Using the Poisson approximation to calculate probability This question was given to me as a review for an upcoming exam.
When a jury clerk is typing the dialogue in a court room, they are expected to make 3 errors in their document. Use Poisson approximation to calculate the probability that a given document has 5 errors. Binomial distribution can be assumed.
What I've attempted thus far:
$X = $number of errors on a document
$np = \langle X\rangle = 3$
$\lambda = np = 3$
$\frac{e^{-\lambda}\lambda ^k}{k!}= \frac{e^{-3}3^5}{5!} = 0.10082$ or 10.082%
Am I missing something? This type of question felt like it had longer steps involved when I was working on the assignments throughout the past weeks.
 A: One question is whether 'has 5 errors' means 'exactly 5' or 'at least 5'.
Poisson model. As you say, if $X \sim \mathsf{Pois}(\lambda = 3),$ then $P(X = 5) =  0.1008.$ But $P(X \ge 5) = 1 - P(X \le 4) = 0.1847.$ [Computations in R, where dpois is a Poisson PDF and ppois is a Poisson CDF.]
dpois(5, 3);  1 - ppois(4, 3)
## 0.1008188
## 0.1847368

Possible normal approximation. For large or moderate $\lambda,$ the distribution $\mathsf{Norm}(\mu = \lambda,\, \sigma=\sqrt{\lambda})$ 
might be used as an approximation to $\mathsf{Pois}(\lambda),$ but $\lambda = 3$ is a little too small.
If $X^\prime \sim \mathsf{Norm}(3, \sqrt{3}),$ then
$P(X^\prime > 4.5) = 0.1932,$ which is not a horrible
approximation of $P(X \ge 5) = 0.1957.$ 
Since you are reviewing for an exam, you might practice the normal approximation for larger values of $\lambda,$ perhaps using standardization and printed normal tables. 
The figure shows PDFs of both the Poisson distribution and (roughly) approximating normal distribution
for the question at hand.
1 - pnorm(4.5, 3, sqrt(3))
0.1932381


Speculative binomial model. I do nut understand why the Binomial distribution is mentioned. Under some circumstances, a binomial probability can be approximated by using a Poisson distribution with $\lambda = np,$  but I don't see
a value of $n$ in the statement of your Question.
If somehow $Y \sim \mathsf{Binom}(n = 1000, p = 0.003),$ then $P(Y \ge 5) = 0.1845 \approx 0.1957$ (from the Poisson distribution). 
This might be a reasonable model if you imagine the document has $n = 1000$ opportunities for errors, each with probability $p = 0.003,$ giving the binomial mean $\mu = np = 3.$ 
In the figure, the open blue dots show the relevant binomial probabilities.
[At the resolution of the figure, about 2 decimal places, it is difficult to
distinguish between the Poisson and binomial models.]
1 - pbinom(4, 1000, 0.003)
## 0.184484


