Assuming that the misprints are independently distributed across the book, and equally likely to be on any page, the number of misprints $X$ on a given page is binomially distributed with parameters $n = 1000$ ($=$ number of misprints) and $p = 1/1000$ ($=$ probability of a misprint ending up on the given page).
Thus, the probability of there being at least $k$ misprints on the page can be calculated using the cumulative distribution function $F_b$ of the binomial distribution:
$$\begin{aligned}
{\rm Pr}(X \ge k)\
&= 1 - {\rm Pr}(X \le k-1) \\
&= 1 - F_b(k-1; n, p) \\
&= 1 - \sum_{i=0}^{k-1} {n \choose i}\, p^i (1-p)^{n-i}, \\
\end{aligned}$$
and thus:
$$\begin{aligned}
{\rm Pr}(X \ge 3)\
&= 1 - F_b(2; 1000, \tfrac1{1000}) \\
&= 1 - \sum_{i=0}^{2} {\textstyle {1000 \choose i}}\, \left(\tfrac{1}{1000}\right)^i \left(\tfrac{999}{1000}\right)^{1000-i}, \\
&= 1
- \left(\tfrac{999}{1000}\right)^{1000}
- \left(\tfrac{999}{1000}\right)^{999}
- \tfrac12 \left(\tfrac{999}{1000}\right)^{999}.
\end{aligned}$$
We can approximate this probability by noting that, for $x \approx 0$, $(1-x)^n \approx e^{-nx}$, and thus:
$$\begin{aligned}
{\rm Pr}(X \ge 3)\
&\approx 1 - e^{-1} - e^{-\frac{999}{1000}} - \tfrac12 e^{-\frac{999}{1000}} \\
&\approx 1 - \tfrac52 e^{-1} \approx 0.08.
\end{aligned}$$
Another way to obtain the same approximation is to note that, for large $n$ and small $p$, the binomial distribution is well approximated by the Poisson distribution with rate parameter $\lambda = np$. The Poisson distribution has the CDF:
$$F_P(k; \lambda) = e^{-\lambda}\sum_{i=0}^k \frac{\lambda^i}{i!},$$
and thus we get:
$$\begin{aligned}
{\rm Pr}(X \ge 3)\
&= 1 - F_b(2; 1000, \tfrac1{1000}) \\
&\approx 1 - F_P(2; 1) \\
&= 1 - e^{-1}\sum_{i=0}^2 \frac{1}{i!} \\
&= 1 - \tfrac52e^{-1}. \\
\end{aligned}$$
\begin{eqnarray*}
\mathsf P(\text{$3$ misprints}\mid \text{$3$ misprints caught})
&=&
\frac{\mathsf P(\text{$3$ misprints}\land \text{$3$ misprints caught})}{\mathsf P(\text{$3$ misprints caught})}
\\
&=&
\frac{\mathrm e^{-1}\cdot0.7^3/3!}{\sum_{k=3}^\infty\mathrm e^{-1}/k!\cdot\binom k3\cdot0.7^3\cdot0.3^{k-3}}
\\
&=&
\frac1{\sum_{k=3}^\infty0.3^{k-3}/(k-3)!}
\\
&=&
\mathrm e^{-0.3}
\\
&\approx&
0.741\;,
\end{eqnarray*}
so the probability of there being further misprints is about $25.9\%$.
Note that the number $3$ of misprints dropped out of the calculation; the result depends only on the proofreader's accuracy; so no matter how many misprints she catches, there's always a one in four chance that there are more. Thus, the conditional probability of having caught all mistakes, given the number of mistakes caught, is simply the unconditional probability of catching all mistakes; the events of catching all mistakes and catching a particular number of mistakes are independent. As the proofreader's accuracy varies between $1$ and $0$, the probability of catching all mistakes varies between $1$ and $\mathrm e^{-1}$ (where the latter is simply the probability of there being no mistakes to catch).
Another way to view this is as a modification of the Poisson process: The proofreader is screening the mistakes with a probability of $0.7$ and thereby producing a process of uncaught mistakes with reduced rate $0.3$, whose probability of producing no uncaught mistakes is $\mathrm e^{-0.3}$. The original process is split into two independent sub-processes, of caught and uncaught mistakes, with rates $0.7$ and $0.3$, respectively, and the number of events in the process of caught mistakes provides no information about the number of events in the process of uncaught mistakes.
Best Answer
Since the probability of one page having at least $2$ misprints is $\frac{7}{400}$, the probability of a page having less than $2$ misprints is $\frac{393}{400}$. Then, the probability of all $250$ pages having less than $2$ misprints is $\big(\frac{393}{400}\big)^{250}$, so the probability of at least one page having at least $2$ misprints is $1-\big(\frac{393}{400}\big)^{250}\approx\boxed{0.98789}$