# Urn I contains 6 whites and 4 blacks balls. Urn II contains 2 white and 2 black balls.

Urn I contains 6 whites and 4 blacks balls. Urn II contains 2 white and 2 black balls. From urn I 2 balls are transferred to Urn II. A sample of size 2 is then drawn without replacement from urn II. What is the probability that the sample will contain exactly 1 white ball?

I don't even know where to start.

• The first step is to divide into cases, with respective probabilities, all the different configurations that urn II can be in after the first transfer. – Arthur Nov 3 '15 at 18:44

Here's a hint.

After the transfer, you'll have either $4W$ and $2B$, $3W$ and $3B$, or $2W$ and $4B$.

So, first calculate the probability of each of these states for Urn II. (For example, to get $4W, 2B$ you'll need to draw $2W$ from Urn I. What is the probability of that happening? The probability of drawing $2W$ from Urn I is $\frac{6}{10} \frac{5}{9} = \frac{1}{3}.$)

Then, calculate the probability of drawing $1W, 1B$ from Urn II, given each of the configurations (and their associated probabilities). So, given $2W$ were drawn from Urn I, the probability of getting $1W,1B$ is

$$P((1W,1B) | 2W) = \frac{1}{3}\left[\frac{4}{6}\frac{2}{5} + \frac{2}{6}\frac{4}{5}\right] = \frac{8}{45}.$$

Then figure out the others.

• I was thinking like that first but I am not sure how to set that up. Am I supposed to use Bayes' theory? – abuchay Nov 3 '15 at 18:51
• Edited my answer. – John Nov 3 '15 at 19:07

If $B_1,...,B_n$ partitons $\Omega$

P(A) = $\sum_{i=1}^n P(A|B_i)P(B_i)$

Let A be the event that exactly one white ball is chosen.

$B_1$ is Urn II contains 4W and 2B

$B_2$ is ... 3W and 3B

$B_3$ is ... 2W and 4B

It can be seen:

P($B_1$)=1/3

P($B_2$)=8/15

P($B_3$)=4/15

Then P(A|$B_1$) = $\frac 46 \frac25$ + $\frac 26 \frac45$

You can move forward from there.

Let $W_1$ be the count of white balls transferred from Urn I($6:4$), and $W_2$ be the count of white balls subsequently drawn from Urn II($2+W_1:4-W_1$). (That is white:black in the urn)

Then by the Law of Total Probability:

\begin{align}\mathsf P(W_2{=}1) & = \sum_{w{=}0}^2 \mathsf P(W_1{=}w)\cdot\mathsf P(W_2{=}1\mid W_1{=}w) \\[1ex] & = \sum_{w=0}^2 \dfrac{\dbinom{6}{w}\dbinom{4}{2-w}}{\dbinom{10}{2}}\cdot\frac{\dbinom{2+w}{1}\dbinom{4-w}{1}}{\dbinom{6}{2}} \\[1ex] & = \dfrac{1}{675}\left(\sum_{w=0}^2(2+w)(4-w)\dbinom{6}{w}\dbinom{4}{2-w} \right)\\[1ex] & = \ldots\end{align}