While revising for my probability test, I saw this question from one of the previous exams:
You flip a fair coin 100 times. What is the probability that you have less than 45 heads?
My question is a simple one. I know that to calculate the probability of a specific amount of $ n $ heads (or tails), we can use the binomial distribution with formula $ P(X = k) = {n \choose k} p^k(1-p)^{n-k} $, with $ n = 100 $ and $ p = 0.5 $ in this case. I also know that we can compute the chance of $ P(X < k) $ as either $ P(X = 0) + P(X = 1) + ... + P(X = k - 10) $ or $ 1 - (P(X = k) + P(X = k + 1) + ... + P(X = n)) $.
However this is a simple question only worth two points out of over 60 total points. I cannot imagine that you need to compute $ P(X = n) $ 44 separate times and sum them together for such a small amount of points.
Is there any way to rewrite the formula or apply a different trick to drastically lower the amount of computations you need to do?
@lulu's comment recommending normal approximation was enough for me to solve the question myself. The trick is to indeed use the central limit theorem since we already have a large $ n $.
Since our distribution is a binomial one, we have $ E[X] = \mu = 0.5 \cdot 100 = 50 $ and $ Var(X) = \sigma^2 = 0.5 \cdot 100 (1 - 0.5) = 25 $.
Then, we can compute $ \frac{45-\mu}{\sqrt{\sigma^2}} = \frac{45-50}{\sqrt{25}} = -1 $. Looking this up in the Z-table gives $ 0.1587 $. This is close to the exact answer of $ 0.14 $ (and indeed, both 0.14 and 0.16 are marked as correct in the answer sheet).