What is the probability of drawing a full house with three aces and two kings?

[Dick]

I understand where both you are coming from now. I guess it was easier to read from a textbook and follow the examples. However the actual question is "in how many ways can three aces and two kings be drawn?"
So i figured out how 1 hand of 3 aces and 2 kings can be arranged=10
I found out how many ways 3 aces can be drawn=24
I found out how many ways 2 kings can be drawn=12
If I were to multiply both the 24 and 12= 288, then I should multiply 288 by the 10(arrangements per hand)
That would equal 2880, thus there are 2880 ways 3 aces and 2 kings can be drawn. Is that sound about right?

And you can also get the same probability answer without considering ordering. The number of unordered ways to pick the cards is C(52,5). The number of unordered ways to pick three kings is C(4,3) and two aces is C(4,2). So C(4,3)*C(4,2)/C(52,4) is the same as your 2880/311875200. That would be the more popular way to solve the probability problem.

 

[Bacle]

Are you playing with a full deck :) ?

But seriously; I don't recognize the 311875200 as an nC5 (i.e., "n choose 5" ).

Are you using replacement?

If you have x cards with 4 suits of 13 cards each ; with 1,2,...,j,q,k in each suit, and

you're dealing without replacement, then the answer would be 4C3*4C2 , i.e.,

select any 3 aces from the four you have in 4C3 ways ,choose 2 kings in 4C2.

If you deal with replacement, you can choose in 4!/(4-3)! , and 4!(4-2)! ways

respectively.

If you give us the original numbers, we may be better able to help.

 

[YODA0311]

Are you playing with a full deck :) ?

But seriously; I don't recognize the 311875200 as an nC5 (i.e., "n choose 5" ).

Are you using replacement?

If you have x cards with 4 suits of 13 cards each ; with 1,2,...,j,q,k in each suit, and

you're dealing without replacement, then the answer would be 4C3*4C2 , i.e.,

select any 3 aces from the four you have in 4C3 ways ,choose 2 kings in 4C2.

If you deal with replacement, you can choose in 4!/(4-3)! , and 4!(4-2)! ways

respectively.

If you give us the original numbers, we may be better able to help.

Yes part (a) is from a whole deck of cards, however I already checked the answers with previous tutors. However maybe you see something wrong.
I was using the permutation formula= n!/(n-r)!

 

[YODA0311]

Are you playing with a full deck :) ?

But seriously; I don't recognize the 311875200 as an nC5 (i.e., "n choose 5" ).

Are you using replacement?

If you have x cards with 4 suits of 13 cards each ; with 1,2,...,j,q,k in each suit, and

you're dealing without replacement, then the answer would be 4C3*4C2 , i.e.,

select any 3 aces from the four you have in 4C3 ways ,choose 2 kings in 4C2.

If you deal with replacement, you can choose in 4!/(4-3)! , and 4!(4-2)! ways

respectively.

If you give us the original numbers, we may be better able to help.

To make it more clearer for you:
(a) in how many ways can 5 cards be drawn from a whole deck?= 52!/(52-5)!=311875200
(b)how many ways can 3 aces be drawn from the total 4 possible aces?=4!/(4-3)!=24
(c)how many ways can 2 kings be drawn from the total 4 possible kings?=4!/(4-2)!=12
(d)how many ways can 3 aces and 2 kings be drawn?=they can be drawn 24x12=288 ways. Then I used n!/n!1*n!2=5!/3!2!=120/12=10, which equals the amount of ways 1 five card hand can be arranged. Then I took 10x288=2880.
Now I need to answer what is the probability of drawing a full house consisting of 3 aces and 2 kings?