# 2 variable binomial distribution?

1. Dec 15, 2011

### Master J

I'm having a bit of trouble understanding a probability distribution of 2 variables.

Take for example taking n cards from a deck, and seeing what is the probability of getting X queens and say Y aces (with replacement). This involves the binomial distribution. The probabilities for the individual events are straight forward, but I'm having trouble getting the binomial coefficients.

Normally, for one variable, the coefficient is just n C X ( n choose X) where n is the total number of trials. This is n! / X! (n-X)! ... but how do you get this now for an X and a Y???

Last edited: Dec 15, 2011
2. Dec 15, 2011

### Stephen Tashi

I think you want a "multinomial distribution" and "multinomial coefficients" rather than a binomial distribution.

The binomial distribution can be understood from an analysis of the algebra used in computing the coefficients of terms in the expression $(p + q)^n$.

The multinomial distribution can be understood from analysing the algebra used in computing the coefficients of terms in expressions like $(p_a + p_q + p_s)^n$.

3. Dec 16, 2011

### Master J

I don't know how searching for "multinomial" escaped me!

I know that the coefficient for my stated problem is as follows:

$\frac{n!}{X! Y! (n - X - Y)!}$

Does that seem correct? I haven't seen that particular form anywhere I've looked.

4. Dec 16, 2011

### Stephen Tashi

That's correct for the coefficient. To get the probability, the coefficient is multiplied by a term of the form $p^X q^Y (1 - p - q)^{(n-X-Y)}$