Box 1 contains 1black ball and 3 white balls; Box 2 contains 2 black balls and 2 white balls; Box 3 contains 2 white balls. Draw a ball form Box 1 and place it in Box 2. Then draw a ball from Box 2 and place it in Box 3. Finally draw a ball from Box 3. What is the probability that the last ball drawn from Box 3 is white?

One way to solve this is to consider the "expected value" of black and white balls in each step. When you select a ball from Box 1 the probability of it being white is 3/4 and the probability of it being black is 1/4. You can consider the ball as being 3/4 white and 1/4 black. Of course the ball must be one color or the other, but if you do this experiment many times over you would expect that 3/4 of the time you add one white ball and 0 black balls to Box 2, and 1/4 of the time you add 0 white balls and 1 black ball to Box 2. Therefore the expected value of the ball, on average, is 3/4 white and 1/4 black. So now the contents of Box 2 becomes (on average) 2 + 3/4 white and 2 + 1/4 black, for a total of 5 balls.

Now repeat this process for the next step - you pull a ball from Box 2 and the probability of it being white is 2.75/5 = 0.55, while the probability of it being black is 2.25/5 = 0.45. Add it to Box 3 and what do you get? With that you can determine the probability of pulling a white ball from Box 3. Post back with what you get for an answer.