# Probability (balls in a bucket)

1. Dec 16, 2007

### kingwinner

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!

2. Dec 16, 2007

### EnumaElish

Are you saying you don't understand a given answer? Or are you saying you don't "see" the solution?

What is the prob. z = 1? (It's zero. Why?)
What's the prob z < 4? (Again 0.)
Prob{z=4} = number of ways 2B+2W balls can be obtained/number of all color combinations in 4 draws

3. Dec 16, 2007

### cepheid

Staff Emeritus
Just work on it systematically. First of all, you know what a pmf is:

pX(x) = P(X = x)

-- pY(y) and pZ(z) are defined similarly.

Now, if X is the number of blue balls drawn, we know that X must be *at least* 2, because we're not going to *stop* drawing balls out until we get two blue ones. Therefore:

pX(x) = 0 for x < 2
likewise for pY(y)

similarly, pZ(z) must be zero for z < 4, because that's the minimum number of draws required to get 2 blue and 2 white. So, that's a start...

4. Dec 16, 2007

### kingwinner

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? :)