A geometric probability question

Question

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

What did I write :

$P(A)$ = (Area measure of set $A$)/(Area measure of set $\Omega$)

The sample space $\Omega = \{(x,y) : 0\le x \le a , 0 \le y \le a \}$

Probability set $A=\{(x,y) : |x-y|\lt ka\}$

Measure of $\Omega$ = $a.a=a^2$

I think I should calculate the area of the rectangle between two lines $y=x+ka$ and $y=x-ka$ and bounded with points $(-a,-ka)$, $(a,ka)$, $(-ka,-a)$ and $(ka,a)$

But I cannot obtain the true answer which is $k(2-k)$

Could someone please fix me please?

Thanks a lot

Answer 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.

$$\frac{a^2 - 2\cdot \frac12 (a-ka)^2}{a^2}=1-(1-k)^2=k(2-k)$$

Remark about your attempt:

$x$ or $y$ won't take negative values.

Answer 2

We need to find the measure of $A$, which is given by \begin{align} \int_A dx dy = \int_0^a \int_{\max(y-ka,0)}^{min(y+ka,a)} dx dy = \int_0^a \min(y+ka,a) - \max(y-ka,0) \, dy. \end{align} Note that \begin{align} \int_0^a \min(y+ka,a) \, dy = \int_0^{a-ka} y+ka \, dy + \int_{a-ka}^a a \, dy. \end{align} Can you proceed?