Poisson Approximation to calculate the probability that at most 2 persons in 500 will have a birthday on Christmas

Question

This is from Hoel's Probability book. Use the Poisson Approximation to calculate the probability that at most 2 persons in 500 will have a birthday on Christmas. Assume 365 days in the year. On average $\frac{500}{365}=1.36$ of people have a birtday every day. This is what I've got at the moment: $$ P\left(X\leq2\right)=\sum_{x=0}^{2}\frac{e^{-\mu}\mu^{x}}{x!}=f\left(0\right)+f\left(1\right)+f\left(2\right)$$

$$P\left(X\leq2\right)=0.2566+0.3490+0.1186=0.7242$$

Answer 1

Following @DanielAdams' Comment:

Binomial probability: $X \sim \mathsf{Binom}(n=500, p=1/365),$ This is exact assuming independence and 365 equally likely birthdays.

Find $P(X \le 2) = 0.8317.$ [Computation in R.]

pbinom(2, 500, 1/355)
[1] 0.8316545

Poisson probability: $ Y \stackrel{aprx}{\sim}\mathsf{Pois}(\lambda = 500/365).$ This is an approximation.

Find $P(Y \le 2) = 0.8407.$

ppois(2, 500/365)
[1] 0.8407328

The three Poisson probabilities summed are as follows:

dpois(0:2, 500/365)
[1] 0.2541418 0.3481394 0.2384517

Note: In R pbinom is a binomial CDF, ppois is a Poisson CDF, and dpois is a Poisson PDF.