# Probability of a typesetter repeating a page

The number of mistakes made by a typesetter per page has an average of 4. If a page has more than 4 mistakes he has to repeat it.

1- What distribution follows the random variable "mistakes made per page"? I believe it is Poisson but what about binomial?

2- What is the probability of not repeating more than two pages if he writes 20? Here I calculated the probability of repeating a page using poisson, but then used binomial to calculate the probability of repeating 0,1 and 2 pages. Is it right?

3- If he writes a book of 1000 pages, what are the probabilities of repeating more than 100 pages? I am struggling a lot with this exercise, I get a probability of a 100% using gauss.

I will be very thankful to any help.

• Clearly it doesn't have to be a Poisson distribution... perhaps there are exactly $4$ errors on each page, or the pages alternate $0$ and $8$ errors, just to name two other possibilities. Poisson is likely to be somewhat more realistic in practice, but there's nothing sure about it. For number three...well, $100$ is awfully low. using a normal distribution I see the mean as about $371$ with $\sigma \approx 15.27746851$ so...
– lulu
Commented Jan 17, 2017 at 19:54

$1.$ For a realistic model, I think you are right to use Poisson. It might be modeled as binomial if there were $n = 40$ lines per page and $p = 0.1$ of making a mistake on any one line. That would give an average of 4 mistakes per page, and $Pois(\lambda = 4)$ is not 'much' different from $Binom(40, 0.1).$ (The main difference is that the binomial model does not envision more than one mistake per line.)

$2.$ Let $X$ be the number of pages that must be repeated out of 20. According to the Poisson model of (1), the probability of a repeat is $p = 0.3711,$ and $X \sim Binom(20, p)$ with $E(X) = 7.4233.$ You want $P(X \le 2) = 0.0074.$

    p = 1 - ppois(4, 4); p
## 0.3711631
20*p
## 7.423261

pbinom(2, 20, p)
## 0.007385354


$3.$ Here it seems you want $P(W > 100),$ where $W \sim Binom(1000, p),$ with the same $p$ as in (2). The exact binomial and (as you say) the normal approximation to binomial both give essentially $1$ as the answer. The expected number of bad pages in 1000 is about 371 with a SD of about 15.3, so it is not surprising that the actual number of bad pages will exceed 100.

1 - pbinom(100, 1000, p)
## 1

mu=1000*p; sg = sqrt(1000*p*(1-p)); mu; sg
## 371.1631
## 15.27747
1 - pnorm(100, mu, sg)
## 1


This guy needs to improve his typesetting to make fewer errors. And so do I, so please proofread everything here.

The figure below shows the distributions and approximations discussed above. (Bars for the Poisson distribution in the right-hand plot are too close together to show individually.)