91Ó°ÊÓ

The expected number of typographical errors is $0.2$.

Such that np$=0.2$

The number of letters is assumed to be very, very high.

Since the distribution of the number of errors is Binomial,

With parameters n and p.

The expected number,

np=0.2

But n is unknown. Thus, p cannot be determined. Then poisson approximation can be used. With parameter

$\mathrm{λ}=np=0.2$

Let Y be the number of errors having approx. Pois (0.2) distribution.

$P(Y=k)=\frac{{\mathrm{λ}}^k{e}^{-\mathrm{λ}}}{k!}$

Thus,

$P(Y=0)=\frac{0.2^0{e}^{-0.2}}{0!}=e^{-0.2}=0.82$

The expected number of typographical errors is $0.2$.

Such that np $=0.2$

The number of letters is assumed to be very, very high.

Since the distribution of the number of errors is Binomial,

With parameters n and p.

The expected number, np$=0.2$

But n is unknown. Thus, p cannot be determined. Then Poisson approximation can be used. With parameter

$\mathrm{λ}=np=0.2$

Let Y be the number of errors having approx. Pois ($0.2$) distribution.

$P(Y=k)=\frac{{\mathrm{λ}}^k{e}^{-\mathrm{λ}}}{k!}$

Thus,

$$\begin{gathered} P\left(Yâ‰\yen 2\right)=1-P\left(Y=0\right)-P\left(Y=1\right) \\ =1-e^{-\mathrm{λ}}-\mathrm{λ}{e}^{-\mathrm{λ}} \\ =1-e^{-0.2}-0.2e^{-0.2} \\ =1-0.82-0.164 \\ =0.016 \end{gathered}$$