# Probability of a 1 in a million event 4 times in a population of 15K?

I'm trying to get to grips with the basics of probability so have been looking out for news articles or comments where I can try to apply my learnings.

I was reading this post concerning the incidence of cancel in a small USA town. I did not read the article they are actually discussing on that thread but I did pick up this comment and wanted to use it to test my knowledge:

Four one in a million events in 15k inhabitants. Yes in any ranked list of stochastic outcomes there must be a highest scorer and a lowest scorer, but in case of Wayland GA it's still off the charts. With these diseases for any given town in the US you'd expect 0 or 1, hardly ever 2, if independently distributed.

Considering only the text highlighted in bold, I attempted to infer the probability and am unsure of my conclusion:

If the probability of the event really is 1 in a million and we assume the events are independent, then is the probability of this occurring $$(1/1000000)^4$$ = $$\frac{1}{1000000000000000000000000}$$ ?

Or, is it 1 in a million times 4 = 1/250000?

No, assuming independence, the number of occurrences is a binomial random variable $$X$$ with $$n=15000$$ and $$p=1/1000000$$. Thus the probability of exactly four occurrences is $$\binom{n}{4}p^4(1-p)^{n-4}.$$ However, you should instead ask for the probability of at least $$4$$ occurrences. This is given by: \begin{align*} P(X \geq 4) &= 1-P(X \leq 3)\\ &=1-(P(X=0)+P(X=1)+P(X=2)+P(X=3))\\ &=1-\left(\binom{n}{0}p^0(1-p)^n + \binom{n}{1}p^1(1-p)^{n-1}+\binom{n}{2}p^2(1-p)^{n-2}+\binom{n}{3}p^3(1-p)^{n-3}\right). \end{align*} The numbers get really big, but this works out to about $$2\times 10^{-9}$$. Notice that even the probability of zero occurrences is: $$\left(1-\frac{1}{1000000}\right)^{15000} \approx 0.985$$ so there is a $$98.5\%$$ chance that no one gets cancer, assuming independence.
If the incidents were independent, then we'd expect them to follow a Binomial Distribution. In that case the probability of seeing exactly $$4$$ incidents would be $$\binom {15000}4\times \left(\frac 1{10^6}\right)^4\times \left(\frac {10^6-1}{10^6}\right)^{14996}\approx 2.077\times 10^{-9}$$
And the probability of seeing at least $$4$$ incidents would be $$1-\sum_{i=0}^3\binom {15000}i\times \left(\frac 1{10^6}\right)^i\times \left(\frac {10^6-1}{10^6}\right)^{15000-i}\approx 2.083\times 10^{-9}$$
• Shouldn’t that be $15000 \choose 4$? – 雨が好きな人 Apr 27 at 16:47