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The number of cracks which are present in a part of an international road has an average value of 2 cracks per kilometer. 1)What is the probability that there are no cracks in a section of road length 5 km? 2)What is the probability that there is at least one flaw in a particular part of 1500 meters? 3)What is the expected number of kilometers to be covered until you find the first crack?

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The usual model is that the number $X$ of cracks in a randomly selected $1$ km stretch has Poisson distribution with parameter $\lambda=2$.

Then the number $Y$ of cracks in a $5$ km stretch has Poisson distribution with parameter $\lambda=(5)(2)$.

The number $W$ of cracks in a $1500$ metre stretch has Poisson distribution with parameter $\lambda=(1.5)(2)$.

This is enough to answer the first two questions. The probability that $Y=0$, by the usual formula for Poisson probabilities, is $e^{-(5)(2)}$.

The probability that $W\ge 1$ is $1$ minus the probability that $W=0$.

For the third question, it is a theorem that if "flaws" have Poisson distribution with parameter $\lambda$, then the "waiting time" between flaws has exponential distribution with parameter $\lambda$, and therefore expectation $\frac{1}{\lambda}$. Using the random variable $X$, we find that the expected length before the first flaw is $\frac{1}{2}$.

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  • $\begingroup$ If you want to use the version you sent, then $\vu=2$. And for Question 1 we use $t=5$, while for Question 2 we use $t=1.5$. We don't need the full formula, since for both questions we only need the probability that the random variable is $0$. $\endgroup$ – André Nicolas Mar 7 '14 at 23:14

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