Probability of island having 8 people born with disease, estimate?

Question

The chances of being born with a certain disease are estimated as $1$ in $1200$. What is a good estimate of the chance that an island with $10000$ inhabitants has precisely $8$ people born with that particular disease? We assume that all $10000$ events of being born with that particular disease are mutually independent.

Answer 1

This random variable has the binomial distribution. Hence, the probability that an island with $10000$ inhabitants has precisely $8$ people born with that particular disease is given by $$ {10000\choose 8}\biggl(\frac1{1200}\biggr)^8\biggl(1-\frac1{1200}\biggr)^{(10000-8)}\approx0.1387. $$

Answer 2

Let $X$ be the number of people with disease. It's clear that $X$ has binomial distribution with $n=10000$ and $p=\frac{1}{1200}$. We can calculate the exact probability using the formula

$$P(X=8) = {10000\choose 8}\biggl(\frac{1}{1200}\biggr)^8\biggl(1-\frac{1}{1200}\biggr)^{(10000-8)}$$

The exact calculation of the above quantity is hard. We can approximate Binomial distribution as Poisson with parameter $\lambda = np$. The larger the $n$ and the smaller the $p$, the approximation is better. Hence,

$$P(X=8) \approx \frac{\exp(-\frac{10000}{1200})(\frac{10000}{1200})^8}{8!} \approx \frac{\exp(-8.33)(8.33)^8}{40320} \approx 0.14$$