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Question: In a book of \(500\) pages, it is found that there are \(250\) typing errors. Assume that Poisson la...

In a book of 500500 pages, it is found that there are 250250 typing errors. Assume that Poisson law holds for the number of errors per page. Then, the probability that a random sample of 22 pages will contain no error, is
A. e−0.3{e^{ - 0.3}}
B. e−0.5{e^{ - 0.5}}
C. e−1{e^{ - 1}}
D. e−2{e^{ - 2}}

Explanation

Solution

Here in this question, it is given itself that it is a question of probability and we have to solve this by Poisson’s method. We have a particular formula to solve this type of question, we just have to put the values of different variables in the formula and then we get our answer.

Formula used:
P(X=r)=e−λnλrr!P\left( {X = r} \right) = \dfrac{{{e^{ - \dfrac{\lambda }{n}}}{\lambda ^r}}}{{r!}}
Where, r=number of errorsr = \text{number of errors}, λ = Total number of pageserror\lambda \, = \,\dfrac{\text{Total number of pages}}{\text{error}} and n =samplesn\, = \text{samples}.

Complete step by step answer:
In the given question, we have to find the probability for a sample of 22 pages that contains no error. So, we will use a formula to find the probability, that is,
P(X=r)=e−λnλrr!P\left( {X = r} \right) = \dfrac{{{e^{ - \dfrac{\lambda }{n}}}{\lambda ^r}}}{{r!}}
From given question, we know that
number of errors=0 Total number of pages=500\text{number of errors} = 0\,\text{Total number of pages} = 500
Errors=250,   samples=2Errors = 250,\,\,\,samples = 2
Therefore,
r=0   ,  λ=500250=2  ,  n=2r = 0\,\,\,,\,\,\lambda = \dfrac{{500}}{{250}} = 2\,\,,\,\,n = 2

Now, putting the values in formula
P(X=0)=e−22λ00!P\left( {X = 0} \right) = \dfrac{{{e^{\dfrac{{ - 2}}{2}}}{\lambda ^0}}}{{0!}}
⇒P(X=0)=e−1λ00!\Rightarrow P\left( {X = 0} \right) = \dfrac{{{e^{ - 1}}{\lambda ^0}}}{{0!}}
We know that if power of any positive number is zero then its value would be 1.1.Also, we know that the value of 0!  is  1.0!\,\,is\,\,1.So, putting these values in the formula.
⇒P(X=0)=e−1(1)(1)\Rightarrow P\left( {X = 0} \right) = \dfrac{{{e^{ - 1}}\left( 1 \right)}}{{\left( 1 \right)}}
∴P(X=0)=e−1\therefore P\left( {X = 0} \right) = {e^{ - 1}}
Therefore, the required probability is e−1{e^{ - 1}}.

Hence, the correct option is (C)\left( C \right).

Note: A Poisson distribution is a probability distribution that can be used to show how many times an event is likely to occur within a specified period of time. In other words, it is a count distribution. For example, suppose that from historical data, we know that earthquakes occur in a certain area with a rate of 22 per month. Other than this information, the timings of earthquakes seem to be completely random. Thus, we can conclude that the Poisson process might be a very good model for earthquakes.