Probability - Book with typos problem 
In a book, $250$ printing errors are randomly and independently distributed on $500$ pages. What is the probability that there will be at least three printing errors on page $317$?

Is binomial distribution a good way to tackle this problem? My second idea was to use Poisson distribution to approximate the probability. What is the better approach? Any help would be much appreciated. 
 A: You can take either approach, as long as you set them up properly (for sufficiently low probabilities, binomial and Poisson distributions resemble each other quite well).
To treat it as a binomial (which will give you the exact answer), you can set $p$ as "the probability of this error appearing on page 317", so you now have 250 independent $Bernoulli(\frac{1}{500})$ trials, so the count of errors on page 317 is $X \sim B(250, \frac{1}{500})$ and you're looking for $P(X \geq 3)$. According to Wolfram Alpha, this probability is roughly 0.01427.
Alternatively, you can look at this as a Poisson distribution, with errors occurring at an average rate of 1/2 per page. If you do that, then $X \sim Poisson(\frac{1}{2})$ and $P(X \geq 3) \approx 0.1439$.
As you can see, there's a difference between the two of about 0.8%, which for most situations is close enough to be ignorable. The Poisson answer is an approximation, albeit a quite good one.
A: A Poisson Distribution is more appropriate for this because if you use a binomial distribution $n$ is the number of independent trials, and $p$ is the individual probability of success for that trial. Here, you are sampling only a single page out of 500 pages, and counting the number of errors in that page. Thus it should be a Poisson with error rate $\lambda = \frac{250}{500}$
$$P(Y \geq 3) = 1 - P(Y \leq 2) = 1 - \sum_{i=0}^2 exp\bigg(-\frac{1}{2}\bigg)\frac{\frac{1}{2^i}}{i!}$$
A: From my perspective, the idea of using binomial distribution can be possible.
For example, 
$$P(\text{only one error on page 317})={250 \choose 1}\times(\frac {1}{500})\times(\frac {499}{500})^{249}$$
Here's my calculation: 
Let X be the number of printing errors on page 317. $X\sim Bin(250,\frac {1}{500})$
$$P(X\geq 3)=1-P(X\lt3)=1-(P(X=0)+P(X=1)+P(X=2))$$
$$P(X\geq 3)=1-((\frac {499}{500})^{250}+{250 \choose 1}\times(\frac {1}{500})\times(\frac {499}{500})^{249}+{250 \choose 2}\times(\frac {1}{500})^{2}\times(\frac {499}{500})^{248})\approx 0.0143\text{ (corr. to 4 d.p.)}$$
