A salesperson has found that the probability of a sale on a single contact is approximately $.03$. If the salesperson contacts $100$ prospects, what is the approximate probability of making at least one sale?
My attempt:
Using the Poisson approximation, $$λ ≈ np = 100(.03) = 3$$ $$P(Y ≥ 1) = 1 – P(Y = 0)$$ $$P(Y ≥ 1) = 1 – \frac{3^0}{0!e^3} = 0.9502$$
I seem to have a conflict with the answer key.
The answer key:
$$λ ≈ np = 100(.03) = 3$$ $$P(Y ≥ 1) = 1 – P(Y = 0)$$ $$P(Y ≥ 1) = 1 – 0.97^{100} = 0.9524$$
The answer key seems to be using the binomial probability distribution. I understand that a Poisson probability distribution may not be necessary here, but the answer also gives the value of $λ$, and the answer key ignores it for the rest of the calculation. Moreover, I found this question in the Poisson probability distribution chapter.
In situations where you have a large $n$ and small $p$, the Poisson distribution can be used to approximate the binomial distribution. The larger $n$ and the smaller $p$, the better the approximation. This due to the Law of Rare Events, also known as the Poisson Limit Theorem.
You have a relatively small $p$ and reasonable $n$ for your problem; as such, one would expect the Poisson distribution to provide an adequate approximation of the Binomial distribution, as evidenced by your results.