Hello mneox Quote:

Originally Posted by

**mneox** Bag A contains 3 black marbles and 1 white marble and Bag B contains 1 black marble and 3 white marbles. If two marbles are randomly taken from Bag A and placed in Bag B, then a marble is randomly selected from Bag B, what is the probability that the marble selected from Bag B is black?

I've tried to draw a tree, but I must be doing something wrong cause I can't get the correct answer. Could someone give me some guidance?

The number of ways of selecting $\displaystyle 2$ marbles from the four in bag A is $\displaystyle \binom42=6$.

The number of ways of selecting the white marble and one of the black marbles from bag A $\displaystyle = 3$.

Therefore the probability that one of the marbles chosen from bag A is white is $\displaystyle \tfrac36 = \tfrac12$; and the probability that both are black is also $\displaystyle \tfrac12$.

So, when a marble is drawn from bag B, half the time it will be chosen from a total of $\displaystyle 2$ black and $\displaystyle 4$ white, and half the time from $\displaystyle 3$ black and $\displaystyle 3$ white. So the probability that it is black is $\displaystyle \frac12\times \frac26 + \frac12\times \frac36 = \frac{5}{12}$

Grandad