Coin toss, probability for less than expected tails?

Question

A fair coin is tossed 500 times. What is the probability that the number of tails differs from expected value ( 250 ) by more than 30?

Answer 1

Let $X$ be the number of tails. Then: $\mathbb E(X)=500\cdot 0.5=250, \sigma =\sqrt{500\cdot 0.5\cdot 0.5}=5\sqrt{5}\approx 11.18$.

You need to find: $$P(|X-250|>30)=P(X<220 \ \cup \ X>280)=P(X<220)+P(X>280)=\\ P(\frac{X-0.5-250}{5\sqrt{5}}<\frac{220-0.5-250}{5\sqrt{5}})+P(\frac{X+0.5-250}{5\sqrt{5}}>\frac{280+0.5-250}{5\sqrt{5}})=\\ P(Z<-2.728)+P(Z>2.728)=\\ 0.003+0.003=0.006.$$ Reference: Normal approximation of binomial distribution.

Answer 2

Let $T$ be the number of tails out of $500$ flips. I would imagine the answer would go something like \begin{align} P(|T-250|>30) &= \frac{ \sum_{i=1}^{219} \binom{500}{i} + \sum_{i=281}^{500} \binom{500}{i}}{2^{500}} \\ &=\frac{ 2\sum_{i=1}^{219} \binom{500}{i}}{2^{500}} \\ \end{align} where we have a total of $2^{500}$ possible outcomes, and we count how many outcomes have $x$ number of tails for every $x$ satisfying $|x-250|>30$.