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If I draw three cards at random (without replacement) from a standard 52-card deck, what is the probability that two of the cards will be black and one of them will be red?

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We have three possible sequences of cards that satisfy your conditions (two black, one red): BBR, BRB and RBB. Let $P(BBR),P(BRB),P(RBB)$ denote the probability that a randomly drawn sequence is of the form BBR, BRB and RBB respectively. The probability of three randomly drawn cards satisfying your condition is $P(BBR)+P(BRB)+P(RBB)$. I will show you how to calculate $P(BBR)$ and leave the rest to you.

When we draw the first card, there are $26$ black cards and $52$ cards total, so there is a $\frac{26}{52}=\frac{1}{2}$ chance that this card is black. If the first card is black, there are $25$ black cards out of $51$ total when we draw the second card, so there is a $\frac{25}{51}$ chance it is black. If the first two cards were both black, then there are $26$ red cards out of $50$ total when we draw the third card, so there is a $\frac{26}{50}=\frac{13}{25}$ chance it is red. Thus $P(BBR)=\frac{1}{2}\cdot\frac{25}{51}\cdot\frac{13}{25}=\frac{13}{102}$.

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There are $\binom{52}{3}$ sets of 3 cards. $\binom{26}{2}\binom{26}{1}$ of them are as you described. The ratio of the two is the probability.

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