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.
Best Answer
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.