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Books from a certain publisher contain on average 1 misprint per page. Suppose those misprints occur according to a Poisson scatter, with a rate on 1 per page. One of the books from this publisher has 200 pages.

The probability that a proofreader catches each misprint is 0.7 (independently for different misprints). If a proofreader catches 3 misprints on a page, what is the probability that there are still misprints that weren't caught on that page?

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  • $\begingroup$ I don't see how it would be possible to solve this without some model of how likely the proofreader is to catch each misprint. $\endgroup$ – Daniel Schepler Jul 9 '18 at 23:34
  • $\begingroup$ I would assume the proofreader would catch all of the misprints so it becomes a probability of their occurrence. $\endgroup$ – Phil H Jul 10 '18 at 0:26
  • $\begingroup$ I'm sorry. I've added the probability the proofreader is to catch each misprint. This question is actually part(d) of the question and I've already solve part (a) till (c) but I did not realized that part (c)'s statement is supposed to be used for part (d) too. My bad $\endgroup$ – ibuntu Jul 10 '18 at 1:17
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\begin{eqnarray*} \mathsf P(\text{$3$ misprints}\mid \text{$3$ misprints caught}) &=& \frac{\mathsf P(\text{$3$ misprints}\land \text{$3$ misprints caught})}{\mathsf P(\text{$3$ misprints caught})} \\ &=& \frac{\mathrm e^{-1}\cdot0.7^3/3!}{\sum_{k=3}^\infty\mathrm e^{-1}/k!\cdot\binom k3\cdot0.7^3\cdot0.3^{k-3}} \\ &=& \frac1{\sum_{k=3}^\infty0.3^{k-3}/(k-3)!} \\ &=& \mathrm e^{-0.3} \\ &\approx& 0.741\;, \end{eqnarray*}

so the probability of there being further misprints is about $25.9\%$.

Note that the number $3$ of misprints dropped out of the calculation; the result depends only on the proofreader's accuracy; so no matter how many misprints she catches, there's always a one in four chance that there are more. Thus, the conditional probability of having caught all mistakes, given the number of mistakes caught, is simply the unconditional probability of catching all mistakes; the events of catching all mistakes and catching a particular number of mistakes are independent. As the proofreader's accuracy varies between $1$ and $0$, the probability of catching all mistakes varies between $1$ and $\mathrm e^{-1}$ (where the latter is simply the probability of there being no mistakes to catch).

Another way to view this is as a modification of the Poisson process: The proofreader is screening the mistakes with a probability of $0.7$ and thereby producing a process of uncaught mistakes with reduced rate $0.3$, whose probability of producing no uncaught mistakes is $\mathrm e^{-0.3}$. The original process is split into two independent sub-processes, of caught and uncaught mistakes, with rates $0.7$ and $0.3$, respectively, and the number of events in the process of caught mistakes provides no information about the number of events in the process of uncaught mistakes.

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