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Getting a Full House Find the probability of getting a full house (3 cards of one denomination and 2 of another) when 5 cards are dealt from an ordinary deck.

Intro Stats / AP Statistics

Chapter 4

Probability and Counting Rules

Section 5

Sampling and Data

Probability Topics

Missouri State University

Piedmont College

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University of St. Thomas

Lectures

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Suppose that 5 cards are d…

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You are dealt five cards f…

02:06

What is the probability th…

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You are dealt one card fro…

02:51

Selecting Cards Find the p…

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01:12

A five-card poker hand is …

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Suppose that 3 cards are d…

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So in this problem, we want to know the probability of getting a full house when we draw five cars from a deck. And ah, full house means we have three cards of one rank and two of another rank. And here is an example. We have three sixes and two kings, and it is a full rank. Mhm. So, um, the probability of getting a full house equals the number of ways Um, we could get a full house divided by the total number of ways we could select five cards out of a deck. And the denominator is relatively easier to get. So first we will focus on the numerator, and to get the numerator, we need two steps. Step one is to determining determining the ranks. So basically, we need to choose to ranks out of 13 and sort them so that the first rank appears three times in the five cards and the second rank appears twice. So, um, we want to know how many the number of ways we could determine the two ranks. So, um, here we should use the permutation because we care about the order. So basically, this is 13 p to and Step two would be to draw cards from each rank. So when we have the two ranks we wanted draw three cards from the first one and two cars from the second rank, Um, so the number of ways we could do that is so here. We don't care about the order. So we used the combination. Um, it is four. Choose three times, um, times for choose to mhm. Um, and so the probability of carrying a full house equals it equals the number of ways we could do step one times the number of ways we could do Step two. And the denominator would be, um, again, we don't care about the order. So there should be 50 to choose five. Oh, and equals 156 times. Four times. Um, sorry. I should be. Should be times six times for Yeah, sorry. It should be four times six. Um, and the denominator would be It's a very large number. And finally, we can get the result. Just 1.44 times turned thio minus three

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