# The probability of at least one of $n$ draws being greater than $y$

Let's say I have a probability distribution $F(x)$, which is a cdf. Then, I have a value $y$ at some place on the support of $F$. Then, there will be some number $n$ random independent draws from $F$, and I want to know the probability that at least one of them is $> y$.
I would think that to find this I would do $n(1-F(y))$, since the pr. of any one draw being $> y$ is $1-F(y)$. But say $F(y)$ is $.5$. Then, is $n=3$, then I would get $1.5$, which of course doesn't make any sense.

What am I doing wrong?

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Try to think about the complementary event: what is the probability that all of the values drawn are smaller (or equal) than $y$? – Fabian Jan 9 '12 at 19:58
If all the values are smaller, then the pr is $F(y)^n$... so I guess the answer is $1-F(y)^n$ ... thanks, @Fabian! (hopefully I'm not still wrong...) – Ac3 Jan 9 '12 at 20:01
you are welcome... – Fabian Jan 9 '12 at 20:02
Ac3: You could write this as an answer and possibly, a little later on, accept it. – Did Jan 9 '12 at 20:07
I think the credit should go to Fabian... but if s/he doesn't add an answer after a while, I will write one so I can close this. – Ac3 Jan 9 '12 at 20:11

The problem is rather easy if one thinks about the complementary event: what is the probability that all of the values drawn are smaller (or equal) than $y$? The probability that a single draw is smaller than $y$ is given by $F(y)$. As the $n$ draws are independent the total probability for the complementary event is $F(y)^n$.
Concluding, the probability that at least one of the draws is larger than $y$ is given by $$1- F(y)^n$$ (which of course is always between 0 and 1).