What's probability of getting head on coin flip If I flip a coin for once, the chances I get the head as the result is $50\%$.
How can I calculate if I flip the coin for $2$ times, $10$ times or $100$ times, what is the chances I get head result for at least once?
I know this question is somehow stupid in this forum. But please show me how to calculate. Thanks.
 A: It can be done a lot easier: instead of calculating the probability of one head, two heads, three heads, ... one just needs to calculate the probability of no heads: that is simply $0.5^n$. If you subtract it from 1, you get the probability you want: it's because that's the chance of not no heads, meaning at least one head.
So, the formula is:
$1-0.5^n$
More formally:
if X is throwing at least one head,
$P(X)=1-P(\neg X)$, where $\neg X$ is throwing zero heads.
A: The probability of getting Head at least once is equal to $1$ minus the probability of never getting Head, which is equal to $1$ minus the probability of getting Tail all the time.
And that's pretty easy to calculate...
Over $N$ coin-flips:


*

*The probability of getting only Tail is $(\frac{1}{2})^N$

*So the probability of getting at least one Head is $1-(\frac{1}{2})^N$

A: I think this can be solved by bernoulli trial
Please see Wikipedia link.
http://en.wikipedia.org/wiki/Bernoulli_trial
Let n be the number of flips then for at least one head your answer should be = (nC1+nC2+nC3+...nCn)*(0.5^n) = 1-(0.5^n)
