Suppose a point $X$ is selected at random from a line segment $AB$ of length $l$ and midpoint $O$. Find the probability that $AX,BX$ and $AO$ form a triangle.
My method and working is:
Case 1: Let $AX<\dfrac{l}{2}$. Then it is obvious that $AX<AO<BX$ and $BX=l-AX$. Then $AX,AO,BX$ will form a triangle iff $AX+AO>BX$ implying $AX+\dfrac{l}{2}>l-AX$ i.e. $AX>\dfrac{l}{4}$. The probability of this event is $P(AX>\dfrac{l}{4},AX<\dfrac{l}{2})=\dfrac{1}{4}$
Case 2: Let $AX>\dfrac{l}{2}$ then obviously $AX>AO>BX$ and we have $BX=l-AX$. Then $AX,AO,BX$ will be a triangle iff $AO+BX>AX$ i.e. $AX<\dfrac{3l}{4}$. The probability of this event is $P(AX<\dfrac{3l}{4},AX>\dfrac{l}{2})=\dfrac{1}{4}.$
So the desired probability is $\dfrac{1}{4}+\dfrac{1}{4}=\dfrac{1}{2}$.
The reason I posted my solution is that I am having certain doubts. If the above is correct, what is wrong with $P(AX>\dfrac{l}{4}|AX<\dfrac{l}{2})P(AX<\dfrac{l}{2})$ in the first case? But this gives different answer. I actually do not know the correct answer. Same for the second case. Actually, I always get confused with conditional probabilities in the continuous case.
A comprehensive evaluation is desired.
You've got the right answer. Easier way to solve this is
Note that three line segments will be $\frac{l}{2}, \mathcal x, l-x$
For these to form a triangle, sum of two sides must be greater than the third. So
$\frac{l}{2} + x > l - x$
Rearranging gives
$x > \frac{l}{4}~~~~~~~~~~~~~~~~~~~~~~~~~~~(1)$
and $\frac{l}{2} + l -x > x$
Rearranging gives
$x < \frac{3l}{4}~~~~~~~~~~~~~~~~~~~~~~~~~(2)$
and
$x + l-x > \frac{l}{2}$
$l > \frac{l}{2}~~~~~~~~~~~~~~~~~~~~~~~~~~~~(3)$ which is obvious ;)
So triangle will be formed if
$\frac{l}{4} <x < \frac{3l}{4}$
The probability of this happening is $0.5$ because $X \sim U(0, l)$
Try a different variant of this problem
You've a stick 10m long. You break it randomly at a point. Then you break the remaining stick randomly at a point again. Find the probability that the three pieces will form a triangle.