An urn has $50$ white and $450$ red balls. Take out $20$ balls with replacement. Find the exact probability that we observe $3$ white balls and the approximate probability via the Poisson distribution.
Here is how I computed the exact probability:
$$P(\text{we observe 3 white balls})= 1-P(\text{at most 2 white balls})$$ $$=1-\frac{\binom{450}{20}+\binom{50}{1}\binom{450}{19}+\binom{50}{2}\binom{450}{18}}{\binom{500}{20}}$$
For the approximate approximation I used $\lambda=50$ and let $X$ be the number of white balls observed. Then $$P(X=3)=\frac{50^3}{3!}e^{-50}.$$
However, these probabilities are not that close. Some help? Thank you!
Since the problem asks you to draw with replacement, then the number of white balls drawn in $20$ trials $N$ follows a binomial distribution with $n=20, p = 50/500 = 0.1$. Then the problem asks for the chance that $N = 3$, $$P(N = 3) = \binom{20}{3} p^3(1-p)^{20-3} \approx 0.190.$$
Since you are being told to use the Poisson approximation, then you should know that $\lambda = np$. Here, $\lambda = 20\cdot0.1 = 2$. So the approximation is $$P(N = 3)\approx e^{-\lambda}\frac{\lambda^3}{3!} \approx0.180.$$