A pack of 36 cards includes 20 numbered cards from 6 to 10 inclusive, 4 aces and 12 picture cards. If a hand of 5 cards is selected at random, how do you find the probability of receiving at least 2 aces?

Let's first see that if I do this, I can say:

Hands with 0 Aces + Hands with 1 Ace + Hands with 2 Aces + Hands with 3 Aces + Hands with 4 Aces = All possible hands

And so we can work this by adding up the hands with 2, 3, and 4 aces, or we can find all possible hands and subtract out hands with 0 and 1 ace. Since they're the same amount of work, I'll do the subtraction method (it being more interesting).

The number of All possible hands is combination of 36 pick 5:

#((36),(5))=(36!)/((31!)(5!))=(36xx35xx34xx33xx32)/120="376,992"#

The number of hands with 0 aces means we have 0 aces from the 4 available and we have 5 cards from the 32 remaining:

#((4),(0))((32),(5))=(1)(32xx31xx30xx29xx28)/120="201,376"#

And the number of hands with 1 ace means we have 1 ace from the 4 available and 4 cards from the remaining 32:

#((4),(1))((32),(4))=(4)(32xx31xx30xx29)/24="143,840"#

This gives:

#"376,992"-"201,376"-"143,840"="31,776"#

~~~~~

Since this might seem like a very small number compared to what we've found so far, let's work out the hands with 2, 3, and 4 aces:

2 aces:

#((4),(2))((32),(3))=(6)(32xx31xx30)/6="29,760"#

3 aces:

#((4),(3))((32),(2))=(4)(32xx31)/2=1984#

4 aces:

#((4),(4))((32),(1))=(1)(32)=32#

#"29,760"+1984+32="31,776"#