Conditional Probability: Black and White Balls Bag A has $3$ white and $2$ black marbles. Bag B has $4$ white and $3$ black marbles.
Suppose we draw a marble at random from Bag A and put it in Bag B. After doing this, we draw a marble at random from Bag B, which turns out to be white. Given this information, what is the probability that the marble we moved from Bag A to Bag B is white?
At first since the probability that white is $1/2$ is $2/5$, and the probability that white is $5/8$ is $3/5$, I thought a sufficient answer would be $$\begin{align}\tfrac 1 2 \cdot \tfrac 2 5 + \tfrac 5 8 \cdot \tfrac 3 5 & = \tfrac 1 5 + \tfrac 3 8 \\ & = \tfrac 8 {40} + \tfrac {15}{40} \\ & = \tfrac {23}{40}\end{align}$$, but that was wrong. 
Can someone provide me with an answer? Thanks

Let $W$ be the event the second ball removed was white, and let $T$ be the event that the ball transferred was white. We want the conditional probability $\Pr(T|W)$. By the definition of conditional probability, we have $$\Pr(T|W)=\frac{\Pr(T\cap W)} {\Pr(W)}.$$ We need to compute the two probabilities on the right.
We first calculate $\Pr(W)$. The event $W$ can happen in two ways (i) We transferred a white ball and then pulled a white or (ii) we transferred a black ball and then pulled a white. The probability of (i) is $\frac{3}{5}\cdot \frac{5}{8}$ and the probability of (ii) is $\frac{2}{5}\cdot \frac{4}{8}$. For $\Pr(W)$, add. By the way, this is the number you computed.
Now we calculate $\Pr(T\cap W)$. This is easy, we already computed it, it is the probability of (i). 
Finally, divide. 
