Two points L and M are chosen at random on a segment AB of…

Two points $L$ and $M$ are chosen at random on a segment $AB$ of length $l$. Find the probability that $L$ is closer to $M$ than to $A$.

Step 1: The model

Put $A$ at $0$ and $B$ at $l$. Let $x=AL$, $y=AM$, independent and uniform on $[0,l]$: the point $(x,y)$ is uniform in the square of area $l^2$.

Step 2: The condition

$LM=|x-y|$ and $LA=x$. We need $|x-y|\lt x$:

• if $y\le x$: $x-y\lt x\iff y\gt0$ — always true;
• if $y\gt x$: $y-x\lt x\iff y\lt 2x$.

So the favourable region is $\{y\lt 2x\}$ — everything to the right of the line $y=2x$.

Step 3: The area

The unfavourable region $\{y\ge 2x\}$ is the triangle with vertices $(0,0)$, $(0,l)$, $(\frac l2,l)$, of area $\frac12\cdot\frac l2\cdot l = \frac{l^2}4$.

\[ \boxed{P = 1-\frac{l^2/4}{l^2} = \frac34.} \]

Step 4: Intuition

$L$ is farther from $M$ than from $A$ only when $M$ lies beyond the point $2x$, i.e. $M$ is to the right of $L$ at a distance greater than $AL$. That requires $L$ in the left half and $M$ far to the right, which happens in only a quarter of the cases.