Bridge hand is made up of 13 cards from a deck of 52. find the probability that a hand chosen at random contains at least 1 nine.
the probability that a bridge hand chosen at random contains at least 1 nine is:

I think the answer in that problem is 10%

C(n,r) = n!/(r!(n-r)!)

You take the number of possible 13 card hands with no 9 in them and subtract that from the total number of possible 13 card hands 9's included.

so C(52,13) - C(48,13)
The number of all possible 13 card hands is:
52!/13!(52-13)! or 52!/13!*39! which is 635,013,559,600

The number of all possible 13 card hands with no 9s is:

48!/13!(48-13)!  or 48!/13!*35! = 192,928,249,296

The difference is 635,013,559,600 - 192,928,249,296 = 442,085,310,304

So 442,085,310,304 out of 635,013,559,600 hands will have at least 1 nine.  The ratio of  442,085,310,304 to 635,013,559,600 is 0.696182472...  So roughly 70% of the time 

27,630,331,894/39,688,347,475 is the best I could do for a whole number ratio.

There are 8,654 left

36,000,00093,000,0002,790,000,000