Probability - Book with typos problem

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 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 $\lambda = \frac{250}{500}$

$$P(Y \geq 3) = 1 - P(Y \leq 2) = 1 - \sum_{i=0}^2 exp\bigg(-\frac{1}{2}\bigg)\frac{\frac{1}{2^i}}{i!}$$

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. $X\sim Bin(250,\frac {1}{500})$ $$P(X\geq 3)=1-P(X\lt3)=1-(P(X=0)+P(X=1)+P(X=2))$$ $$P(X\geq 3)=1-((\frac {499}{500})^{250}+{250 \choose 1}\times(\frac {1}{500})\times(\frac {499}{500})^{249}+{250 \choose 2}\times(\frac {1}{500})^{2}\times(\frac {499}{500})^{248})\approx 0.0143\text{ (corr. to 4 d.p.)}$$