Probability of distance between two points in an interval

Question

On interval [-2,4] we choose random numbers x and y.

a)What is the probability, that distance between x and y is greater then distance between y and number 4.

I did this: $$|y-x|>|4-y|$$$$\frac{x+4}{2}<y$$ I draw a line and 'paint' the surface above the line, in range [-2,4] for both x and y in the coordinate system. The areas's surface is 9, surface of the whole area between [-2,4] for both x and y is 36. So, the probability is: $$\frac{9}{36} = \frac{1}{4}$$ which is a correct answer.

My question is: If I switch the above starting formula to $$|y-x|>|y-4|$$ I will get $$\frac{x+4}{2}>y$$ and the end result will be area under the graph. That would give me a wrong answer of $$\frac{3}{4}$$ How to approach this kind of problem?

Answer 1

Regardless of whether you use $|4-y|$ or $|y-4|$ you should get the same $x+4 < 2y$, so you did something wrong in between.

\begin{align} |y-x| &> |y-4|\\ |y-x| &> 4-y & \text{you know $y \le 4$}\\ x > y + (4-y) & \text{ or } x < y - (4-y)\\ x &< 2y - 4 & \text{the first case is impossible} \end{align}