine attention to the subpopulation of people who tested positive. 
$(2)$: Use the defining formula 
$$\Pr(S|P)=\frac{S\cap P}{\Pr(P)}.$$
The two numbers on the right are not hard to compute. I can give further help if they pose difficulty. 
A: Here is a relatively plain English explanation for Bayesian statistics.
Here, you have an evidence which says a user tested positive and the hypothesis you want to test against is whether steroids were used.
P(H | E) = P(E|H) x P(H) / [ P(E|H) x P(H) + P(E|~H) x P(~H) ]
from your numbers, P(H) = 8% so P(~H) = 92%, P(E|H) = 96% and P(E|~H) = 9%. Plug in all these numbers and you get ~ 48% which is quite low.
