Question

In a book of 500 pages, there are 500 misprinting errors, find the probability of at most three misprinting 3 pages selected at random the book.

Answer 1

Hint: This problem will be solved using Poisson distribution, with $\lambda ={\displaystyle \frac{500}{500}}$=1 . A random variable X is to follow a Poisson Distribution if it assumes only non-negative values and its probability mass function is given by: $P(X=x)={\displaystyle \frac{e^{-\lambda }{\lambda }^x}{x!}};x=0,1,2.....;\lambda >0$. Using this find the probability of at most three misprinting 3 pages is given by: $P(X\le 3)=P(X=0)+P(X=1)+P(X=2)+P(X=3)$.

Complete step by step answer:
It is given in the question that in a book 500 pages are there out of which 500 misprinting errors are there. So the average number of errors per page in the book is given by lambda, $\lambda ={\displaystyle \frac{500}{500}}$=1. We need to find the probability of at most three misprinting in three pages that is we have to calculate, $P(X\le 3)=P(X=0)+P(X=1)+P(X=2)+P(X=3)........(1)$
Hence using Poisson probability law, the probability of $x$ errors per page is given by:
$P(X=x)={\displaystyle \frac{e^{-\lambda }{\lambda }^x}{x!}}$
  $={\displaystyle \frac{e^{-(1)}{(1)}^x}{x!}};x=0,1,2,3$
$={\displaystyle \frac{e^{-1}}{x!}};x=0,1,2,3$ as ${(1)}^x=1$
Now calculating $P(X=0),P(X=1),P(X=2),P(X=3)$ separately by using the above expression and then substituting these values in equation (1) we will get the required probability.
$\begin{array}{rl}& P(X=0)={\displaystyle \frac{e^{-1}}{0!}}\\ & ={\displaystyle \frac{e^{-1}}{1}}\\ & =e^{-1}\end{array}$as factorial of zero is 1.
$=0.3678$(approximately)
Therefore, $P(X=0)=0.3678...........(2)$
Now we will calculate the value of $P(X=1)$
$\begin{array}{rl}& P(X=1)={\displaystyle \frac{e^{-1}}{1!}}\\ & =e^{-1}\end{array}$
$=0.3678$
Thus, $P(X=1)=0.3678...........(3)$
Now we will calculate the value of $P(X=2)$
$\begin{array}{rl}& P(X=2)={\displaystyle \frac{e^{-1}}{2!}}\\ & ={\displaystyle \frac{e^{-1}}{2}}\\ & ={\displaystyle \frac{0.3678}{2}}\\ & =0.1839\end{array}$
Thus we get $P(X=2)=0.1839...........(4)$
Now we will calculate the value of $P(X=3)$
$\begin{array}{rl}& P(X=3)={\displaystyle \frac{e^{-1}}{3!}}\\ & ={\displaystyle \frac{e^{-1}}{6}}\\ & ={\displaystyle \frac{0.3678}{6}}\\ & =0.0613\end{array}$
Thus we get $P(X=3)=0.0613...........(5)$
Putting the values of $P(X=0),P(X=1),P(X=2),P(X=3)$from equation (2), (3), (4) and (5) in equation (1) we get,
$\begin{array}{rl}& P(X\le 3)=0.3678+0.3678+0.1839+0.0613\\ & =0.9808\end{array}$
Therefore, $P(X\le 3)$= 0.9808. Hence, the probability of at most three misprinting in three pages is 0.9808 approximately.

Note:
A random variable X is to follow a Poisson Distribution if it assumes only non-negative values and its probability mass function is given by: $P(X=x)={\displaystyle \frac{e^{-\lambda }{\lambda }^x}{x!}};x=0,1,2.....;\lambda >0$. In Poisson distribution $\lambda >0$ and in this question $\lambda ={\displaystyle \frac{500}{500}}$=1 which is greater than 0, that is why we have used Poisson distribution.