Binomial/Geometric Distribution explanation

I've found the following exercise in my Stats coursework. I only have solutions to it, but no explanation. And I would really like to know how to get to the answer.

An urn holds 5 white and 3 black marbles

If two marbles are drawn at random without replacement and X denotes the number of white marbles find the probability distribution of X.

Could someone please explain this to me?

Thanks!

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Name the marbles $W_1, W_2, \ldots, W_5$ and $B_1, B_2, B_3$. You are drawing a subset of size $2$ from this collection of $8$ marbles. There are $\binom{8}{2}= 28$ such sets. List them all, and then count how many subsets have $0$ white marbles, how many have $1$ white marble, and how many have $2$. Then, $$P\{X=1\} = \frac{\text{number of subsets with one white marble}}{28}$$ etc. –  Dilip Sarwate Jul 18 '12 at 21:42

The possible outcomes are $X=0,X=1$, and $X=2$. In order to get no white marbles ($X=0$), you must draw two black marbles. There are $\binom82$ pairs of marbles altogether, and there are $\binom32$ pairs of black marbles, so the probability of drawing a black pair is

$$\frac{\binom32}{\binom82}=\frac3{28}\;.$$

That is, $\Bbb P(X=0)=\frac3{28}$.

There are $\binom52$ pairs of white marbles, so the probability of drawing one of those pairs is

$$\frac{\binom52}{\binom82}=\frac{10}{28}=\frac5{14}\;.$$

That is, $\Bbb P(X=2)=\frac5{14}$.

The total probability of all three possible outcomes must be $1$, so we must have

$$\Bbb P(X=1)=1-\frac3{28}-\frac5{14}=1-\frac{13}{28}=\frac{15}{28}\;,$$

and we have our probability distribution:

\begin{align*} \Bbb P(X=0)&=\frac3{28}\\ \Bbb P(X=1)&=\frac{15}{28}\\ \Bbb P(X=2)&=\frac5{14}\;. \end{align*}

We can double-check this by calculating $\Bbb P(X=1)$ directly. Each of the $5$ white balls can be paired with any of the $3$ black balls to make a pair with one white ball, so there are $15$ such pairs, and as we’ve seen already, there are $\binom82=28$ pairs altogether, so the probability of drawing a pair with one ball of each color is indeed $\frac{15}{28}$.

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Thanks a lot for the thorough explanation! –  Sorin Cioban Jul 18 '12 at 21:56

Consider an outcome of the first drawing. Let $U_1$ be one if the first marble is white, and zero otherwise, and $U_2$ be a similarly defined variable for the second draw.

You know $\mathbb{P}(U_1 = 1) = \frac{5}{8}$ and $\mathbb{P}(U_1 = 0 ) = \frac{3}{8}$. Given that the first marble is white, the composition of the urn is 4 whites and 3 blacks, thus: $$\mathbb{P}(U_2 = 1 | U_1 =1) = \frac{4}{7} \quad \mathbb{P}(U_2 = 0 | U_1 =1) = \frac{3}{7}$$ Similarly: $$\mathbb{P}(U_2 = 1 | U_1 =0) = \frac{5}{5+2} = \frac{5}{7} \quad \mathbb{P}(U_2 = 0 | U_1 =0) = \frac{2}{7}$$ Now you are interested in $X= U_1 + U_2$, thus: $$\mathbb{P}(X=0) = \mathbb{P}(U_2=0|U_1=0) \cdot \mathbb{P}(U_1=0) = \frac{2}{7} \cdot \frac{3}{8} = \frac{3}{28}$$ and $$\mathbb{P}(X=2) = \mathbb{P}(U_2=1|U_1=1) \cdot \mathbb{P}(U_1=1) = \frac{4}{7} \cdot \frac{5}{8} = \frac{5}{14}$$ Then $$\mathbb{P}(X=1) = 1 - \mathbb{P}(X=0) - \mathbb{P}(X=2) = \frac{15}{28}$$ Indicentally, variable $X$ follow hypergeometric distribution $\operatorname{Hyp}(2,5,8)$.

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I appreciate the help! –  Sorin Cioban Jul 18 '12 at 21:56