What is the probability of selecting $2$ red queens when $5$ cards are selected from the $12$ face cards in a standard deck?

Question

I have $12$ face cards ($4$ kings, $4$ queens, $4$ jacks). What is the probability that I’ll pick $2$ red queens if $5$ cards are chosen?

My answer is $\frac{\binom{10}{3}}{\binom{12}{5}} = \frac{5}{33}$, since the number of ways to choose the other 3 cards once the 2 red queens are already chosen is $\binom{10}{3}$, and the number of ways to choose the 5 cards is $\binom{12}{5}$.

I seem to be missing out something here, particularly on the part when choosing the $2$ red queens.

Can anyone help me point out my mistake? Thank you!

Answer 1

Your answer is okay and applies hypergeometric distribution:$$\frac{\binom22\binom{10}3}{\binom{12}5}$$

Maybe the expression $\binom22$ is what you are missing in your own approach.

Fortunately that does not harm here because it equals $1$ hence is neutral wrt multiplication.

Answer 2

Hint: Try to answer the following questions.

How many events you have?

How many of them are ''good''?

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Yes, your answer is correct.