The probability mass function of drawing two red kings out of a standard deck of 52 cards when 4 cards are dealt? Can someone tell me how to get the probability of drawing two red kings out of a standard deck of $52$ cards when drawing $4$ cards? My TA said it was $\dfrac{1}{{}^{50}\mathrm C_2}$, but I don't think that is correct.
 A: The probability of selecting two from two red kings (and two from fifty other cards) when selecting four from all fifty-two cards, is:
$$\newcommand{\ch}[2]{\hspace{.25ex}{^{#1}\mathrm C_{#2}}\hspace{.25ex}}
\dfrac{\ch 22 \ch {50}2}{\ch {52}4} = \dfrac{\ch {50}2}{\ch {52}4}
$$
A: Do you mean "the probability that, if you draw four cards from a regular deck of 52 cards, you draw exactly two Kings"? That is a single number, not a "probability mass function". There are, initially, 52 cards in the deck, 4 of them kings, 48 of them not kings.  The probability that the first card drawn is a king is 4/52= 1/13.  If that happens, there are 51 cards left in the deck, 3 of them kings.  The probability the second card drawn is a king is 3/51= 1/17.  Then there are 50 cards left, 4 of them non-kings.  The probability the third card drawn is not a king is 4/50= 2/25.  Finally, there are 49 cards in the deck, 47 of them non-kings.  The probability that the fourth card drawn is a non-king is 47/49.  The probability that we draw "king, king, non-king, non-king", in [b]that[/b] order, is (1/13)(1/17)(2/25)(47/49).  Writing "K" for a king and "N" for anything other than a king, we could have 4!/(2!2!)= 6 different hands: KKNN, KNKN, KNNK, NKKN, NKNK, and NNKK.  Using the same argument as above, we can show that the probability of any one of those is also (1/13)(1/17)(2/25)(47/49) so the probability of any of those is 6(1/13)(1/17)(2/25)(47/49).
