Suppose 5 cards are drawn from the standard deck of 52 cards. Denote random variables $X$ to be the number of red cards in the hand and $Y$ to be the number of spades in the hand.
How do I calculate $P(X=3,Y=2)$, the probability that a poker hand contains 3 red cards and 2 spades?
My attempt:
There are ${52 \choose 5}$ ways to draw a 5 card hand.
The number of ways to draw 3 red cards and 2 black cards from the deck is ${26\choose 3}{26\choose 2}$.
The number of ways to draw 2 spades is ${13 \choose 2}{39\choose3}$ where ${12 \choose 2}$ is the number of ways to choose ranks for 2 spades and ${39\choose3}$ is the number of ways to choose the remaining 3 cards, excluding all spades.
The probability that a poker hand contains 3 red cards and spades is therefore
$$P(X=3,Y=2)=\frac{{26\choose 3}{26\choose 2}{13 \choose 2}{39\choose3}}{{52 \choose 5}}$$
Is that right? I am not sure if $X$ and $Y$ are independent.
Best Answer
You are putting too many factors in your formula
There are $26$ red cards and $13$ spades in the deck, so using your notation,
$P(X=3, Y=2) = \dfrac{\binom{26}3\binom{13}2}{\binom{52}5}$
Your numerator can't contain $10$ cards against $5$ in the denominator