Black and white balls in 2 boxes and probability that I pick the white ball. Maybe there already is solution for that and if it is so, then maybe someone can tell me where I can find it.
I have 2 boxes. In first box there are 3 white and 2 black balls. In second box there are 4 black and 4 white balls. Then I randomly pick one ball from first box and put it in second box. What is the probability that I pick white ball from the second box?
I think that I need to calculate combinations for black balls and then for white balls (for second box after I put in random ball from first box) and then multiply them.
 A: Case 1
You picked a white ball from the first box.
This happens with a chance of $3/5$. Putting it into the second box, now there are $5$ white and $4$ black balls. Now the chance to pick a white ball is $5/9$. So the total chance in case 1 is $3/5 \cdot 5/9 = 1/3$.
Case 2
You picked a black ball from the first box.
This happens with a chance of $2/5$. Putting it into the second box, now there are $4$ white and $5$ black balls. Now the chance to pick a white ball is $4/9$. So The total chance in case 2 is $2/5 \cdot 4/9 = 8/45$.
Combining the two cases, the chance is $$1/3 + 8/45 = 23/45 \approx 51\%.$$
A: We know there are $9$ balls in the second box when you draw from it. Of these, $4+\frac25$ are expected to be black and $4+\frac35$ are expected to be white. Since the denominator is known and only the numerator is random, we can apply linearity of expectation and find the probability of picking a white ball as the expected number of white balls picked:
$$
\frac{4+\frac35}9=\frac{23}{45}\;.
$$
A: HINT: You have to look at the two possible cases, what is the probability of taking a white ball of the first box (3/5) and taking a white ball of the second box (there are now 5 balls in the second box, so 5/9), than look at the second case, where you take a black ball of the first box. 
Combine these two cases with addition. 
