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This is one of the question asked in my textbook I am able to solve this using Poisson distribution but can't solve it using exponential distribution The distance between major cracks in a highway follows an exponential distribution with a mean of five miles. (a) What is the probability that there are no major cracks in a 10-mile stretch of the highway? (b) What is the probability that there are two major cracks in a 10-mile stretch of the highway?

So what I tried is for the first question :

Since the mean 5 miles/per crack so λ=0.2crack/miles

So using this formula 1-e^(-λx)=(cumulative function of exponential distribution)

P(X<=10)=1-e^(-λx)=1-e^(-0.2*10)=0.864

which was obviously false. The solution used P(X>10) instead of P(X<=10) which I don't understand since they asked to find the probability IN a 10 miles radius.

For question 2, I'm lost, I don't understand where im suppose to plug the 2 major cracks. I used Poisson distribution but I was wondering if its possible to solve it using exponential distribution instead.

user788888
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  • Hint: ...that there are $\color{red}{\textrm{no}}$ major cracks in a 10-mile... Does it help? – callculus42 Oct 20 '20 at 03:35
  • no not really... im still confused , how is have no crack mean that we should find P(X>10 miles)? – user788888 Oct 24 '20 at 19:44
  • The pmf is $f(x)=e^{-0.2}\cdot \frac{0.2^x}{x!}$. That is right. There is no major crack in a 10-mile stretch, which means that there are no major crack in two consecutive 5-mile stretches. $e^{-0.2}\cdot \frac{0.2^0}{0!}\cdot e^{-0.2}\cdot \frac{0.2^0}{0!}=e^{-0.4}$ – callculus42 Oct 24 '20 at 20:39
  • I think I get it now thank you – user788888 Nov 08 '20 at 09:28

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