Poisson Probability Distribution with big number A lot of $5000$ parts is received. The probability that a part is defective is $0.05$.
What is the probability that the total number of defective parts is exactly $200$?
I thought the solution will be
$$P(X=200) = e^{-0.05} \cdot \frac{ 0.05^{200} }{ 200!}$$
 A: The Binomial Distribution is a better fit here.
Recall that Poisson has support on all non-negative integers. Since the maximum value for $X$ is $5000$, we know that the Poisson distribution is (technically) inappropriate.

Binomial
The "most correct" approach, is to use the Binomial distribution with $n = 5000$ and $p=0.05$. Recall, the probability mass function for Binomial distribution is:
$$P(X=x\mid n,p) = \binom{n}{x}p^x(1-p)^{n-x}$$
For $x = 0, 1, 2, \ldots, n$. Hence you should simply plug in the appropriate values.

Normal approximation
For large values of $n$ (i.e. 5000), calculating this probability exactly may be difficult since binomial coefficients are hard to calculate with large numbers. Since $np > 10$, we may use the Normal approximation to the Binomial.
That is, $X$ is approximately equal in distribution to $Y \sim N(np, np(1-p))$. We can compute the approximate probability as:
$$P(X=200) \approx P(199.5 < Y < 200.5) = P(Y < 200.5) - P(199.5)$$
This answer is not exact, but will get you a pretty decent approximation.

Poisson Approximation
Another alternative is to use the Poisson Approximation to the Binomial (See Ross' answer). Most naive computations of the factorial will still have a problem computing $200!$.
With that said, the approximation appears to be better than the Normal approximation here. Of course, in my opinion if you're going to use mathematical software to compute your answer, you may as well stick with the Binomial distribution...

Computations with R
Binomial (The exact probability)
 dbinom(200, 5000, 0.05)
 [1] 0.0001029124

Normal Approximation (Can be computed without software. i.e. Normal table)
 pnorm(200.5, 5000*.05, sqrt(5000*.05*.95)) - pnorm(199.5, 5000*.05, sqrt(5000*.05*.95))
 [1] 0.0001342899

Poisson Approximation
 dpois(200, 5000*.05)
 [1] 0.0001310668


It also occurred to me, that it might be of interest to visualize the approximations. Note that the Normal Approximation is better than the Poisson near the mode of the distribution, but there is almost no difference between the two approximations in the tail (Poisson is slightly better).

A: The $\lambda$ in the Poisson distribution is the expected number, which here is $0.05 \cdot 5000=250$, so it should be $\dfrac {250^{200}e^{-250}}{200!}$
A: This is the sort of situation for which the normal approximation was first introduced (by Abraham de Moivre in the 18th century). I have up-voted the answer by Big Agness, but I want to give some detail here that was not in that answer.
The difficulty is that computing the exact answer arithmetically is prohibitively expensive in this kind of situation, and would gain nothing of practical import over the normal approximation.
We have
\begin{align}
\text{expected value} & = np = 5000 \times 0.5 = 250 \\
\text{variance} & = npq = 5000 \times 0.5 \times 0.95 = 237.5 \\
\text{so standard deviation} & = \sqrt{npq} \approx 15.411\ldots \\[12pt]
\Pr(199.5 < X < 200.5) & \approx \Pr\left( \frac{199.5-250}{15.411} < \frac{X-250}{15.411} < \frac{200.5 - 250}{15.411} \right) \\[10pt]
& \approx \Pr\left( -3.277 < Z < -3.212 \right) = \Phi(-3.212) - \Phi(-3.277).
\end{align}
You get those numbers from a table or from standard software packages (unless you want to work with numerical algorithms used in creating those tables or software).
Not surprisingly, you get a very small probability because $200$ is so far from the mean.
If the binomial distribution were not a better model than the Poisson distirbution in this case, you would use the fact that the expected value and variance of $\operatorname{Poisson}(\lambda)$ are both $\lambda,$ and the standard normal distribution would approximate the distribution of
$$
\frac{X-\lambda}{\sqrt{\lambda}}.
$$
The Poisson approximation to the binomial should be used when $\lambda$ is small and $n$ is big.
