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This is a nice solution if you are required to give a quick answer to the question. The binomial distribution may be well approximated by Normal is n is fairly large (~100), which would be N(np, sqrt(npq)) = N(200, 10) here. The Z score of 220 is (220 - 200) / 10 = 2. According to Normal distribution, P(|X| < 2) = 0.95, thus the probability of X > 200 would be P(X > 200) = (1 - 95%) / 2 = 2.5%, which is indeed a good approximation.
- xlzhang.iop March 27, 2015