Applying the Bayes theorem a) A research institute associated with the Olympics claims that its drug test will detect steroid use (that is, show a positive result for an athletic who uses steroids) 95% of the time. Your friend on Canada’s hockey team has just tested positive. 
In order to determine the probability that your friend uses steroids given that he tested positive, we need two additional pieces of information. The P(athletic on the team uses steroids; the prevalence) and the P(test positive │ athletic does not use steroids; the false positive result)
After considerable effort, the research institute finally concedes that 15% of all steroid-free individuals also test positive (the false positive result) and that 10% of the hockey team members use steroids (the prevalence result).
So now with this information, determine
a)  The sensitivity; P(tested positive │ athletic uses steroids) 
b)  The specificity; P(tested negative │ athletic does not use steroids), 
c)  Using Bayes Theorem, determine the probability that your friend uses steroids given that he has just tested positive for steroids; i.e., P(athletic uses steroids │ test positive). 
So what I did was
Let $P$ = tested positive and $S$ = athletic uses steroids and from the question I know $P(s) = 0.1$ and $P( p|s') = 0.15$  so
For a) it is $P(p|s) = \frac{P(p,s)}{P(s)}$ I don't know how to get the numerator I was assuming $0.95 \cdot 0.1$? Would that be right?
b) $P(p' | s') = \frac{P(p',s')}{P(s')}$ and for the numerator could I use $0.05 \cdot 0.9$?
c) by using the Bayes theorem $P(s|p) = \frac{P(p|s)\cdot P(s)}{P(p)}$ and that would be using the answer from a but is $P(p) = 0.95$
please help out
 A: I will work out the answer in my style, but using more or less your notation. The idea you were proposing for (c) is fine. The answers given for (a) and (b) show some misunderstanding of these questions. 
(a) We want $\Pr(P|S)$. We have been told this, it is $0.95$.
(b) We want $\Pr(P'|S')$. We have been almost told this. The probability of a false positive is $0.15$, so $\Pr(P'|S')=0.85$.
(c) We want $\Pr(S|P)$. I think the easiest way to do this is to use the fact that
$$\Pr(S|P)=\frac{\Pr(S\cap P)}{\Pr(P)}.\tag{$1$}$$
Let us first calculate the denominator $\Pr(P)$ in $(1)$. A positive can happen in two ways: (i) We use steroids and test positive or (ii) We do not use steroids and test positive. 
In symbols, we want to find (i) $\Pr(S\cap P)$ and (ii) $\Pr(S'\cap P)$ and then add them.
For (i), the result is $(0.1)(0.95)$. I think of this as immediate, but if you want to use a formula, $\Pr(S\cap P)=\Pr(P|S)\Pr(S)$.   For (ii), the result is $(0.9)(0.15)$. Thus
$$\Pr(P)=(0.1)(0.95)+(0.9)(0.15).$$
Calculating the numerator $\Pr(S\cap P)$ of $(1)$ is easy, we have already done it when we computed (i). We have $\Pr(S\cap P)=(0.1)(0.95)$.
Finally, for the conditional probability, divide.  
