Probability Distribution

Question

In cartesian co-ordinate system,along the x axis two points p and q are selected uniformly at random in $\left [ 0,L \right ]$ where L > 0.

What is the probability of $\text{distance(p,q)} \leq \frac{L}{4}$.

Answer 1

Let X axis of the graph represent the point p

and let Y axis of the graph represent the point q in the randomly chosen point [0,L].

2 condition to be satisfied by the chosen point 

$q-x\leq \frac{L}{4}; q-x\geq 0;$

$P(q-p)\leq \frac{L}{4}=\frac{shaded area in violet}{total area of square of side L}//=\frac{L^{2}-\frac{1}{2}*L*L-\frac{1}{2}*\frac{3L}{4}*\frac{3L}{4}}{L^{2}}//=1-\frac{1}{2}-\frac{9}{32}//=\frac{7}{32}$

Comments

distance(p,q) is always positive so if co-ordinate of p is ($x_1$,0) and q is ($x_2$,0) then, |$x_1$ - $x_2$|$\leq \frac{L}{4}$.

This condition provides us with two lines $x_1 - x_2 = \frac{L}{4}$ and $x_2 - x_1 = \frac{L}{4}$, answer is area between them. bounded by $0\leq x_1.x_2 \leq L$

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ok..ans is 7/32 if we are given point q is ahead of p

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yes ! true, I was about to mention that