Question

Potholes on a highway can be a serious problem and are in constant need of repair. With a particular type of terrain and make of concrete, past experience suggests that there are, on the average, 2 potholes per mile after a certain amount of usage. It is assumed that the Poisson process applies to the random variable "number of potholes" (a) What is the probability that no more that one pothole will appear in a section of one mile? (b)What is the probability that no more that 4 potholes will occur in a given section of 5 miles?

Answer 1

Answer:

If a Poisson process applies, we have the probability that k potholes are find in t miles as:

$P(x=k)=\frac{(rt)^ke^{-rt}}{k!}$

The probability that no more that one pothole will appear in a section of one mile is P=0.1353.

The probability that no more that 4 potholes will occur in a given section of 5 miles is P=0.0293.

Step-by-step explanation:

We have the rate of potholes:

$r=2\,pothole/mile$

If a Poisson process applies, we have the probability that k potholes are find in t miles as:

$P(x=k)=\frac{(rt)^ke^{-rt}}{k!}$

The probability that no more that one pothole will appear in a section of one mile is P=0.1353:

$t=1\\\\P(x<1)=P(0)=\frac{(2*1)^0e^{-2*1}}{0!}=\frac{1*e^{-2}}{1}= 0.1353$

The probability that no more that 4 potholes will occur in a given section of 5 miles is P=0.0293:

$t=5\\\\P(x\leq4)=P(0)+P(1)+P(2)+P(4)\\\\P(0)=\frac{(2*5)^0e^{-2*5}}{0!}= 0.000045 \\\\P(1)= \frac{(2*5)^1e^{-2*5}}{1!}= 0.000454 \\\\ P(2)= \frac{(2*5)^2e^{-2*5}}{2!}= 0.002270 \\\\P(3)= \frac{(2*5)^3e^{-2*5}}{3!}= 0.007567 \\\\P(4)= \frac{(2*5)^4e^{-2*5}}{4!}= 0.018917 \\\\\\P(x\leq4)=P(0)+P(1)+P(2)+P(4)\\\\P(x\leq4)=0.000045+0.000454+0.00227+0.007567+0.018917\\\\P(x\leq 4)= 0.029253$

Answer 2

Following are the solution to the given points:

Assuming that X denotes the number of potholes

Given:

$\to \text{per 1 mile} \ \lambda = 2 \\\\\to \text{per 5 mile} \ \lambda= (2)(5) = 10$

So, the pmf of X:

$\to P(X=x) =\frac{e^{-2} (2)^2}{x!}\ \ \ \ \ x=0,1,2,... per 1 mile$

For point a)  

The probability having only one pothole occurring in a one-mile stretch is low:

$\to P(X \leq 1)= P(X = 0)+P(X =1)$

                    $=\frac{e^{-2} (2)^0}{0!}+ \frac{e^{-2} (2)^1}{1!} \\\\ =0.135335+0.270671\\\\ =0.406006$

For point b)

The probability with no more than four potholes occurring in a five-mile stretch of road is: $\to P (X \leq 4) = P(X =0)+P(X =1)+P(X =2)+P(X = 3)+P(X =4)$

                     $= \frac{e^{-10} (10)^0}{0!}+ \frac{e^{-10} (10)^1}{1!}+ \frac{e^{-10} (10)^2}{2!}+ \frac{e^{-10} (10)^3}{3!}+ \frac{e^{-10} (10)^4}{4!} \\\\= 0.00045+0.000454+0.00227+0.007567+0.018917\\\\ =0.029253$

Therefore, the final answer is "0.406006 and 0.029253"

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