# Solved (Free): The potholes on a major Chicago highway occur at an average rate of 3.4 per mile. Suppose one mile of highway is randomly selected

#### ByDr. Raju Chaudhari

Apr 4, 2021

The potholes on a major Chicago highway occur at an average rate of 3.4 per mile. Suppose one mile of highway is randomly selected.

(a) What is the probability of more than 1 pothole occurring?
(b) What is the probability of fewer than 3 potholes occurring?

#### Solution

Let $X$ denote the number of potholes on a one mile of highway.

The potholes on a major Chicago highway occur at an average rate of 3.4 per mile, i.e., $\lambda =3.4$.

Thus the mean number of potholes per one mileof highwayis $\lambda = 3.4$.

$X\sim P(3.4)$.

The probability mass function of $X$ is Poisson distribution with $\lambda =3.4$ is

 \begin{aligned} P(X=x) &= \frac{e^{-3.4}(3.4)^x}{x!},\\ &\quad x=0,1,2,\cdots \end{aligned}

(a) The probability of more than 1 pothole occurring is

 \begin{aligned} P(X> 1) &= 1- P(X\leq 1)\\ &= 1- \big(P(X=0)+P(X=1)\big)\\ &=1-\bigg(\frac{e^{-3.4}3.4^{0}}{0!}+\frac{e^{-3.4}3.4^{1}}{1!}\bigg)\\ &= 1-\big(0.0334+0.1135\big)\\ &= 1-0.1468\\ &= 0.8532. \end{aligned}

(b) The probability of fewer than 3 pothole occurring is

 \begin{aligned} P(X < 3) &= P(X\leq 2)\\ &=P(X=0)+P(X=1)+P(X=2)\\ &= \frac{e^{-3.4}3.4^{0}}{0!}+\frac{e^{-3.4}3.4^{1}}{1!}+\frac{e^{-3.4}3.4^{2}}{2!}\\ &= 0.0334+0.1135+0.1929\\ &= 0.3397. \end{aligned}