Q.4.51 - A First Course in Probability

Question

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

Step-by-Step Solution

Answer

The probability that the next page you read contain zero is $0.8187$. The probability that the next page you read contain $2$ or more typographical errors is $0.0175.$

Step 1: Given Information (Part-a)

the probability that the next page you read contains zero. The expected number of typographical errors on a page of a certain magazine is $0.2.$

That is, $\lambda =0.2$

Let $X$ is a random variable representing the number of typos per page.

Since, we have a large number of letters per page, and the number of errors in one page will be small, we use Poisson distribution. The probability function of the Poisson distribution is defined as,

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

Here, $\lambda =$parameter.

Step 2: Solution of the Problem (Part-a)

Find the probability that the next page you read contains zero.

That is, find $P(X=0).$

$P(X=0)=\frac{e^{-\lambda }(\lambda {)}^k}{k!}$

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

We get $=e^{-0.2}$

$=0.8187$.

Step 3: Final Answer (Part-a)

The probability that the next page you read contain zero is $0.8187.$

Step 4: Given Information (Part-b)

Given the question that the probability that the next page you read contains $2$ more typographical errors. The expected number of typographical errors on a page of a certain magazine is $0.2.$

That is, $\lambda =0.2$

Let $X$is a random variable representing the number of typos per page.

Step 5: Solution of the problem (Part-b)

Find the probability that the next page you read contains $2$more typographical errors.

That is, find $P(X \ge 2).$

$P(X\ge 2)=1-P(X\le 1)$

$=1-\left(P\right(X=0)+P(X=1\left)\right)$

Step 6: Solution of the Problem (Part-b)

Insert the value into the equation

$=1-\left(e^{-0.2}+\frac{e^{-0.2}(0.2{)}^1}{1!}\right)$

We get,

$=1-1.2e^{-0.2}$

Find the value,

$=1-0.982477$

$=0.0175.$

Step 7: Final Answer (Part-b)

The probability that the next page you read contains $2$ or more typographical errors is $0.0175.$