Previous studies show that there is an average of 0.2 breakdowns on each on kilometre of the Wellington motorway on any particular day.
If a towing service wants to be available to tow vehicle that has broken down, what length of the motor way would they have to patrol to have a probability of 0.75 of being close to at least one breakdown.
Confusion
Since we are given the average (0.2) intuitively it almost feels like I can do this
$$ \frac{0.75}{0.2} . 1 = 3.75km$$
as if we double the amount of kilometres then we double the probability. We can't and I don't understand why. Because it seems like a rectangular distribution where the probability is the same throughout each kilometre.
Another way to frame the question is to say that, on average, a towing service would need to patrol $5$ kilometers to see one breakdown. The random number of breakdowns $X$ observed in $k$ kilometers is modeled as a Poisson distribution with rate $\lambda = 0.2k$; that is to say, $$\Pr[X = x] = e^{-\lambda} \frac{\lambda^x}{x!}, \quad x = 0, 1, 2, \ldots.$$ So the probability of observing at least one breakdown in $k$ kilometers of patrolled road is $$\Pr[X \ge 1] = 1 - \Pr[X = 0] = 1 - e^{-\lambda} = 1 - e^{-0.2k}.$$ All that's left is to note that we want to find the smallest $k$ for which this probability is at least $0.75$; i.e., to solve $$1 - e^{-0.2k} = 0.75.$$
The reason why your solution doesn't work is because if $0.75$ were replaced with $1$, then it would mean that you would be guaranteed to find a breakdown within a $5$ kilometer patrol region. But this is not what the question stipulates. Rather, it simply says that on average, there is one breakdown in a given $5$ kilometer stretch of road.