Standard probability deck of cards problem

Question

A hand of 8 playing cards is selected at random from a standard pack of 52. Calculate

the probabilities that the hand contains:

(a) no royalty (Jack, Queen, King);

(b) exactly 2 clubs;

(c) at least 3 of the 4 aces.

Answer 1

The key to this problem is the idea of Combinations: The number of ways you can pick a set of "r" things from a group of "n" things. rCn is one notation for this. For example, the total # of ways you can pick any 8 cards from a set of 52 is 52C8. The formula for this example would be (52!)/((8!)*(52-8)!), where 52! means "52 factorial", or 52*51*50*49...*3*2*1. It's easy to get caught up in the arithmetic.

a) Since you only get to pick from the non-royalty cards, there are 52-12 = 40 of those.

So it's (40 C 8)/(52 C 8). 40 choose 8 divided by 52 choose 8. Or, the numerator would be (40*39*38*37*36*35*34*33)/(1*2*3*4*5*6*7*8), and the denominator would be (52*51*50*49*48*47*46*45)/(1*2*3*4*5*6*7*8).

b)You have to get 6 non clubs and 2 clubs: (39 C 6)*(13 C 2) / (52 C 8)

c) You need 3 or 4 aces, and 5 or 4 non-aces.

So (4 C 3) * (48 C 5) / (52 C 8) + (4 C 4) * (48 C 4) / (52 C 8)