Breaking The Rod!!

Imagine a square of side length $6$ . Each point $(x,y)$ in the square corresponds to the distance of $A$ and $B$ from the left of the rod (i.e. $A$ is $x$ metres away and $B$ is $y$ metres away). The set of points for which AB is at least $1$ m corresponds to two right angle isosceles triangles at the bottom right and top left of the square, of side length $6-1 = 5$ . The probability is thus $\frac{2(0.5)(5)(5)}{6 \times 6}$ = $\frac{25}{36}$ . The answer is thus $25 + 36$ = $\boxed {61}$

Let us call the first point $A$ , and its distance from left end of the rod $x$ .

$\textbf{First case}$ - $0<x<1$

In this case the required probability is $\displaystyle\int_{0}^{1} \dfrac{5-x}{36} \mathrm{d}x=\dfrac{1}{8}$ .

$\textbf{Second case}$ - $1<x<5$

In this case the required probability is $\displaystyle\int_{1}^{5} \dfrac{4}{36} \mathrm{d}x=\dfrac{4}{9}$ .

$\textbf{Third case}$ - $5<x<6$

In this case the required probability is $\displaystyle\int_{5}^{6} \dfrac{x-1}{36} \mathrm{d}x=\dfrac{1}{8}$ .

Therefore, our final probability is the sum of the probabilities in all the three cases, IE,

$\dfrac{1}{8}+\dfrac{4}{9}+\dfrac{1}{8}=\boxed{\dfrac{25}{36}}$ .