how to solve this conditional probability Manufacture A and B produce one type of electrical element, given that the probability of produced element being faulty is $0.05$ for A and $0.01$ for B.
If two of these elements has been picked, from A or B (both from one manufacturer with equal probability) and first one is faulty, what is the probability of second element being faulty?
I know I should use
$P(Second\;is\;Faulty\;|\;First\;is\;Faulty)$ = $\frac{P(Second\;is\;Faulty\;\cap \;First\;is\;Faulty)}{P(First\;is\;Faulty)}$ 
but don't know how to find these numbers from the given info.
 A: Let $X$ be the event of a faulty element (and $X_n$ the $n^{th}$ element being faulty). And let $A$ and $B$ be the event of picking both elements from the manufacture A and B, respectively. We look for $P(X_2\mid X_1)$.
Since nothing's said, let's assume that all events are independent, since otherwise we miss information. Note that since $A$ and $B$ are disjoint, $P(X)=P(A)P(X\mid A)+P(B)P(X\mid B)$.
$$(PX_2\mid X_1)=\frac{P(X_2\cap X_1)}{P(X_1)}=\frac{P(A)P(X_2\cap X_1\mid A)+ P(B)P(X_2\cap X_1\mid B)}{P(A)P(X_1\mid A)+P(B)P(X_1\mid B)}=\frac{\frac{1}{2}(0.05^2+0.01^2)}{\frac{1}{2}(0.05+0.01)}\approx 0.0433
$$
A: Define events:
$$F_1 = \text{"first product is faulty"} \\
F_2 = \text{"second product is faulty"} \\
A = \text{"products chosen from $A$"} \\
B = \text{"products chosen from $B$"}.$$
Then, we require
\begin{eqnarray*}
P(F_2\mid F_1) &=& \dfrac{P(F_2\cap F_1)}{P(F_1)} \\
&& \\
&=& \dfrac{P(F_2\cap F_1 \mid A)P(A) + P(F_2\cap F_1 \mid B)P(B)}{P(F_1 \mid A)P(A) + P(F_1 \mid B)P(B)} \qquad\text{by Law of Total Probability} \\
&& \\
&=& \dfrac{0.05^2\times 0.5 + 0.01^2\times 0.5}{0.05\times 0.5 + 0.01\times 0.5} \\
&& \qquad\qquad\qquad\text{assuming conditional independence of $F_1,F_2$ given $A$ or given $B$} \\
&=& 13/300 \\
&& \\
&\approx& 0.043.
\end{eqnarray*}
