There are two local factories that produce microwaves. (Conditional probability and total probability) There are two local factories that produce microwaves. Each microwave produced at a factory $A$ is defective with probability $0.05$ where as each microwave at factory $B$ is defective with $0.01$.Suppose you purchase two microwaves that were both produced at the same factory which is equally likely to have been either factory $A$ or $B$. If the first microwave you check is defective, what is the probability that second one will also be defective? I tried the following: Let $X$ be the random variable denoting the number of defective radios. 
\begin{align}
P(X=x)&=P(X=x|A)P(A)+P(X=x|B)P(B)\\
&=\frac{1}{2}\binom 2k 0.01^{k}(1-0.01)^{2-k}+\frac{1}{2}\binom2k 0.05^k(1-0.05)^{2-k},
\end{align}
and then I compute $\displaystyle \frac{P(X=2)}{1-P(X=0)}$. Is this reasonable?
 A: You are being asked by a conditional probability. Let me first solve the problem using just notation of events. Let's define $D_i$ as the event that the $i$-th checked microwave is defective, and $A$ and $B$ the events associated with the type of factory the microwaves were bought. The desired probability is 
\begin{align}
P(D_2 \mid D_1) &\stackrel{a}{=} \frac{P(D_1 \cap D_2)}{P(D_1)}\\
&\stackrel{b}{=} \frac{P(A \cap D_1 \cap D_2)+P(B \cap D_1 \cap D_2)}{P(A \cap D_1)+P(B \cap D_1)}\\
&\stackrel{c}{=} \frac{P(A) P(D_1 \mid A) P(D_2 \mid A \cap D_1)+P(B) P(D_1 \mid B) P(D_2 \mid B \cap D_1)}{P(A \cap D_1)+P(B \cap D_1)}\\
&\stackrel{d}{=} \frac{P(A) P(D_1 \mid A) P(D_2 \mid A)+P(B) P(D_1 \mid B) P(D_2 \mid B)}{P(A) P(D_1 \mid A)+P(B) P(D_1 \mid B)}\\
&\stackrel{e}{=} \frac{P(D_1 \mid A) P(D_2 \mid A)+P(D_1 \mid B) P(D_2 \mid B)}{P(D_1 \mid A)+ P(D_1 \mid B)}\\
&\stackrel{}{=}  \frac{0.05^2+0.01^2}{0.05+ 0.01}\\
&= \frac{13}{300}
\end{align}
(a) Definition of conditional probability. (b)Total probability theorem. (c) Multiplication rule. (d) Given $A$ (or $B$), $D_1$ and $D_2$ are conditionally independent. (e) $P(A) = P(B)$.
I am not sure how the expression
$$\frac{P(X=2)}{1-P(X=0)} = \frac{13}{199}$$
could represent a conditional probability. The use of a random variable like $X$, instead of just events, could lead us to wrong conclusions. For example, $X=1$ when one of the following to possibilities occur: $D_1 \cap D_2^c$ or $D_1^c \cap D_2$. But $D_1^c \cap D_2$ is not possible because we already know that $D_1$ has occured!
