Question

A given line segment is divided at random into three parts. What is the probability that they form sides of a possible triangle$?$

Answer 1

Let $AB$ be the line segment of length $l.$
Let $C$ and $D$ be the points which divide $AB$ into three parts.
Let $AC=x,CD=y.$ Then $DB=l-x-y.$
Clearly $x+y<l.$
$\therefore$ the sample space is given by
the region enclosed by $△OPQ,$ where $OP=OQ=l.$
Area of $△OPQ=\frac{l^2}{2}$
Now if the parts $AC,CD$ and $DB$ form a triangle, then
$x+y>l-x-y$ i.e. $x+y>\frac{l}{2}$ (i)
$x+l-x-y+y>y$ i.e. $y<\frac{l}{2}$ (ii)
from (i), (ii) and (iii), we get
the event is given by the region closed in $△RST$
$\therefore$ Probability of the event$=\frac{ar\left(△RST\right)}{ar\left(△OPQ\right)}=\frac{\frac{1}{2}.\frac{l}{2}.\frac{l}{2}}{\frac{l^2}{2}}=\frac{1}{4}$