# 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!

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$P$(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.

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Thanks! I presume if the question expanded to say we had 9 coins which were thrown at the same time (so we threw 9 coins 10 times) would this change the answer? My guess is that it wouldnt be "a coin" is still being thrown 90 times.... – user997112 Mar 26 '14 at 16:42
It does not change the answer, because what matters is that individual outcomes are independent and identically distributed. That is to say, for any given trial of a single coin toss, the outcome does not depend on any other trials that have occurred, and the probability of the outcome being heads/tails does not change from trial to trial. – heropup Mar 26 '14 at 16:48
Thank you for replying. Could I just ask one last Q- what about if the question was finding a) N successive heads from M tosses and b) at least N successive heads from M tosses? – user997112 Mar 26 '14 at 17:00