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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.

StubbornAtom
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David Antonio
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  • You say you think you will be using the binomial distribution. Cool. What exactly led you to that conclusion? Also, have you tried actually using it? What answer did you get? – Arthur Nov 26 '19 at 15:53
  • How many inmates in total? – Alex Nov 26 '19 at 15:59
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    Given that the total number of inmates is not mentioned, I can only imagine that there are a "large" number of inmates in this "large prison." In such a case, using the hypergeometric distribution although more correct, will not provide a noticeable difference in the final answer compared to using a binomial distribution to approximate the answer, massively simplifying calculations. I agree that binomial is the way to go, not hypergeometric like Alex's answer would otherwise suggest. – JMoravitz Nov 26 '19 at 16:03
  • I'm assuming you calculate the probability of $0, 1,$ or $2$ innocent inmates and subtract from $1$. I'm also assuming that by "large" prison they are implying we can assume if one prisoner is innocent than the probability of another is innocent, which for a "small" prison would be $\frac {0.1\times X-1}{X-1}$ where $X$ is the size of the prison, is still presumed to be $10%$. – fleablood Nov 26 '19 at 16:04
  • The point between using binomial versus hypergeometric, the event that one inmate is innocent and the event that a different inmate is innocent are technically dependent events. This is incredibly noticeable when the prison was small. If the prison had only 20 inmates total, there are only two total innocent inmates among them and it would be impossible to have picked 3 innocent inmates. As the number of inmates grows however, these events get closer and closer to being independent to the point where we might as well have treated them as though they were independent, simplifying calculation – JMoravitz Nov 26 '19 at 16:07
  • I don't see how you can use Binomial here, since it is sampling without replacement. Once a prisoner was 'eliminated', it changes the probability of sampling the next one – Alex Nov 26 '19 at 21:28
  • @Alex: the fact that the prison is large, with no further indication of size, would seem to indicate we are supposed to assume the prison is essentially infinite with respect to our sample. If this is the case, the event of finding an innocent or guilty person does not affect the probability of guilt or innocence of the other people. – robjohn Nov 26 '19 at 21:40
  • @robjohn I still don't see what's wrong with my answer – Alex Nov 26 '19 at 22:45

3 Answers3

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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

$P(n \ge 3) = 1-P(n\lt 3) = 1-(P(0)+P(1)+P(2))$

gandalf61
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Since we are randomly selecting $20$ inmates from a large population we can view the number of innocent inmates $N$ among the $20$ as a binomial random variable with probability of success $p=0.1$. (Each inmate is independently either innocent or guilty). Implicitly we are using the binomial approximation to the hypergeometric distribution. Hence $$ P(N\geq 3)=1-P(N=0)-P(N=1)-P(N=2) $$
where $$ P(N=k)=\binom{20}{k}(0.1)^k(0.9)^{20-k};\quad (k=0,\dotsc,20). $$

Sri-Amirthan Theivendran
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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?

Alex
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  • Suppose that there are $10n$ inmates, of which $n$ are innocent. In a sample of $20$ inmates, the probability of getting exactly $k$ innocent is $\frac{\binom{10n-20}{n-k}\binom{20}{k}}{\binom{10n}{n}}$. – robjohn Nov 26 '19 at 23:49
  • $\begin{align}\frac{\binom{10n-20}{n-k}\binom{20}{k}}{\binom{10n}{n}} &=\frac{\overbrace{(n-k+1)\cdots(n)}^{\sim n^k},,\overbrace{(9n-20+k+1)\cdots(9n)}^{\sim(9n)^{20-k}}}{\underbrace{(10n-19)\cdots(10n)}_{(10n)^{20}}}\binom{20}{k}\ &\sim\left(\frac1{10}\right)^k\left(\frac9{10}\right)^{20-k}\binom{20}{k}\end{align}$ – robjohn Nov 27 '19 at 01:36
  • I see that Binomial approximates Hypergeometric. Why does this make my solution incorrect? – Alex Nov 27 '19 at 08:42
  • I was addressing your question about why we can use the binomial distribution when $n\to\infty$. I never said anything was wrong with your solution; in fact, your formula is equal to mine. I cannot comment on the downvotes. – robjohn Nov 27 '19 at 10:16