If q=1-p=0.03, the probability of failure, then we have the binomial distribution:
(p+q)n=pn+npn-1q+n(n-1)pn-2q2/2!+... where n=the number of independent trials=65. The sum of these terms =1 because (p+1-p)n=1 for all n.
Each term in this series represents an individual exact probability:
(65 successes)+(64 successes)+(63 successes)+...
So we need the third term: 65×64/2×0.9763×0.032=0.2747 approx (27.47%).