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There are 6 players and 18 cards. Each of the 18 cards is numbered 1-18 (each is unique). Each player is dealt 3 cards. Players A and B are the first two players in the deal. The deal was a uniform distribution. There are two conditions:

Condition 1 Player A has card 1 or 2. She could have both.

Condition 2 Player B has card 1 or 3. She could have both.

What is the probability that both conditions are true? i.e, find $$P(C_1 \cap C_2)$$

My solution at this point

$$P(C_1 \cap C_2) = P(C_1 | C_2) * P(C_2)$$

$C_1 = event\ that\ player\ A\ has\ card\ 1\ or\ 2 $ $C_2 = event\ that\ player\ B\ has\ card\ 1\ or\ 3 $

$J = event\ that\ player\ B\ has\ card\ 1\ = P(J) = 1/6$

$K = event\ that\ player\ B\ has\ card\ 3\ = P(K) = 1/6$ $$P(C_2) = P(J \cup K) = P(J) + P(K) - P(J \cap K)$$

$$P(C_2) = 1/6 + 1/6 - nPr(3,2)/nPr(18,2)$$

$L = event\ that\ player\ A\ has\ card\ 1 = P(L) = 1/6$ $H = event\ that\ player\ A\ has\ card\ 2 = P(H) = 1/6$ $M = event\ that\ player\ B\ does\ not\ have\ card\ 1 = P(M) = 1- P(J) = 5/6$

We can agree here that $P(C_1) = P(C_2)$

$$P(H | J) = 3/17$$

$$P(C_1 | M) = P(L | M) + P(H | M) - P(L \cup H | M)$$ $$P(C_1 | M) = 1/5 + 14/(15*17) - nPr(3, 2) / nPr(18,2)$$

$$P(C_1 | C_2) = \frac{P(H|J)*P(J) + P(C_1 | M)*P(M)}{P(C_2)}$$

$$P(C_1 \cap C_2) = P(H|J)*P(J) + P(C_1 | M)*P(M)$$

$$P(C_1 \cap C_2) = .22549$$

My question:

does my math check out? Is there a more elegant way to solve this type of problem? (For example, if I was to add a third condition such that Player C has card 1 or 4, is there a simpler and more concise way to solve this problem?)

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