Find the probability that a hand of 13 cards dealt from a well shuffled pack of 52 contains:
a) Exactly two kings and one ace
b) Exactly one ace given that it contains exactly two kings
c) Exactly 10 spades
d) At least two diamonds given that it contains exactly 3 spades and 4 hearts
Solution. Define C(n,k)=n!/[k!(n-k)!]. For example, C(3,2)=3!/2!=3 and C(5,2)=5!/(3!*2!)=10.
If we choose a hand of 13 cards dealt from a well shuffled pack of 52, there are C(52,13) possible cases.
(a) Exactly two kings and one ace.
There are C(4,2)*C(4,1)=6*4=24 possible case to choose exactly two kings and a ace. Then the rest of 10 cards must be chosen from those 11*4=44 cards. So there are C(44,10) possible case. So the total possible case for "exactly two kings and one ace" is ...
Combinations are used to calculate the probability of picking groups of cards.