I need help with the following problem:

A brighe hand is made up of the 13 cards from a deck of 52. What is the probability of a hand chosen at random containing 6 of one suit, 5 off another, and 2 of another?

THe first card can be anything.
It seems to me that the probabality is this:
13*4/52 *12/51*11/50*10/49*9/48*8/47 *
13/46*12/45*11/44*10/43*9/42*
13/41*12/40

check my thinking.

I presume, D, you know the formula for calculating n-choose-x, where you have n things and need to choose x of them. = n!/(x!*(n-x)!)

So, lets start with the easy part, the denominator. It is 52-choose-13 -- the total of combinations of 13 cards from a deck of 52.

Now for the numerator. Of the 4 suits, one has 6 cards, one has 5, one has 2 and one has 0. How many permutations does this represent? The answer is n! or 4*3*2*1 = 24.

Now then, for the first suit, there are 13 cards in a suit, you need 6 -- so 13-choose-6. For the second suit its 13-choose-5, for the third its 13-choose-2.
So the numerator is 24 * (13-C-6) * (13-C-5) * (13-C-2).

check my thinking.

This may or may not agree with the other answers posted. It was provided by my son, who is an actuary and runs the website www.wizardofodds.com. He makes his living doing this sort of thing.

There are 24 ways to arrange the suits (4 possible suits for the long one, 3 for the medium, and 2 for the short suit).

There are 13!/(6!*7!) ways to pick 6 ranks out of 13 for the long suit.

There are 13!/(5!*8!) ways to pick 5 ranks out of 13 for the medium suit.

There are 13!/(2!*11!) ways to pick 2 ranks out of 13 for the short suit.

So the total number of combinations fitting the criteria is 4134297024.

The total number of all ways to pick 13 cards out of 52 is 52!/(13!*39!) = 635013559600

So the probability is 4134297024/635013559600 = 0.006510565

Your son's method and my posted method are the same. Your post goes further by actually showing the calculations.

Thanks. I see tnat now that your method is the same. I requested mu son's help last night before any answers were posted, and thought it would be a shame not to use his answer.

Thanks everyone! I've been stressing over this and it really helps seeing the different explanations.

how do i do probability trees when using replacement

fyunoeht ncqklmy kavyfrgp qrvfhig efcdnbvp fexc ijzwaq

fyunoeht ncqklmy kavyfrgp qrvfhig efcdnbvp fexc ijzwaq