10% of inmates are innocent. We randomly select 20 inmates. Find the probability that we will find at least 3 innocent inmates. Suppose that 10% of inmates in a large prison are known to be innocent. A non-profit group randomly selects 20 inmates from this prison. Find the probability the group will find at least 3 innocent inmates.
I assume that I will be using Binomial Distribution. Can someone confirm and give the value of the answer to this.
Suppose $p$ is the probability of an inmate being innocent. And suppose we can assume that the probability of each inmate being innocent is independent of the innocence or guilt of the other $19$ inmates (which is approximately true if the $20$ inmates are chosen from a large population). Then the probability of finding exactly $n$ innocent inmates in $20$ is

$P(n)=p^n(1-p)^{(20-n)}\binom{n}{20}$

Now you could work out $P(3), P(4), P(5) \dots$ etc. and add then up to find $P(n \ge 3)$. But is is easier to use the fact the $P(n \ge 3)+P(n \lt 3)=1$ (how do we know this ?) and so

$P(n \ge 3) = 1-P(n\lt 3) = 1-(P(0)+P(1)+P(2))$
How many inmates in total? Let's say it's $n$. So $0.1n$ of them are innocent. The total number to select 20 inmates is $\binom{n}{20}$. There's $\binom{0.9n}{20}$ to select no innocents at all and $\binom{0.9n}{19} \binom{0.1n}{1}$ to select exactly one. Can you take it from here?