91影视

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

That is, ${位}_1=3$and ${位}_2=4.2$

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

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

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

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

We get,

$=e^{-4.2}$

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

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,

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

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

We get,

$=0.0324.$

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