If you have a pair for your hole cards. Is the probability that you will flop a set:

(2/50) + (2/49) + (2/48)

Thanks.

First off, I'm going to assume that by flopping a set, you really mean a set or better (so quads would count also, as would flopping a set when the other two cards are a pair giving you a full house). Anyhow the way you did it is a common mistake, which I'll explain later but first: The easiest way to do this is to calculate the probability you don't flop a set, and then subtract that from 1.

p = 1 - (48/50 * 47/49 * 46/48) = 1 - 1081/1225 = 144/1225 = 11.8%

You can also calculate it directly, but you have to be careful. Probability the first card gives you a set is 2/50. Then if the first card didn't give you a set (48/50), the probability the second did is 2/49. Then if the first and second cards both failed to give you a set (48/50 * 47/49), the probability the third one did is 2/48. So the sum is:

p = 2/50 + (48/50 * 2/49) + (48/50 * 47/49 * 2/48)
p = 1/25 + 48/1225 + 47/1225 = 144/1225

You can also solve this using combinations.

There are 50 remaining cards, two of which will give you a set or quads. The number of combinations for flopping a set or quads is then C(2,2)C(48,1) + C(2,1)C(48,2). The total number of flops is C(50,3). Dividing the number of success combinations by the number of flops gives you 11.8%. For a set only, use just C(2.1)C(48,2) in the numerator and the result is 11.5%