# Probability with two boxes and balls

There are $10$ white and $10$ black balls in the first box, and $8$ white and $10$ black balls in the second box. Two balls are put from the first to the second box, then we choose one ball from the second box. What is the probability that the chosen ball from the second box is white?

From the first box, we can choose two balls in ${20\choose 2}=190$ ways. Possible combinations are $\{BB,BW,WB,WW\}$.

After putting two balls from the first to the second box, there are total $18$ balls in the first, and $20$ in the second.

From the second box, we can choose one ball in ${20\choose 1}=20$ ways.

How can we find the probability that the chosen ball from the second box is white?

• You could use conditional probability. If the two balls moved to the second box were both black... – Fabio Somenzi Jan 11 '17 at 19:34

Case $1$: $2$ white balls are put from first box to second and one ball chosen from the second box is white

probability $=\dfrac{\dbinom{10}{2}}{\dbinom{20}{2}} \times \dfrac{10}{20}$

Case $2$: $2$ black balls are put from first box to second and one ball chosen from the second box is white

probability $=\dfrac{\dbinom{10}{2}}{\dbinom{20}{2}} \times \dfrac{8}{20}$

Case $3$: $1$ white ball and $1$ black balls are put from first box to second and one ball chosen from the second box is white

probability $=\dfrac{\left[\dfrac{\dbinom{10}{1}\dbinom{10}{1}}{2!}\right]}{\dbinom{20}{2}} \times \dfrac{9}{20}$

Sum of these three cases will give required probability

I don't see an answer more elegant than just checking all four cases. It's actually just three cases, because if a black and a white ball are moved, the order in which they are chosen doesn't matter.

If two black balls are moved (which has probability $\frac{10}{20} \cdot \frac{9}{19} = \frac{9}{38}$), there is an $\frac{8}{20} = \frac{2}{5}$ chance of drawing a white ball on the final step. So this case has a $\frac{9}{38} \cdot \frac{2}{5} = \frac{9}{95}$ chance of occuring.

If two white balls are moved (which has the same probability $\frac{10}{20} \cdot \frac{9}{19} = \frac{9}{38}$), then the probability of drawing a white ball from the second box is $\frac{1}{2}$. This case has a probability $\frac{9}{76}$.

The probability of moving a black and a white ball from the first box is $1 - \frac{9}{38} - \frac{9}{38}$. You can caclulate the probability of this happening and then drawing a white ball. Once you've done that, add the probabilities of the three cases to get your final answer.

After the transfer, each ball in the second box has a probability of $$\frac{8}{18}\times \frac{18}{20}+\frac{10}{20}\times \frac{2}{20}=\frac{9}{20}$$ of being white. They are not independent, but this does not matter here. So the solution is $\dfrac9{20}$

A simple way is to see that the expected break-up of balls transferred is $1W, 1B$

Thus P(white from second box after transfer) $= \dfrac{8+1}{18+2} = \dfrac9{20}$