Drawing a second ball from a randomly selected box Suppose I have 3 boxes. Box $i$ has $b_i$ black balls and $w_i$ white balls (all the boxes contain the same number of balls).
A box is selected randomly out of the three, and a ball is selected randomly.
Given that the randomly selected ball is white, and given that the ball is not returned to the box, what is the probability to draw a second white ball from the same box?
I want to solve this using only conditional probability.
My attempt:
The probability to draw a first white ball ($W_1$) is given by the law of total probability:
$$
P(W_1) = P(A_1)P(W_1|A_1) + P(A_2)P(W_1|A_2) + P(A_3)P(W_1|A_3) = \frac{1}{3}\left(\frac{w_1}{w_1+b_1} + \frac{w_2}{w_2+b_2} + \frac{w_3}{w_3+b_3}\right)
$$
For the second white ball, $W_2$, its probability is given by
$$
P(W_2|W_1) = \frac{P(W_2\cap W_1)}{P(W_1)}
$$
but I am having troubles calculating $P(W_1\cap W_2)$
 A: Try
$$ P(W_2\cap W_1)=\frac{1}{9}\left(\frac{w_1}{w_1+b_1}\frac{w_1-1}{w_1+b_1-1} + \frac{w_2}{w_2+b_2}\frac{w_2-1}{w_2+b_2-1} + \frac{w_3-1}{w_3+b_3-1}\frac{w_3}{w_3+b_3}\right)$$
A: It's fairly obvious what the final result will be. Calculating it
two ways gives you an opportunity to check how well you apply Bayes's Theorem, at least in principle, though it's not clear that this
particular problem is a particularly good test of that ability.
In order to compute the probability of two white balls, you could use
\begin{align}
P(W_1 \cap W_2) & = P(W_1 \cap W_2 \mid A_1) P(A_1)\\ 
&\qquad {} + P(W_1 \cap W_2 \mid A_2) P(A_2) \\
&\qquad {} + P(W_1 \cap W_2 \mid A_3) P(A_3).
\end{align}
The probability $P(W_1 \cap W_2 \mid A_1)$ is the probability that
both balls will be white if you pull two balls from a box 
containing $b_1$ black balls and $w_1$ white balls.
One way to look at this is that there are
There are $\binom{b_1 + w_1}{2}$ equally likely pairs of balls to draw,
of which $\binom{w_1}{2}$ pairs are two white balls, so
$$
P(W_1 \cap W_2 \mid A_1) = \frac{\binom{w_1}{2}}{\binom{b_1 + w_1}{2}}
= \frac{w_1 (w_1 - 1)}{(b_1 + w_1)(b_1 + w_1 - 1)},
$$
which is the same result you get by using the "obvious" values
of $P(W_2 \mid A_1 \cap W_1)$ and $P(W_1 \mid A_1)$ in
$$
P(W_2 \cap W_1 \mid A_1) = P(W_2 \mid W_1 \cap A_1) P(W_1 \mid A_1).
$$
(The "obvious" values are 
$P(W_2 \mid A_1 \cap W_1) = \frac{w_1 - 1}{b_1 + w_1 - 1}$
and $P(W_1 \mid A_1) = \frac{w_1}{b_1 + w_1}$,
as already indicated in other answers.)
