What is the probability that a hand of five cards will contain at least one king?

Another way to get CPhill's answer:

Still using his idea of subtracting the probability of getting no kings from 1, the total possible probability.

The only way that you can get no kings is:

to get no king for the first card:  probability = 48/52    

to get no king for the second card:  probability = 47/51

to get no king for the third card:  probability = 46/50

to get no king for the fourth card:  probability = 45/49

and, finally:

to get no king for the fifth, and last, card:  probability = 44/48

Multiplying all these probabilities together:  48/52 x 47/51 x 46/50 x 45/49 x 44/ 48  =  0.6588

Subtracting this from 1.000:  1.000 - 0.6588  =  0.341

2 jokers,  4 aces, 4 2's, 4 3's, 4 4's, 4 5's, 4 6's, 4 7's, 4 8's, 4 9's, 4 10's, 4 jacks, 4 queens, and 4 kings

I'm not good with probabiltiy...but in 54 cards....there are 4 kings,  so maybe 4/25 because their are 5 cards in the hand....like I said,  not good, giving my best try

What's up with all these probability problems?

This equals..... [1 - probability of no kings ]

So, the ways to choose no kings is to just select 5 cards from the remaining 48

This is ginen by  C(48,5) = 1712304

And the total way of selecting 5 cards from 52 is C(52,5) = 2598960

So....the probability of at least one king is

1 - 1712304/2598960 = about 34.1%