Probability of making triangles

Let the stick be divided into three parts having length $x, y$ and $1-(x+y)$

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Sample space:

Constraints: All parts must have length in $(0, 1)$ .

This gives: $x > 0, y > 0, x + y < 1$

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Favourable outcome:

Constraint: The sides must satisfy the triange inequality.

1. $x + y > 1 - (x+y)$ $\implies x + y > \frac12$

2. $x + 1 - (x+y) > y \implies y < \frac12$

3. $y + 1 - (x+y) > x \implies x < \frac12$

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As $x$ and $y$ are uniformly distributed, the probability can be calculated by taking the ratio of the area of the favourable outcome to the area of the sample space when the constraints are plotted in the $\text{x-y plane}$ .

The sample space is a triangle with vertices: $(0, 0), (1, 0)\text{ and }(0, 1)$

The favourable outcome region is a triangle with vertices: $(\frac12, \frac12), (\frac12, 0)\text{ and }(0, \frac12)$

$\therefore \text{Required probability= } \boxed{\frac14}$