# Finding joint prob mass function

Suppose two balls are chosen from a box containing 3 white, 2 red and 5 blue balls. Let X = the number of white balls chosen and Y = the number of blue balls chosen. Find the joint pmf of X and Y.

### Attempt

We want $$p_{XY}(x,y) = P(X = x \cap Y =y )$$

Let's count the number of ways to select $$X$$ white balls from $$10$$ given balls. $${3 \choose x}$$ gives all possible ways to pick $$x$$ white balls. Now, to pick the remaining we can do that in $${7 \choose 7-x}$$ ways. Similarly for the number of blue balls chosen, same reasoning gives $${5 \choose y } \times {5 \choose 5-x}$$. Now, by multiplication princliple one has

$$p_{XY}(x,y) = \frac{ {3 \choose x}{7 \choose 7-x}{5 \choose y } {5 \choose 5-x} }{{10 \choose 2 } }$$

Is this a correct joint pmf?

## 1 Answer

Numerator should be white's, blue's and red's:

$$p_{XY}(x,y) = \frac{ {3 \choose x}{5 \choose y } {10-5-3 \choose 2-x-y}}{{10 \choose 2 } }$$

Our answers are the same if

$${7 \choose 7-x} {5 \choose 5-y} = {10-5-3 \choose 2-x-y}$$

which is not the case (Note: I guess you meant ${5 \choose 5-y}$ and not ${5 \choose 5-x}$).

Yours is wrong for 2 reasons.

1. You seem to assume independence. We don't compute $P(X=x)$ and $P(Y=y)$ separately and then multiply them.

2. The individual computations are wrong. Observe the additions below:

$$P(X=x) = \frac{{3 \choose x}{7 \choose 2-x}}{{10 \choose 2}}$$

Here, we see that $3+7=10$ and $x+2-x=2$

So, ${3 \choose x}{7 \choose 7-x}$ should instead be ${3 \choose x}{7 \choose 2-x}$

Next:

$$P(Y=y) = \frac{{5 \choose y}{5 \choose 2-y}}{{10 \choose 2}}$$

Here, we see that $5+5=10$ and $y+2-y=2$

So, ${5 \choose y } \times {5 \choose 5-y}$ should instead be ${5 \choose y } \times {5 \choose 2-y}$

Verify that the following is wrong as well:

$${7 \choose 2-x} {5 \choose 2-y} = {10-5-3 \choose 2-x-y}$$

Okay, so how did I get ${3 \choose x}{5 \choose y } {10-5-3 \choose 2-x-y}$? I first chose among the 3 x's. Instead of choosing among the 7 non-x's, I then chose among the 5 y's (the order doesn't matter for choosing x's and y's). Now instead of choosing among the 7 non-x's or the 5 non-y's (note that 7 non-x's and 5 non-y's overlap), I choose among the remaining 2 balls which are not x and not y.