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A standard deck of playing cards consists of 52 cards. Each card has a rank and a suit. There are 4 possible suits (spades, clubs, hearts, diamonds) with 13 cards each. Assume that the deck is perfectly shuffled (that is, all outcomes are equally likely).

What is the probability that a hand of 5 cards dealt from the deck contains only diamonds given that the first 3 cards in the hand are the ace of diamonds, the queen of diamonds, and the king of diamonds?

$n|S|=\binom{52}{5}$

$n|E|=4*\binom{13}{3}*\binom{48}{2}$

=> $\frac{4*\binom{13}{3}*\binom{48}{2}}{\binom{52}{5}}$

Please verify this..

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  • $\begingroup$ Your answer is incorrect. The favorable cases consists of hands that include five diamonds. Since you are told that the hand includes the $\color{red}{A\diamondsuit}, \color{red}{Q\diamondsuit}, \color{red}{K\diamondsuit}$, you need two more diamonds. $\endgroup$
    – N. F. Taussig
    Commented May 5, 2018 at 20:13

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There are 49 cards left, of which 10 are diamonds. The probability that the next card is a diamond is $\frac{10}{49}$ and probability that the next card after is also a diamond is $\frac{9}{48}$. Therefore what you want is $\frac{10\times 9}{49\times 48}$.

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