Setting
A book of 1000 pages contains 1000 misprints. Estimate the chances that a given page contains at least three misprints.
Solution
My solution is
$$\binom{1000}{1}\left(\frac{1}{1000}\right)^3\left(\frac{999}{1000}\right)^{1000 - 3}$$
Please confirm?
On
The usual model here is a Poisson model: the number $X$ of misprints in a randomly chosen page has roughly Poisson distrobution, mean (and therefore parameter) $\lambda=\frac{1000}{1000}$. We have $\Pr(X\ge 3)=1-\Pr(X\le 2)$.
And $\Pr(X\le 2)\approx e^{-1}\frac{1^0}{0!}+e^{-1}\frac{1^1}{1!}+e^{-1}\frac{1^2}{2!}$.
On
The number of ways 1000 mistakes can happen in 1000 pages is $1000^{1000}$. Assuming all of them are equally likely, we calculate the required probability as follows.
The number of ways given page can have at most 2 mistakes is $999^{1000} + 1000(999)^{999}+\frac{1000*999}{2}999^{998}$.
Therefore the required probability is $$1-\frac{999^{1000} + 1000(999)^{999}+\frac{1000*999}{2}999^{998}}{1000^{1000}}$$
On
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:
$$\begin{aligned} {\rm Pr}(X \ge 3)\ &= 1 - F_b(2; 1000, \tfrac1{1000}) \\ &\approx 1 - F_P(2; 1) \\ &= 1 - e^{-1}\sum_{i=0}^2 \frac{1}{i!} \\ &= 1 - \tfrac52e^{-1}. \\ \end{aligned}$$
No, your expression has a number of problems: First of all, even if you assume a binomial distribution applies to your situation, your expression only calculates the probability of exactly three errors, not at least three.
Second, the binomial coefficient is incorrect: it should be $\binom{1000}{3}$ if you were to apply such a formula.
Third--and this more or less makes the previous two comments irrelevant: a binomial distribution is not an appropriate model for this problem. Instead, a Poisson distribution is the intended approach. This is because if you use a binomial distribution, what is your $n$ and what is your $p$? $n$ 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 $1000$ pages, and counting the number of errors in that page. We cannot choose $n = 1000$ nor can we choose $p = 1/1000$. Such a choice of parameters would be equivalent to a situation where you were to look at every page in the book, in which each page has a probability of $1/1000$ of having an error, and getting the probability that $x$ pages with errors are found among those $n$ pages (but you would be counting pages, not errors, so in particular, such pages may contain more than one error on each!). That isn't what you want.
Instead, what we want to do is use a Poisson distribution: here, we have an error rate of $1000$ errors per $1000$ pages, or equivalently, $\lambda = 1$ error per page. Thus, if we look at a single page, the random number of errors found in that page is Poisson distributed with parameter $\lambda = 1$, and the probability of finding $k$ errors on that page is $$\Pr[X = k] = e^{-\lambda} \frac{\lambda^k}{k!}.$$ Thus the probability that there are at least $3$ errors is $$\Pr[X \ge 3] = 1 - \Pr[X \le 2] = 1 - \sum_{k=0}^2 e^{-\lambda} \frac{\lambda^k}{k!} = 1 - e^{-1} (1 + 1 + \tfrac{1}{2}).$$