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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 $$ P(x, y) = \frac{(_{x}^{6})(_{y}^{8})}{(^{24}_4)} $$

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    $\begingroup$ 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. $\endgroup$ – jbowman Nov 9 '18 at 1:58
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As jbowman correctly points out in the comments, your probability mass function should be:

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

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

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