Originally Posted by

**badgerigar** We have an additional piece of information that we can use: a white ball was selected from bag N.

Before we take the ball out, there is Sengnee's 2/7 chance of there being 5 red balls and 3 white balls (case one), and a 1-2/7=5/7 chance of there being 4 red balls and 4 white balls (case 2).

So taking a white ball out of case one has probability $\displaystyle \frac {3}{5+3} * \frac {2}{7} = \frac {3}{28}$ and the probability of taking a white ball out of case 2 is $\displaystyle \frac {5}{7} * \frac {1}{2} = \frac {5}{14}$

So knowing we took a white ball out, the probability that a red ball was transferred is the probability of taking a white ball out of case one divided by the total probability of taking out a white ball ie. $\displaystyle \frac {\frac {3}{28}}{\frac{3}{28}}+\frac {5}{14}} = \frac {3}{13}$