X and Y - independent random variables, which are uniformly distributed on [0, 1]. How to find P{max(2*X,Y)−min(2*X,Y)<1/3}?
I've spent a lot of time searching information about possible ways of solution, but all of them did not give me the full picture of the technique. I suppose at first we need to find distribution function(df) of 2*X, then df of max(2*X, Y), then df of min(2*X, Y), and at last df of max-min of them. And then we could use the interval [0, 1/3] for calculation of the probability. But I don't understand how to make all calculations from the beginning to the end.
It can be computed geometrically.
The desired probability is the area of the subregion $D$ of the standard unit square in the $xy$-plane such that $$\max(2x,y)-\min(2x,y)<{\small{\frac{1}{3}}}$$ Consider two cases . . .
Case $(1)\;\colon\,y \ge 2x.\;$Then $$\max(2x,y)-\min(2x,y) = y - 2x$$ so we want the area of the subregion of the unit square between the lines $y = 2x$ and $y-2x = 1/3$. By elementary geometry, this area is $5/36$.
Case $(2)\;\colon\,y \le 2x.\;$Then
$$\max(2x,y)-\min(2x,y) = 2x - y$$
so we want the area of the subregion of the unit square between the lines $y = 2x$ and $2x-y = 1/3$. By elementary geometry, this area is $1/6$.
Hence the area of $D$ is
$$\frac{5}{36} + \frac{1}{6} = \frac{11}{36}$$
which is the desired probability.