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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.

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    "When four cards" what... are drawn? It seems the sentence has been cut short. Also, don't you just want the probability? Why the mass function? – 2017-02-16
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    It is a question we have for homework. I don't know why they want the PMF. – 2017-02-16
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    I meant your are dealt four cards out of 52. so your hand is four cards – 2017-02-17
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    The number $X$ of red kings in a randomly dealt hand of four cards has PMF $P(X=k) = {2 \choose k}{50 \choose 4-k}/{52 \choose 4},$ where $k = 0, 1, 2.$ Is it possible the TA asked you to find the PMF and then, in particular, $P(X=2)=0.0045?$ (This is a _hypergeometric_ distribution.) – 2017-02-17

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The probability of selecting two from two red kings (and two from fifty other cards) when selecting four from all fifty-two cards, is:

$$\newcommand{\ch}[2]{\hspace{.25ex}{^{#1}\mathrm C_{#2}}\hspace{.25ex}} \dfrac{\ch 22 \ch {50}2}{\ch {52}4} = \dfrac{\ch {50}2}{\ch {52}4} $$

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    Write {}_nC_k to get ${}_nC_k$. Please excuse me if your notation was intended, I have just never seen it before. – 2017-02-16
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    @Lovsovs there are [many](https://en.wikipedia.org/wiki/Binomial_coefficient#History_and_notation) different notations for binomial coefficients, the three most common being $\binom{n}{r},~_nC_r,~^nC_r$ but still several more can be seen in use. When attempting to write an answer, people will often try to match the notation used by the original question as best as possible. In many cases though, the original asker does not correctly use $\LaTeX$ to format the mathematics making it difficult to distinguish between the two mentioned notations involving the use of a C. – 2017-02-16
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    Do you mean ${50 \choose 2}/{52 \choose 4}?$ Or did I miss a turn? – 2017-02-17
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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).

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    Thank you for the answer. I am only interested in the probability of drawing 2 red kings from the deck. – 2017-02-16