# Probability question about sample proportion

This question goes,

"According to a study, 4% of a population in a region have lung disease. Of those people without lung disease, 13% are smokers. Of those people with lung disease, 92% are smokers.

(a) If 20 people are randomly selected from the region, what is the probability that at least one of them has lung disease?

For (a) I tried to solve by using $$\mu = np$$ and $$\sigma=\sqrt{npq}$$, I got $$\mu=0.8$$ and $$\sigma=0.4$$. So to find at least 1 of them has lung cancer, I let $$P(X \ge 1)=P(z \ge (\frac{1-0.8}{0.4/\sqrt{20}})=0.0125$$. But from my lecture notes it says that to use $$\mu=np$$ it is only for approximating binomial distribution to normal if $$n$$ is large enough. However for this question, since they are asking "at least one of them has lung cancer", can I still use this? If not, another way I thought to solve this is to use sample proportion where $$P(X \ge 1) = P(Z \ge \frac{0.05-0.04}{\sqrt{0.04 \times 0.96/20}} )$$ where $$0.05$$ comes from $$1/20$$.

$P(X \ge 1)$ is the same as $1-P(X=0)$. (If you're not greater or equal than 1, you have to be 0.)
• That's better, thanks. I'd be tempted to mention why the original poster (OP) has "overthought" the problem - since it's only necessary to find $P(X=0)$, there's no need to use a normal approximation. – Silverfish Nov 18 '15 at 17:47