# Calculate the PMF of drawing balls from urn

Question:

There are nine black balls and two white balls in an urn. You draw balls from the urn one at a time. Note that this drawing of balls is without replacement. You keep all the black balls that are drawn between the two white balls. Denote the black balls that are drawn between the two white balls with the random variable $X$. Calculate the PMF for $X$.

Attempt at Solution:

I am thinking of this as arranging the balls in 11 slots. Thus there are $(\frac{11!}{9!2!}) = 55$ total ways to arrange the balls.

Now, for the case when there are 0 black balls between the 2 white balls, there would be 10 ways to arrange this configuration. I set the 2 white balls as one block and I move them around across the 11 slots. Thus, $P(X=0) = \frac{10}{55}$.

Similarly, $P(X=1) = \frac{9}{55}$ as we can move the block of 3 balls (2 white with a black ball in between) in 9 ways across the 11 slots.

Generalizing, the PMF would be $\frac{10-x}{55}$ for $0\leq x \leq 9$.

Is this correct?

## 1 Answer

Quite so.

There are indeed $\binom {11}{2}$ equally probable ways to select places for the two white balls.   That is $55$.

The event labeled $\{X=x\}$ consists of some arrangement in this outcome set, where the white balls have a block of $x$ black balls between them.   There are $(10-x)$ outcomes that form such an arrangement.

Then the PMF is: $$p_X(x) = \begin{cases} (10-x)/55 &:& 0\leq x\leq 9~,~ x\in\Bbb N\\ 0 &:& \text{elsewhere}\end{cases}$$

As a reality check, quickly confirm in your head that: $\sum_{x=0}^9 (10-x)=55\quad(\checkmark)$