The probability of at least 3 successes means he could have exactly 3, 4, 5, or 6 successes.Let p = probability of success and q = probability of failure (.5 for both in this case).
The formula for exactly x successes in N trials is NCxpxq(N-x)
So P(At least 3 successes) = 6C3p3q(6-3) + 6C4p4q(6-4) + 6C5p5q(6-5) + 6C6p6q(6-6)
P(at least 3) = .3125 + .2344 + .0938 + .0156 = .6563