Cards In Exercises 59-62, you are dealt a hand of five cards f...

Question

Cards In Exercises 59-62, you are dealt a hand of five cards from a standard deck of 52 playing cards.

62. Find the probability of being dealt three of a kind (the other two cards are different from each other).

Accepted Answer

Welcome back, everyone. A hand of cards is built from a standard 52 card deck. What is the probability that your hand contains exactly one pair with the other three cards all of different ranks and different from the pair? AS is 0.4226, B is 0.5048, C 0.3766, and the D 0.4317. Now here in this problem we want one pair, same rank and different suits. 3 other cards of different ranks are not matching the pair, and each of these 3 cards must also be of different suits since suit doesn't matter for ranks unless they are duplicates. How can we find the probability that we're looking for using this information? Well, recall that the probability of an event. OK, is equal. To the number of successes in this case the total favorable hands. OK. Divided by the number of outcomes. In this case, the total number of hands, OK. So if we can find both, then we should be able to solve for this probability. Now, let's start with the total number of hands, OK? Now, we know that, that from the standard 52 card deck, OK, from the deck. Then the total number. Of hands, OK. The total number of hands we can choose. In other words, all the possible ways that we can get 5 hands from a 52 card deck will be equal to 52, sorry, 5 from 52, OK, as a combination, which is 2,598,960. OK? In other words, these are all the possible combinations that we can get. Now, if we choose the rank for the peer, OK, choosing a rank, choosing one rank. OK. Then we know that it's gonna be one rank from 13 different possibilities, and that means it's gonna be 13 ways, OK. Next, to choose 22 suits out of 4 for this rank, OK? 2 suits out of 4. OK, for this rank. OK. Can be done in 6 ways, OK. Choosing 3 different ranks excluding the peer, the peer's rank. Since there are 12 remaining ranks, we'll choose we choose 3, sorry. So choosing 3 different ranks. OK, for the other cards, so that's excluding the peers rank. OK. And we know that that's going to be equal to 3 out of 12 possible or comes here 220 ways, OK. And then for each of these three ranks, we want to choose one suit, OK? One suit for each rank, OK? And now one suit for each rank can be done in 4 multiplied by 4 multiplied by 4, or 64 ways. So no, this implies then that the total favorable hands, OK. To out favor of her hands. Can be found by multiplying all of the ways that we found. OK. In other words, choosing one rank 13 ways multiplied by 2 suits out of 4 for this rank, 6 ways multiplied by choosing 3 different ranks, 220 ways multiplied by 1 suit for each rank, 64 ways, and that gives us 198,240. So no, that means then that our probability. Now that we have the total favorable hands and the total number of hands, it's gonna be 198,240 divided by 2,598,960, which is gonna be equal to 0.4226, OK? Thus, this is the probability. Of the probability that our hand contains exactly 1 pair with the other 3 cards all of different ranks and different from the pier. If we look back at our answer choices, that means A is the correct answer. Thanks a lot for watching everyone. I hope this video helped.

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Cards In Exercises 59-62, you are dealt a hand of five cards from a standard deck of 52 playing cards.
62. Find the probability of being dealt three of a kind (the other two cards are different from each other).

Key Concepts

Probability

Combinatorics

Standard Deck of Cards

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Cards In Exercises 59-62, you are dealt a hand of five cards from a standard deck of 52 playing cards.62. Find the probability of being dealt three of a kind (the other two cards are different from each other).