Conditional Probability drug testing 
I understand part a), i'm just a bit confused by part b).
Let $E_1=$ The event the first test is positive,
$E_2=$ The event the second test is positive and
$F=$ The event drugs are present in the sample
To work out $Pr(F|E_1 \cap E_2)$ first I need $Pr(E_1 \cap E_2|F)$
By conditional probability $Pr(E_1 \cap E_2|F)= Pr(E_1|F)Pr(E_2|F)$
However, the solution then states $ Pr(E_1|F)Pr(E_2|F)={Pr(E_1|F)}^2$ i.e $ Pr(E_1|F)=Pr(E_2|F)$ Why is this? Surely the fact that the first test was positive, affects the probability that the second is also?
 A: 
However, the solution then states $Pr(E_1\mid F)Pr(E_2\mid F)=Pr(E_1\mid F)^2$ i.e $Pr(E_1\mid F)=Pr(E_2\mid F)$  Why is this? Surely the fact that the first test was positive, affects the probability that the second is also?

You need to be careful what is meant by event $E_2$. It helps to assume the second test will be done even if the first test was negative. Then $E_1$ and $E_2$ are conditionally independent given $F$ and are identically distributed (so that $P(E_1\mid F)= P(E_2\mid F)$). If you don't have this assumption then $E_1$ must occur for $E_2$ to occur; that is, $E_2 \subseteq E_1$, and then $E_1$ and $E_2$ are no longer conditionally independent.
This allows us to proceed as follows:
\begin{eqnarray*}
P(F\mid E_1\cap E_2) &=& \dfrac{P(E_1\cap E_2\mid F)P(F)}{P(E_1\cap E_2\mid F)P(F) + P(E_1\cap E_2\mid F^c)P(F^c)} \\
&& \\
&=& \dfrac{P(E_1\mid F)P(E_2\mid F)P(F)}{P(E_1\mid F)P(E_2\mid F)P(F) + P(E_1\mid F^c)P(E_2\mid F^c)P(F^c)} \\
&& \\
&=& \dfrac{P(E_1\mid F)^2P(F)}{P(E_1\mid F)^2P(F) + (1-P(E_1^c\mid F^c))^2(1-P(F))} \\
&& \\
&=& \dfrac{0.99^2\times 0.0002}{0.99^2\times 0.0002 + (1-0.98)^2(1-0.0002)} \\
&& \\
&\approx& 0.3289 \\
\end{eqnarray*}
