Question

The expected number of typographical errors on a page of a certain magazine is $.2$. What is the probability that an article of $10$pages contains (a) $0$and (b) $2$or more typographical errors? Explain your reasoning!

Short answer

(a) Probability: P$(Y=0)=0.82$

(b) Probability:$P\left(Y\ge 2\right)=0.016$

Step-by-step solution

Given information (part a)

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.

Explanation (part a)

The expected number,

np=0.2

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

$\lambda =np=0.2$

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

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

Thus,

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

Given information (part b)

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.

Explanation (part b)

The expected number, np$=0.2$

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

$\lambda =np=0.2$

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

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

Thus,

$$\begin{gathered} P\left(Y\ge 2\right)=1-P\left(Y=0\right)-P\left(Y=1\right) \\ =1-e^{-\lambda }-\lambda e^{-\lambda } \\ =1-e^{-0.2}-0.2e^{-0.2} \\ =1-0.82-0.164 \\ =0.016 \end{gathered}$$