Problem about marginal pmfs: I can't understand why it is.

Question

(It's the problem of the book 'Introduction to mathematical statistics'. 2.3.9) Five cards are drawn at random and without replacement from an ordinary deck of cards. Let $X_1$ and $X_2$ denote, respectively, the number of spades and the number of hearts that appear in the five cards. (b) Find the two marginal pmfs. I found the joint pmf of $X_1$ and $X_2$, which is $$ P(X_1 = x_1, X_2 = x_2) = \frac{\binom{13}{x_1} \cdot \binom{13}{x_2} \cdot \binom{26}{5 - x_1 - x_2}}{\binom{52}{5}} $$ So I tried to get marginal pmf of $X_1$ by solving summation $P(X_1 = x_1, X_2 = x_2)$ with respect to $x_2$. But I found it's insane.. and then I looked for solution, and its $$ P(X_1 = x_1) = \frac{\binom{13}{x_1} \cdot \binom{39}{x_1}}{\binom{52}{5}}$$ Could you please explain about this problem?

Answer 1

I think you mean the marginal pmf is $$ P(X_1 = x_1) = \frac{\binom{13}{x_1} \cdot \binom{39}{5 - x_1}}{\binom{52}{5}}$$ Rather than summing out the $x_2$ (which will work, but is harder) it's easier to derive this from first principles. Each combination of 5 cards is equally likely to be drawn, and there are $\binom{52}{5}$ combinations. We want to find the number of combinations which contains $x_1$ spades. There is $\binom{13}{x_1}$ ways to choose the spades, then $\binom{39}{5 -x _1}$ ways to choose the remaining cards. From this we can derive the formula.