Suppose the number of spelling error on any given page in particular book can be modeled by a Poisson distribution with $\lambda=2$, and assume that the number of errors on different pages is independent. Find the probability that exactly three of the first ten pages have exactly one spelling errors.
Proof: we know that $$p(x)=\frac{\lambda^x}{x!} e^{-\lambda}$$ thus, the solution is $$p(3)=\frac{2^3}{3!}e^{-2}$$
Is this correct?
On
You calculated the probability that one page has $3$ errors.
The trials are independent so you need to calculate the probability that one page has one error. Then cube it in order to account for three such pages. Finally, you need to take care of the other seven pages. Multiply by the seventh power of the probability of not having one error. That is $(1-p(1))^7$. Now you have $p(1)^3(1-p(1))^7$, which is the probability that any fixed three pages have exactly one error each and the other seven do not have exactly one.
However, you do not care about the order of occurrence of the mistakes. All you have left to do is calculate the number of favorable outcomes and multiply by it. Can you take it from here?
To justify this algorithm look at Bayes' theorem.
If the trials were not independent, Bayes' formula might have done the work.
No. $p(3)$ is the probability of one page containing $3$ errors, not the probability of three pages containing one error each.