Finding n from a Poisson Approximation

Question

Q. In a certain population, 1 per cent of sheep are brown. How large must a random sample be to ensure that the probability of including at least brown sheep is at least 0.95? (Assume that sampling is with replacement, and use a Poisson approximation.)

I am familiar with using the Poisson distribution to obtain a probability, but this seems to be asking me to do the reverse and I can't wrap my head around it. My thought process was:

success = brown sheep

Pr(>=1 success out of n trials) >= 0.95

x >= 1

lambda = np = 0.01n

Solve for n using the Poisson equation. But this is algebraically tricky (at my level anyway) and this question was worth very few marks. What am I missing?

Thanks so much.

Answer 1

Using a binomial distribution, we'd be looking at the complement of the probability of getting no brown sheep in $n$ trials. That is, we'd be looking for $n$ such that

$$\binom n 0 p^0 (1-p)^n < 0.05,$$

where $p = 0.01$ is the probability of seeing a brown sheep. With the Poisson approximation, we'd be looking at

$$\frac{\lambda^0 e^{-\lambda}}{0!} = e^{-\lambda} < 0.05,$$

where $\lambda = np$. That amounts to $-\lambda< \log(0.05)$, or $n > 100\log(20) \approx 299.57$. So $n=300$ should do.

---

Contrast this with the direct calculation from the binomial distribution:

$$n\log(1-p) < \log(0.05) \implies n > 298.07$$

In other words, you'd be off by $1$ or $2$ samples using the approximation.

Answer 2

Suppose you are drawing $n$ sheeps randomly, with replacement. If $X$ be the no. of brown sheeps in your random sample, then $X\sim \text{Bin}(n, 0.01).$ And we wish to find (minimum) $n$ such that $\Pr(X\ge 1) \ge 0.95,$ i.e. $\Pr(X=0)\le 0.05.$ It seems that $n$ should be large enough, so we may approximate the distribution of $X$ by a Poisson distribution with mean $\lambda = 0.01n.$ Now, $\Pr(X=0)=e^{-\lambda},$ so we require $e^{-0.01n}\le 0.05$ which gives $n\ge -\frac{\ln(0.05)}{0.01}\approx 299.57,$ i.e. $n\ge 300.$