When to use binomial versus poisson distribution?

Question

I was doing the following question:

The probability that Bhim loses each game of badminton, independently of all others is $0.05$. Bhim and Joe play $60$ games. Use a suitable approximation, calculate the probability that Bhim loses more than 4 games.

I thought that the number of games Bhim loses would be distributed following: $X$~$B(60, 0.05)$. I then calculated $1-P(X\le4)=0.180$.

However, the answer said the distribution was: $X$~$Po(3)$. Doing a similar calculation that I did, and got $0.185$. Why is my answer incorrect, and the poisson distribution correct?

Many thanks.

Answer 1

Technically speaking, your approach is the correct one. The information we are given matches up perfectly to a binomial distribution. I believe the question's motive is to show you how good of an approximation the Poisson distribution is to the Binomial when we are dealing with relatively small values of $p$.

I think you understand that if we use a Poisson approximation, we get that $\lambda = 60(0.05) = 3$. And when you compare the answers you get that they are obviously similar.

But to answer your question, your answer is the exact answer, the Poisson approach is the approximation.

Answer 2

As has already been noted, in this case Binomial is exact but Poisson is a convenient approximation. It saves you having to sum a lot of terms (well, in this case a few we then subtract from $1$, but still). But as for the more general title question, there is a case where you can't use a Binomial distribution: when the mean is known, but can't be written as $np$ in an obvious way. For example, the number of goals a football team scores in a random game shouldn't be thought of as having a Binomial distribution, because you can't really say how many goals they try to score ($n$).