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Imagine a deck that consists of two types of cards. There are 2 type A cards and 9 Type B cards, totaling 11 cards. Assuming that 1 card will be dealt to 5 players. What are the probabilities that Type A will be dealt zero, one, and two times?

I hope I've asked that clearly. Any reply would be appreciated.

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Please do not use signature in you post. – Sasha Jan 25 '12 at 17:54
Sorry about that. I presume that is a common newbie mistake. – Bub Jan 25 '12 at 18:08
up vote 2 down vote accepted

Note that three events $A_0$, $A_1$, and $A_2$ of players receiving 0, 1 or 2 cards of type A are exclusive and partition the entire event space (there are only 2 cards of type A). This means that $$ \mathbb{P}(A_0) + \mathbb{P}(A_1) + \mathbb{P}(A_2) = 1 $$

There total $\binom{11}{5}$ ways to deal 1 card to each of 5 players. There are $\binom{11-2}{5}$ ways to deal 5 cards with no type A card in it (just throw out 2 type A card out of the deck), that makes $$ \mathbb{P}(A_0) = \frac{\binom{11-2}{5}}{\binom{11}{5}} = \frac{\frac{9!}{4! 5!}}{\frac{11!}{6! 5!}} = \frac{9!}{11!} \frac{6!}{4!} = \frac{6 \cdot 5}{11 \cdot 10} = \frac{3}{11} $$

To compute $\mathbb{P}(A_2)$, deal type A cards first. What remains is to deal 9 type B cards to 3 remaining players, thus $$ \mathbb{P}(A_2) = \frac{\binom{9}{3}}{\binom{11}{5}} = \frac{\frac{9!}{6! 3!}}{\frac{11!}{6! 5!}} = \frac{9!}{11!} \cdot \frac{5!}{3!} = \frac{5 \cdot 4}{11 \cdot 10} = \frac{2}{11} $$

Thus $$ \mathbb{P}(A_1) = 1- \mathbb{P}(A_0) -\mathbb{P}(A_2) = \frac{6}{11} $$

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To find the probablity that Type $A$ is dealt exactly once, split this event up into cases according to which player receives the Type $A$ card. There are 5 possibilities, each occuring with probability ${2\over11}\cdot {9\over10}\cdot{8\over9}\cdot{7\over8}\cdot{6\over7}$. So, the probability that Type A is dealt exactly once is $$ 5\cdot {2\over11}\cdot {9\over10}\cdot{8\over9}\cdot{7\over8}\cdot{6\over7}={6\over11}. $$ (or, more simply, you could compute the probability that type $A$ is dealt exactly once as ${2\choose1}{9\choose4}\over{11\choose5} $).

I'll leave it for you to do justify the easy case: type $A$ is dealt zero times with probability $3\over11$.

To do the last case, the probability that Type $A$ is dealt exactly twice, you could note that the sum of the three probabilities that you ask for must be 1...

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