Probability that a bridge hand contains at least 2 fours...

A bridge hand is made up of 13 cards from a deck of 52. Find the probability that a hand chosen at random contains at least 2 fours. How would you find this out? I know how to do it if it's asking if it contains EXACTLY 2 fours, but how do you find the probability for AT LEAST 2 fours? Thanks!

You can calculate it 2 ways. Either P(2 fours), + P(3 fours) + P(4 fours), or
1 - (P(0 fours) + P(1 four))

P(0 fours) is the number of ways we can pick 0 fours C(4,0) times the probability of picking 13 cards from the non-fours C(48,13), divided by the total ways to pick cards C(52,13).
P(0 fours) = C(4,0) C(48,13) / C(52,13)
P(0 fours) ~ 0.3038

P(1 four) = C(4,1) C(48,12) / C(52,13)
P(1 four) ~ 0.4388

P(0 fours) + P(1 four) ~ 0.7426
P(2+ fours) = 1 - (P(0 fours) + P(1 four))
P(2+ fours) ~ 0.2574