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

Two points $M$ and $N$ are chosen at random on a segment $AB$ of length $l$. Determine the probability that none of the three segments obtained is longer than $a$, where $l\gt a\gt\frac l3$.

Step 1: The model

The points are at $x,y$, independent and uniform on $[0,l]$, so $(x,y)$ is uniform in the square $[0,l]^2$. Let $D_1,D_2,D_3$ be the left, middle and right parts (in order along the segment).

Step 2: One part longer than $a$

Take $x\lt y$ (the other half of the square is symmetric).

• $D_1=x\gt a$: then $a\lt x\lt y\lt l$ — a triangle with legs $l-a$, area $\frac12(l-a)^2$.
• $D_3=l-y\gt a$: $0\lt x\lt y\lt l-a$ — again area $\frac12(l-a)^2$.
• $D_2=y-x\gt a$: $y\gt x+a$ — again a triangle with legs $l-a$, area $\frac12(l-a)^2$.

With both halves of the square, each event $\{D_i\gt a\}$ has probability

\[ P(D_i\gt a) = \frac{2\cdot\frac12(l-a)^2}{l^2} = \left(1-\frac al\right)^2. \]

Step 3: Two parts longer than $a$

Two parts longer than $a$ require $2a\lt l$. For $D_1\gt a$ and $D_3\gt a$ (with $x\lt y$): $a\lt x\lt y\lt l-a$, a triangle with legs $l-2a$. The same for the other pairs. So

\[ P(D_i\gt a,\ D_j\gt a) = \left(1-\frac{2a}l\right)^2 \quad (a\lt\tfrac l2), \qquad 0 \quad (a\ge\tfrac l2). \]

All three parts cannot exceed $a\gt\frac l3$, since they sum to $l$.

Step 4: Inclusion–exclusion

\[ P(\text{some part}\gt a) = 3\left(1-\frac al\right)^2 - 3\left(1-\frac{2a}{l}\right)_+^2. \]

Hence the probability that no part exceeds $a$:

\[ \boxed{P = \begin{cases} 1-3\left(1-\dfrac al\right)^2, & \dfrac l2\le a\lt l,\\[10pt] \left(\dfrac{3a}{l}-1\right)^2, & \dfrac l3\lt a\le\dfrac l2. \end{cases}} \]

For $a\lt\frac l2$: $1-3(1-\frac al)^2+3(1-\frac{2a}l)^2 = 1-\frac{6a}{l}+\frac{9a^2}{l^2} = (\frac{3a}l-1)^2$.

Step 5: Checks

• $a\to\frac l3$: $P\to0$ (all three parts would have to equal $\frac l3$ exactly).
• $a=\frac l2$: both formulas give $\frac14$ — the triangle probability of Problem 81.
• $a\to l$: $P\to1$.

A simulation with $a=0.4\,l$ gave $0.0402$, against $(1.2-1)^2=0.04$.