Probability of salesperson making a sale off one person is $3\%$. If 100 people are contacted (and the distribution of sales is Poisson), what is the probability of making at least one sale?
I used $P(X\ge1)=1-P(X=0)=1-e^{-3}=0.95021$ with $n=100, p=0.03$ hence $\lambda=3$
The book gave $0.9524$. What did I do wrong?
The book's answer apparently uses a more straightforward method, assuming the Binomial distribution. That's because with a binomial distribution, the answer is $$ 1- P(X=0) = 1- 0.97^{100} \approx 0.952~447 $$ This is a simpler approach, not using the Poisson approximation. The Poisson approximation is easier to calculate but it applies only to "large $n$". In this case $n=100$ which is not too large. In a way, I would say that the book's answer is closer to the "real world" answer, but on the other hand, if the book is trying to teach the Poisson approximation, your solution with $$ 1- P(X=0) \approx 1- e^{-3} $$ would be the correct way to calculate it with the Poisson approximation.