A point $X$ is selected at random from a line segment $AB$ with mid point $O$. The probability that the line segments $AX, XB$ and $AO$ can form a triangle is :
Step-by-Step Solution
Key Concept: Apply the triangle inequality constraint and use symmetry about the midpoint to determine the favorable region.
For a triangle to form with sides $AX$, $XB$, and $AO$ where $X$ lies on segment $AB$ of length $2a$, the triangle inequality requires $AX + AO > XB$, which gives $x < \frac{a}{2}$. Similarly, if $x$ lies in section $OB$, then $x < \frac{a}{2}$. For the favorable condition, $X$ should lie symmetrically about $O$ such that $OX < \frac{a}{2}$. The probability is $\frac{a}{2a} = \frac{1}{2}$.
Correct Answer: 1