Q. 4.55 - A First Course in Probability

Question

A certain typing agency employs $2$ typists. The average number of errors per article is $3$ when typed by the first typist and $4.2$ when typed by the second. If your article is equally likely to be typed by either typist, approximate the probability that it will have no errors.  

Step-by-Step Solution

Answer

The probability that the typist will have no errors is$0.0324.$

Step 1: Given Information

It is given that a typing agency employs $2$ typists. The average number of errors per article is$3.$

That is, ${\lambda }_1=3$ and ${\lambda }_2=4.2$

Step 2: Solution of the Problem

The average number of mistakes follows Poisson distribution. The probability mass function of Poisson distribution is,

$P(X=x)=\frac{{\mathrm{e}}^{-\lambda }(\lambda {)}^x}{x!} x=0,1,2,\dots$

Step 3: Calculation of the Value

The probability that the second typist write the test there are no mistakes hit is given by$P(X=0)=\frac{e^{-{\lambda }_1}{\left({\lambda }_2\right)}^0}{0!}$

$=\frac{e^{-4.2}4.2^0}{0!}$

We get,

$=e^{-4.2}$

$=0.015\left(\begin{array}{l}\text{ Using excel finction }\\ =\exp (-4.2)\end{array}\right)$

Step 4: Computation of Probability

It is given that the article is equally likely to be typed by either typist.

That $P\left(\text{ First typist }\right)=P\left(\text{ Second typist }\right)=\frac{1}{2}$

The probability that the typist will have no errors is,

$P\left(\text{ No errors }\right)=\left\{\begin{array}{l}P(\text{ no errors }\mid \text{ first typist })P\left(\text{ first typist }\right)+\\ P(\text{ no errors }\mid \text{ second typist })P\left(\text{ second typist }\right)\end{array}\right\}$

$=0.0498\times \frac{1}{2}+0.015\times \frac{1}{2}$

We get,

$=0.0324.$

Step 5: Final Answer

The probability that the typist will have no errors is $0.0324.$