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During an article revision the authors found, in average, 1.6 errors by page. Assuming the errors happen randomly following a Poisson process, what is the probability of finding 5 errors in 3 consecutive pages?

Please explain your methodology, as the main purpose of the question is not getting the answer but the "how to". Please consider this as homework if you will.


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@Queops, are the three consecutive pages assumed to be predetermined or is it the probability of finding errors on any set of three consecutive pages out of a total of $n$? –  cardinal Feb 7 '11 at 2:00
I realize this question will be answered by someone wanting to rack up points, but I want to point out that answering trivial homework question dilutes the usefulness of this forum to professionals. –  Leo Alekseyev Feb 7 '11 at 3:00
@Leo, I, for one, am not looking to "rack up" points. I am interested in learning from others and also good pedagogy. My understanding is that this forum's target audience is fairly broad and students are welcome. Probability and statistics can both seem daunting at first. That said, when looking at homework questions, it's always nice to see that the OPs have already spent some time on their own thinking about it. –  cardinal Feb 7 '11 at 3:19
@cardinal, I don't know what you mean by predetermined, but the problem indicates it's not some specific set of them, just the probability of finding 5 errors in 3 consecutive pages. @Leo, are you suggesting there isn't space for everyone? Sure most professionals probably can figure out this by themselves but we have to start somewhere right? I for one, am very new to this and I find it hard to follow most articles on the internet and I find this community most helpful and friendly. –  Queops Feb 7 '11 at 3:23
@Queops, I think the problem probably means given a set of three prespecified pages, e.g., pages 13-15, what's the probability of five errors on these three pages. This is very different from the asking, e.g., what's the probability of finding 5 errors on some set of three consecutive pages between, say, pages 1--50. See the difference? –  cardinal Feb 7 '11 at 3:27
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2 Answers

up vote 8 down vote accepted

The two most important characteristics of a Poisson process with rate $\lambda$ are

  1. For any interval $(s, t)$, the number of arrivals within the interval follows a Poisson distribution with mean $\lambda (t-s)$.
  2. The number of arrivals in disjoint intervals are independent of one another. So, if $s_1 < t_1 < s_2 < t_2$, then the number of arrivals in $(s_1, t_1)$ and $(s_2, t_2)$ are independent of one another (and have means of $\lambda (t_1 - s_1)$ and $\lambda (t_2 - s_2)$, respectively).

For this problem let "time" be denoted in "pages". And so the Poisson process has rate $\lambda = 1.6 \text{ errors/page}$. Suppose we are interested in the probability that there are $x$ errors in three (prespecified!) pages. Call the random variable corresponding to the number of errors $X$. Then, $X$ has a Poisson distribution with mean $\lambda_3 = 3 \lambda = 3 \cdot 1.6 = 4.8$. And so

$$ \Pr(\text{$x$ errors in three pages}) = \Pr(X = x) = \frac{e^{-\lambda_3} \lambda_3^x}{x!}, $$ so, for $x = 5$, we get $$ \Pr(X = 5) = \frac{e^{-\lambda_3} \lambda_3^5}{5!} = \frac{e^{-4.8} 4.8^5}{5!} \approx 0.175 $$

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I understand now! You were most helpful. Much appreciated. I believe I'm ready for some real "Poissonic" action hehe. –  Queops Feb 7 '11 at 3:38
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If I were his instructor I'd prefer the following explanation, which obviously is equivalent to that given by @cardinal.

Let $N_t$ be the Poisson process of counting the number of errors on consequtive pages, of rate $\lambda=1.6/page$. One supposes that we observe the process at integer moments t (here t being assimilated with number of pages). Because $P(N_t=k)=e^{-\lambda t} (\lambda t)^k/k!$, we have $P(N_3=5)=e^{-(3*1.6)}(4.8)^5/5!$.

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Small correction: 1.6^5 should be (3*1.6) ^5. Thanks! –  Queops Feb 7 '11 at 14:18
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