Explain conditional probability? Consider 3 baskets. Basket A contains 3 white and 5 red marbles. Basket B contains 8 white and 3 red marbles. Basket C contains 4 white and 4 red marbles. An experiment consists of selecting one marble from each basket at random. What is the probability that the marble selected from basket A was white, given that exactly 2 white marbles were selected in this process.
I know P(AB) = 33/176 and I understand how to get there.
However I'm not sure why P(B) = 73/176.
Here is the answer, but can you explain why?
 A: By Bayes rule, the required probability is equal to
$$P(A_w|W=2)=\frac{P(W=2|A_w)P(A_w)}{P(W=2)}=\frac{\frac{1}{2}\frac{3}{8}}{\frac{73}{176}}=\frac{33}{73}$$ since $$P(W=2|A_w)=P(B_w)P(C_r)+P(B_r)(C_w)=\frac{1}{2}$$ and $$\begin{align*}P(W=2)&=P(A_w)P(B_w)P(C_r)+P(A_w)P(B_r)P(C_w)+P(A_r)P(B_w)P(C_w)\\\\&=\frac{3}{8}\frac{8}{11}\frac{1}{2}+\frac{3}{8}\frac{3}{11}\frac{1}{2}+\frac{5}{8}\frac{8}{11}\frac{1}{2}=\frac{73}{176}\end{align*}$$ 
A: Let $A,B,C$ be the event of selecting a white marble from the relevant container.
$$\mathsf P(A) = 3/8, \mathsf P(B) = 8/11, \mathsf P(C) = 1/2$$
We want:  $$\require{cancel} \begin{align}
\mathsf P(A\mid ABC^c\cup AB^cC\cup A^cBC)
 & = \frac{\mathsf P(ABC^c\cup AB^cC)}{\mathsf P(ABC^c\cup AB^cC\cup A^cBC)}
\\[1ex] & =\frac{\mathsf P(A)\Big(\mathsf P(B)\mathsf P(C^c)+\mathsf P(B^c)\mathsf P(C)\Big)}{\mathsf P(A)\Big(\mathsf P(B)\mathsf P(C^c)+\mathsf P(B^c)\mathsf P(C)\Big)+\mathsf P(A^c)\mathsf P(B)\mathsf P(C)}
\\[1ex] & = \frac{\frac 3 8\big(\frac 8 {11}\frac{1}{2}+\frac{3}{11}\frac 1 2\big)}{\frac 3 8\big(\frac 8 {11}\frac 12+\frac 3 {11}\frac 1 2\big)+ \frac 5 8\frac 8 {11}\frac 1 2}
\\[1ex] & = \frac{33\cancel{/176}}{73\cancel{/176}}
\end{align}$$
