# Thread: Joint Probability Question

1. ## Joint Probability Question

Hi, I am having trouble with the following question:
Five balls are drawn without replacement from an urn containing 7 blue balls, 4 green balls, and 8 yellow balls. Let B, G, and Y be the total number of blue, green and yellow balls drawn, respectively. Find
a) The joint probability mass function of B, G, and Y.
b) The joint probability mass function of Y and B
c) THe marginal probability mass function for B, G, and Y separately. Do these marginal probabilities look familiar? ID them.

For a) I tried modelling the joint probability mass function using a combinatorial approach.
c(7 choose B)c(4 choose G)c(8 choose Y)/c(19 choose 5).
And B+G+Y=5 (because you can only choose 5 balls). However, after this I'm stuck.

2. these are all hypergeometrics

joint is $\displaystyle P(B=b,G=g,Y=y)={ {7 \choose b} {4 \choose g} {8 \choose y}\over {19 \choose 5}}$.

b is really the same...

$\displaystyle P(B=b,G=g)={ {7 \choose b} {4 \choose g} {8 \choose 5-b-g}\over {19 \choose 5}}$.

I'll only do B's distribution. It's B vs the world so...

$\displaystyle P(B=b)={ {7 \choose b} {12 \choose 5-b} \over {19 \choose 5}}$.

you do the other two.

3. I understand how you got part c, but
I don't really understand how you got part b. Is it because you are focusing on B and G, so Y is just what's left over?

4. I was in a rush.
I wrote the joint distribution of B and G.
Here's the joint of Y and B...

$\displaystyle P(B=b,Y=y)={ {7 \choose b} {8 \choose y} {4 \choose 5-b-y}\over {19 \choose 5}}$.

This is because b+y+g=5.
So, if you know two of them, you know the third one.