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.