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Current time:0:00Total duration:2:29

Probability with combinations example: choosing cards


Video transcript

- We're told that a standard deck of 52 playing cards includes four aces, four kings, and 44 other cards. Suppose that Luis randomly draws four cards without replacement. What is the probability that Luis gets two aces and two kings, in any order? So like always, pause this video and see if you can work through this. All right. Now, to figure out this probability, we can think about this, it's going to be the number of, let's call them draws with exactly two aces and two kings, two aces and two kings. And that's going to be over the total number of possible draws of four cards. So number of possible draws of four cards. Now, for many of y'all, this bottom, the denominator here, might be a little bit easier to think about. We know that there's 52 total cards, of which we are choosing four. So we could say 52 choose four, and that will tell us the total number of possible draws of four cards. How many combinations of four cards can we get when we're picking from 52? But the top here might be a little bit more of a stumper. We can think we have exactly two spots for aces, so we're choosing two aces out of how many possible aces? Well, there's four total aces. So if we say four choose two, this is the total number of ways, when you don't care about order, that you can have two out of your four aces picked. And then separately, we can use similar logic to say, all right, there's also four choose two ways of picking two kings out of four possible kings. And now the total number of draws with two aces and two kings, this is going to be the product of these two. And if you're wondering why you can just multiply it, think about it. For every scenario that you have these two aces, you have four choose two scenarios of which kings you're dealing with. So you would take the product of them. And we've already done many examples of computing combinatorics like this, so I will leave you there. If you're so motivated, I encourage you to be, you can actually calculate this value.