A publisher assumes that when printing a typical book, the probability for a misprint on a page is $1.25$%.
I'm trying to find out these two things:
- What is the probability that on a book with $200$ pages three misprints are made at most?
- How can one calculate the above using the Poisson limit theorem?
I think the probability that each misprint appears on a given page is $p = \frac{1}{n}$.
Exactly three would be $\binom{m}{3}p^3(1-p)^{m-3}$.
$$\binom{200}{3}(\frac{1}{200})^3 (1-\frac{1}{200})^{200-3} = 0.0612 = 6,12 \text{%}$$
Is that correct?
I know that the poisson limit theorem is defined like this:
$$\lim_{n \to \infty} \binom{n}{k}p^k_n(1-p_n)^{n-k} = e^{-\lambda} \frac{\lambda^k}{k!}$$ where $p_n$ is a sequence of real numbers in $[0,1]$ so that the sequence $np_n$ converges to a finite limit $\lambda$.
I tried
$$P(X \leq 3) = \frac{e^{-1}1^0}{0!} + \frac{e^{-1}1^1}{1!} + \frac{e^{-1}1^2}{2!} + \frac{e^{-1}1^3}{3!}$$
but this doesn't give the answer as above.
Any help is appreciated!
This might be a useful approach to consider, based on my reading of the information:
The probability of a page containing a misprint is given as 1.25%. I interpret this to imply $p=0.0125$.
To determing the probability of having exactly 3 misprinted pages out of 200, using the binomial distribution, one then has:
$$ P(k=3)=\binom{200}{3} p^{3}(1-p)^{\left(200-3\right)} $$
$$ \binom{200}{3} = \frac{200!}{3!(200-3)!} = \frac{200 \times 199 \times 198}{6}=1313400 $$
$$ P(k=3)=\binom{200}{3} p^{3}(1-p)^{\left(200-3\right)} = 1313400 \times 0.0125^{3} \times (1-0.0125)^{197} = 0.215246 $$
Comparing to the Poisson distribution,
$$ P(k=3)=\frac{\lambda^{k}e^{-\lambda}}{k!} $$
let $\lambda=pn= 0.0125 \times 200=2.5$, we have
$$ P(k=3)=\frac{2.5^{3}e^{-2.5}}{6}=0.2137630 $$
To compute the "at most 3", I believe it requires adding the contributions of $k=0,k=1,k=2$ to the one for $k=3$.
I hope this helps.