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: $$|y-x|>|4-y|$$$$\frac{x+4}{2}
My question is: If I switch the above starting formula to $$|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?
Probability of distance between two points in an interval
0
$\begingroup$
combinatorics
graphing-functions
1 Answers
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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}