Q: 2 blue balls and 1 white ball are in a bucket. You draw one ball at a time, record the color of the ball and put it back to the bucket until you get at least 2 blue balls and 2 white balls. Let X be the number of blue balls you drew, Y be the number of white balls you drew, Z be the number of balls you drew. What is the probability mass functions of X,Y, and Z? (i.e. give the probabilities of all the possible outcomes for X,Y,Z, respectively)

I understand the question, but don't understand the solution. Can someone please explain/help me out?

Thanks a lot!

I don't understand a given answer...

According to the solutions:

P(X=2)=(3C1) (2/3)2 (1/3)2 + (2C1) (2/3) (1/3)2 + (1/3)2
P(X=3)=(4C1) (2/3)2 (1/3)2
...

P(Z=4)=(4C3) (2/3)2 (1/3)2
P(Z=5)=P(X=5-2)+P(Y=5-2)
...


But I don't get it...can someone please explain? :)

This is sampling with replacement, for which the probabilities do not change. P(Draw Blue) = 2/3 & P(Draw White) = 1/3

The probabilities associated with drawing z balls can be determined by binomial expansion of (B + W)z: B = 2/3, W = 1/3, and z = the total number of balls drawn.

The following are expansions for various values of z (I have omitted multiplication operators).1: B + W 2: B2 + 2BW + W2 3: B3 + 3B2W + 3BW2 + W3 4: B4 + 4B3W + 6B2W2 + 4BW3 + W4 5: B5 + 5B4W + 10B3W2 + 10B2W3 + 5BW4 + W5 6: B6 + 6B5W + 15B4W2 + 20B3W3 + 15B2W4 + 6BW5 + W6

The coefficients in the above expansions are combinations of z items taken 1, 2, 3, 4, 5, or 6 at a time.

For example, in the expansion for z = 5, the third term (10B3W2 = 10*8/243) is the probability of drawing 5 times, getting 3 Blue balls & 2 White balls. In this term, 10 = 5C3 In the context of the problem, this term represents z = 5, x = 3, & y = 2

There seems to be some ambiguiity, missing information, or other discrepancy in the description of this problem. You are supposed to draw with replacement until you have at least 2 Blue balls and (at least or exactly) 2 White balls.Is it at least or exactly? Is there any limit to how many drawings you make? There is a small probability (2n/3n) that you will draw n Blue balls without drawing any White balls or n White balls (1/3n) without drawing any Blue balls, where n could be 10, 20, 500, one million, whatever. The first & last terms in each expansion provide probabilities for these possibilities. If the problem specified at least 2 White balls and at most 6 drawings, evaluating & adding up the bold terms (above) would give the correct probability.