Forgive me if this is simple, but I've been twisting around this problem for a bit.
I know how to calculate if a given event happens exactly $x$ times over $n$ trials (where $p$ is the probability of the event occurring during a single trial):
$$p^x (1-p)^{n-x}$$
It seems like I could get the result of it occurring at least $x$ times by doing a sum…
$$\sum_{q=0}^{n-x} p^{x+q} (1-p)^{n-(x+q)}$$
… But I'm assuming there's a simpler way, mathematically, to go about calculating this. Can anyone enlighten me?
Best Answer
No simpler general formula than $$\sum_{k=x}^{n}{n\choose k} p^{k} (1-p)^{n-k}$$ Of course, if $n$ and $x$ go to infty, this is another story. For example, if $n\to\infty$, $x\to\infty$, and $x/n\to r$ for some fixed $r$ in $[0,1]$, then the sum converges to $1$ for every $r\lt p$, to $\frac12$ for $r=p$, and to $0$ for every $r\gt p$.
Edit (to answer a comment): Recall that $0!=1$ hence, in the sum above, the $k=n$ term is $$ {n\choose n} p^{n} (1-p)^{n-n}=1\cdot p^n\cdot(1-p)^0=p^n. $$