How Do You Calculate the Probability of Drawing 3 Aces and 1 King from a Deck?

[Frank69]

Hello people. I m new to probability and now I have to give an exam at uni about it. One of the sample question is like this:

Four cards are drawn from a pack of 52 well-shuffled cards. Find the probability that 3 are aces and 1 is a king.The way I see it, it should be like:

A = { (ace)(ace)(ace)(king), (ace)(ace)(king)(ace), (ace)(king)(ace)(ace), (king)(ace)(ace)(ace) }

P(A) = P((ace)(ace)(ace)(king)) + P((ace)(ace)(king)(ace)) + P((ace)(king)(ace)(ace)) + P((king)(ace)(ace)(ace))

Where:
P((ace)(ace)(ace)(king)) = (4/52*3/51*2/50*4/49)
P((ace)(ace)(king)(ace)) = (4/52*3/51*4/50*2/49)
P((ace)(king)(ace)(ace)) = (4/52*4/51*3/50*2/49)
P((king)(ace)(ace)(ace)) = (4/52*4/51*3/50*2/49)

is this correct?

if not, would you show me the proper way to solve such problems?

Sorry for the very basic question, but I m real noob.

Thanks in advance

 

[mathman]

The answer is correct. A slightly clearer approach would be to consider the king first. It can be in any of 4 positions and for each position the probability is 4/52. Then the aces take up the remaining 3 positions with probability (4/51)(3/50)(2/49). Put it all together and you will have the same result you got.