Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?


share|cite|improve this question
up vote 2 down vote accepted

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 $P(BBR)=\frac{1}{2}\cdot\frac{25}{51}\cdot\frac{13}{25}=\frac{13}{102}$.

share|cite|improve this answer

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.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.