clovnerie0q

2022-10-08

A geometric probability question

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

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

Elliott Rollins

Beginner2022-10-09Added 8 answers

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

$\frac{{a}^{2}-2\cdot \frac{1}{2}(a-ka{)}^{2}}{{a}^{2}}=1-(1-k{)}^{2}=k(2-k)$

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

$\frac{{a}^{2}-2\cdot \frac{1}{2}(a-ka{)}^{2}}{{a}^{2}}=1-(1-k{)}^{2}=k(2-k)$

solvarmedw

Beginner2022-10-10Added 3 answers

Step 1

We need to find the measure of A, which is given by $\begin{array}{r}{\int}_{A}dxdy={\int}_{0}^{a}{\int}_{max(y-ka,0)}^{min(y+ka,a)}dxdy={\int}_{0}^{a}min(y+ka,a)-max(y-ka,0)\phantom{\rule{thinmathspace}{0ex}}dy.\end{array}$.

Step 2

Note that $\begin{array}{r}{\int}_{0}^{a}min(y+ka,a)\phantom{\rule{thinmathspace}{0ex}}dy={\int}_{0}^{a-ka}y+ka\phantom{\rule{thinmathspace}{0ex}}dy+{\int}_{a-ka}^{a}a\phantom{\rule{thinmathspace}{0ex}}dy.\end{array}$

We need to find the measure of A, which is given by $\begin{array}{r}{\int}_{A}dxdy={\int}_{0}^{a}{\int}_{max(y-ka,0)}^{min(y+ka,a)}dxdy={\int}_{0}^{a}min(y+ka,a)-max(y-ka,0)\phantom{\rule{thinmathspace}{0ex}}dy.\end{array}$.

Step 2

Note that $\begin{array}{r}{\int}_{0}^{a}min(y+ka,a)\phantom{\rule{thinmathspace}{0ex}}dy={\int}_{0}^{a-ka}y+ka\phantom{\rule{thinmathspace}{0ex}}dy+{\int}_{a-ka}^{a}a\phantom{\rule{thinmathspace}{0ex}}dy.\end{array}$

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