# Find the probability of distance of two points, which are selected in [0,a] closed interval, is less than ka k<1

A geometric probability question
Find the probability of distance of two points ,which are selected in [0,a] closed interval, is less than ka $k<1$.
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Elliott Rollins
Step 1

The region of $|y-x|\le ka$ looks similar to the picture that I have attached. We just have to use the total area to subtract away the area of the area of the two triangles.
Step 2
$\frac{{a}^{2}-2\cdot \frac{1}{2}\left(a-ka{\right)}^{2}}{{a}^{2}}=1-\left(1-k{\right)}^{2}=k\left(2-k\right)$
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solvarmedw
Step 1
We need to find the measure of A, which is given by $\begin{array}{r}{\int }_{A}dxdy={\int }_{0}^{a}{\int }_{max\left(y-ka,0\right)}^{min\left(y+ka,a\right)}dxdy={\int }_{0}^{a}min\left(y+ka,a\right)-max\left(y-ka,0\right)\phantom{\rule{thinmathspace}{0ex}}dy.\end{array}$.
Step 2
Note that $\begin{array}{r}{\int }_{0}^{a}min\left(y+ka,a\right)\phantom{\rule{thinmathspace}{0ex}}dy={\int }_{0}^{a-ka}y+ka\phantom{\rule{thinmathspace}{0ex}}dy+{\int }_{a-ka}^{a}a\phantom{\rule{thinmathspace}{0ex}}dy.\end{array}$