Find the probability that a pump would have cost more to rent than to buy We have a continuous random variable $X$ modelled by $f(x)=\frac{k}x$.   For $1<x<9$ this is the length of time in years that water pumps last for. 
(Therefore $k$ can be solved to get $k=\frac{1}{2\ln3}$)
Then there are two very difficult concepts that come with this question. 
First part: The farmer is offered a guarantee to cover the cost of replacing a pump that fails during the second year, at a cost of $300$. Given the pump will cost $1000$ to replace if it fails during the year, what advice would you give as to the offer?
Second part: pumps can be rented for an installation charge of $200$ plus $250$ per year payable in advance. The yearly payment is not refundable if the pump fails before the end of the year. The farmer does not purchase the guarantee. Find the probability that the pump, at the end of its life, would have cost more to rent than to buy for $1000$. 
To be honest I don't even know where to start. In fact I made this account because nobody I talk to knows where to start. I hope that if somebody can explain this to me then I will be able to have a much greater understanding of the concepts underpinning this question 

For the first part, he can accept a loss of $300$ or take his chances on losing $1000$.  Buying the insurance will be a good deal if the probability of the pump failing is greater than $0.3$.  Integrate the probability distribution from $1$ to $2$ to get the chance it fails during the second year.  For the second part, if it fails during the second year you pay $700$.  If it fails during the third year you pay $950$.  If it fails after the third year, you pay more than $1000$ to rent it.  Compute the chance it lasts more than three years.
