Suppose that two cards are dealt from a standard 52-card poker deck. let a be the event that the sum of the two cards is 8 (assume that aces have a numerical value of 1). how many outcomes are in a?

The Combinations that equal 8 are: A+7, 2+6, 3+5, 4+4, 5+3, 6+2, 7+A

For instance there are 4 Aces and 4 sevens therefore = 4*4 = 16

So there are 16 possible combinations of A, 7. Another 16 for 6,2 and another 16 for 3,5.

16*3=48

4C2=6 a special case for 4,4

Therefore:

48+6=54

a = 54