# Probability of more than 50 heads from 90 coin tosses?

How would I work out the probability of getting more than 50 heads from 90 coin tosses? I thought I needed to use the binomial distribution but on wikipedia that mentions "successes in a sequence", whereas here the heads may not be in a sequence.

To be brutally honest I have no idea which basic rules to use for whichever "coin tossing" question so if somebody could give me the answer but also some general advice how to know which technique to use for slightly-differing questions that would be great!

-

$P$(more than 50 heads) = $\sum_{n=51}^{90} P$ (exactly n heads) = $\sum_{n=51}^{90} {90 \choose n} (1/2)^{90} = (1/2)^{90}\sum_{n=51}^{90} {90 \choose n} .$ You are right. If $X$ denotes the number of heads achieved in $90$ tosses then $X \sim Binomial(n,1/2)$, assuming the coins are fair.