# For m white and m black balls, to show probability both ball has same colour is $\frac{m-1}{2m-1}$ when m random balls in bag A and remaining in bag B

Question is that: From a total of $$m$$ white balls and $$m$$ black balls ($$m>1$$), $$m$$ balls are selected at random and put into a bag A, the remaining $$m$$ balls are put into bag B. A ball is then drawn randomly from each bag. Show that the probability that the two balls have the same colour is $$\frac{m-1}{2m-1}$$.

I have been working on this problem for one hour, but I have not clue.

• Please clarify your specific problem or provide additional details to highlight exactly what you need. As it's currently written, it's hard to tell exactly what you're asking.
– Community Bot
Oct 2 at 4:28

There are then $$2m - 1$$ remaining balls, each equally likely to be chosen as the second ball. Of those, $$m-1$$ are the same color as the first ball.
Hence: $$P=\frac{m-1}{2m-1}$$
• Maybe, then, an accept ($\checkmark$) is in order? Oct 2 at 12:55