# Probability of drawing 3 red marbles?

Here is one more marble question...... I think I have the answer, but am confused.

An urn contains 6 red marbles and 4 blue marbles. Draw 3 marbles at random. $x$ is the number of red marbles. Find $P(x=0), P(x=1), P(x=2), P(x=3)$.

Here is the solution I decided on... \begin{align} P(x=3)= \frac{\binom{6}{3}\binom{3}{0}}{\binom{10}{3}} && P(x=2)= \frac{\binom{6}{2}\binom{4}{1}}{\binom{10}{3}} \end{align} etc. for $x= 1, 0$.

However, I believe the sum of all the answers should equal $1$? If so, my sum does not equal $1$, so I'm confused about the solution. Thank you for any help!!!

• Thank you, this is very helpful. I realized I was just making a simple factorial mistake, otherwise my answers would have been correct. It was useful to be able to compare it to your answers. – jlajla Oct 7 '13 at 18:23

The probabilities that all the marbles have the same color are simple:

$$P(X=0) = \frac{4}{10}\frac{3}{9}\frac{2}{8} = \frac{1}{30}$$

and

$$P(X=3) = \frac{6}{10}\frac{5}{9}\frac{4}{8} = \frac{1}{6}$$

the other two are just a little bit trickier:

$$P(X=1) = \frac{4}{10}\frac{3}{9}\frac{6}{8} + \frac{4}{10}\frac{6}{9}\frac{3}{8} + \frac{6}{10}\frac{4}{9}\frac{3}{8} = 3 \frac{4 \times 3 \times 6}{10 \times 9 \times 8} = \frac{3}{10}$$

and, similarly,

$$P(X=2) = 3 \frac{4 \times 5 \times 6}{10 \times 9 \times 8} = \frac{1}{2}$$

which indeed add up to 1.

In general, if you have $r$ red and $b$ blue balls, then the probability or pulling $k$ red balls out of $n$ selections $X(k,n)$ is

$$P(X(k,n)) = \binom{n}{k} \frac{r\dots(r-k+1)b\dots(b-n+k+1)}{(r+b)\dots(r+b-n)}$$

The idea is that you select the balls one by one, and the binomial coefficient is the number of orders in which the $k$ red balls can appear in the selection of the $n$ balls, while the fraction is the product of the probabilities of each next ball having the required color.