Filling station is being supplied with drinking water once a week. If its weekly volume of sales in thousands of gallons is R.V with $pdf$ given by PROBLEM: A filling station is being supplied with drinking water once a week. if its weekly volume of sales in thousands of gallons is random variable with pdf given by:
$f(x)= \begin{cases}
5(1-x)^{4}, \text {} 0<x<1 \\
\ 0, \text{} otherwise 
\end{cases}
$
What must the capacity of the tank be so that the probability of the supply's being exhausted in a given week is $0.01?$
MY WORKING:
As far as I have understood the problem is asking for value of $x$ where $P(X)=0.01$, but I am not sure. Can anyone guide me or give a hint?

You are looking for a value of $x$ that has a chance of $0.01$ of being exceeded.  You are given a pdf.  What you really want is a cdf, which is the integral of the pdf from $-\infty$ to $x$ of the pdf and represents the chance that the random variable is less than $x$.  You want the $x$ that makes the cdf equal to $0.99$
