# Joint probability function of discrete variable (combinatorics)

There's a box with three types of objects: A, B, and C

There are 6 of A, 8 of B, and 10 of C. At random, we remove four objects from the box

I'm trying to find the joint probability $$P(x, y)$$ where $$x$$ is the number of class A that is selected and $$y$$ is the number of class B that is selected

My current solution is this. However I'm finding all probabilities to be quite low $$P(x, y) = \frac{(_{x}^{6})(_{y}^{8})}{(^{24}_4)}$$

• If you sum your distribution over $x = 0, \dots, 4$ and $y = 0, \dots, 4-x$, you'll find out it doesn't sum to one. That's because you haven't included the term $10 \choose 4-x-y$ in the numerator. – jbowman Nov 9 '18 at 1:58

## 1 Answer

As jbowman correctly points out in the comments, your probability mass function should be:

$$p(x,y) = \frac{{6 \choose x} {8 \choose y} {10 \choose 4-x-y}}{{24 \choose 4}}\quad \quad \quad \text{for } x \geqslant 0, y \geqslant 0 \text{ and } x+y \leqslant 4.$$

• Is the probability mass function the same as the joint probability in this case? – Alter Nov 9 '18 at 2:26
• @Alter: Yes, it is. – Ben Nov 9 '18 at 2:37