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Question: Suppose 300 misprints are distributed randomly throughout a book of 500 pages. The probability that ...

Suppose 300 misprints are distributed randomly throughout a book of 500 pages. The probability that a given page contains at least one misprint is
A) 1.e−0.61.{e^{ - 0.6}}
B) 1−e−0.61 - {e^{ - 0.6}}
C) (0.6)e−0.6\left( {0.6} \right){e^{ - 0.6}}
D) (0.06)e−0.6\left( {0.06} \right){e^{ - 0.6}}

Explanation

Solution

First, find the average number of times that the event occurs λ\lambda . Then find the probability that the page contains no misprint by Poisson distribution probability P(x)=e−λλXX!P\left( x \right) = \dfrac{{{e^{ - \lambda }}{\lambda ^X}}}{{X!}}. After that for the probability that the page contains at least one misprint subtract the probability of the page containing no misprint from 1. The value obtained is the desired result.

Complete step-by-step answer:
If the average number of time that the event occurs is λ\lambda and X is the actual number of success, then the Poisson’s distribution probability is given by,
P(X)=e−λλXX!P\left( X \right) = \dfrac{{{e^{ - \lambda }}{\lambda ^X}}}{{X!}}
Given: - 300 misprints are distributed randomly throughout a book of 500 pages.
The average number of time that the event occurs is given by,
λ=number of misprintsnumber of pages\lambda = \dfrac{{{\text{number of misprints}}}}{{{\text{number of pages}}}}
Substitute the values of the number of misprints and pages in the above equation,
⇒λ=300500\Rightarrow \lambda = \dfrac{{300}}{{500}}
Cancel out the common factors from the numerator and denominator,
⇒λ=0.6\Rightarrow \lambda = 0.6
Now, the probability that no misprints occur in the page by Poisson distribution is,
⇒P(X=0)=e−0.6(0.6)00!\Rightarrow P\left( {X = 0} \right) = \dfrac{{{e^{ - 0.6}}{{\left( {0.6} \right)}^0}}}{{0!}}
Simplify the terms,
⇒P(X=0)=e−0.6\Rightarrow P\left( {X = 0} \right) = {e^{ - 0.6}}
Now, subtract the probability from 1 to get the probability that at least one misprints occur on the page,
⇒P(X⩾1)=1−P(X=0)\Rightarrow P\left( {X \geqslant 1} \right) = 1 - P\left( {X = 0} \right)
Substitute the values,
∴P(X⩾1)=1−e−0.6\therefore P\left( {X \geqslant 1} \right) = 1 - {e^{ - 0.6}}

Hence, option (B) is the correct answer.

Note: The Poisson’s distribution is the discrete probability distribution of the number of events occurring in a given period given the average number of times the event occurs over that time of period.
For instance, in a binomial distribution, if the number of trials gets larger and larger as the probability of success gets smaller and smaller, we obtain a Poisson’s distribution.