If five cards are selected at random from a conventional 52 map deck, what is the probability of getting a full house.

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This is what ns am thinking.$(52*\binom43*\binom42)/_52C_5$

Is that right?

A full house has three cards the one kind and two the another, so think about it favor this: first you pick a kind of card (13 choices), then you choose three the end of 4 of those cards, climate you choose a second form of card, and finally you select two of those 4 cards. Thus you have actually $13\choose 14\choose 312\choose 14\choose 2$ possible full home hands. For this reason the probability is then

$$13\choose 14\choose 312\choose 14\choose 2\over52\choose 5=(13)(4)(12)(6)\over2598960=3744\over2598960\approx0.00144$$

Arthur Skirvin"s answer is great, but I just want to administer another thinking process for the numerator:

$$(\textrmchoose 2 kinds)\cdot(\textrmchoose the type of three cards\cdot\textrmchoose cards)\cdot(\textrmchoose the type of cards\cdot\textrmchoose cards)\\= 13\choose2\cdot2\choose14\choose3\cdot1\choose14\choose2.$$

You could also think about it this way, wherein I i think the card selections to it is in order dependency in both the numerator and also the denominator.

The total variety of possible selections is $52\times51\times50\times49\times48$.

To uncover the number of full home choices, very first pick 3 out that the 5 cards. For the 3 cards you have actually $52\times3\times2$ situations (once you pick the very first card, the rest need to have the exact same number), and also for the 2 cards you have $48\times3$ cases (you cannot pick from the previously selected class).

Thus the probability would be$\fracC(5,3)\times52\times3\times2\times48\times352\times51\times50\times49\times48\approx0.00144$.

Not quite. I would certainly break the count up into the adhering to steps.

How plenty of ways are there to pick the confront value because that the triple?How numerous ways space there to pick the suits for the triple? (You currently have this.)How plenty of ways room there to select the face value for the pair?How numerous ways are there to choose the suits because that the pair? (You currently have this.)
Unfortunately, it"s a little difficult to parse your question without LaTeX formatting, i m sorry compounds with the truth that I never use that particular notation because that binomial coefficients. For those who don"t know, a full house is a hand that $5$ cards such that $3$ of lock share the exact same rank and the continuing to be $2$ additionally share the exact same rank.

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You have $13$ choices (2-10, J, K, Q, A) for the location of the triple, and once that has actually been chosen, you have actually $12$ remaining options for the rank of the pair. Mental you also have to select $3$ suits that the possible $4$ because that the triple, and similarly $2$ of the feasible $4$ because that the pair. Express this information in regards to binomial coefficients to acquire the total number of possible full house hands.

Then just divide by the total number of unrestricted $5$-card hands because that the probability.