JEE Main · 2024 · 05 Apr 2024 · Shift 2 · Probability

The equation $a{x}^2+bx+c=0$ has "one real root bigger than the other" precisely when it has two distinct real roots, i.e. $b^2>4ac$, with $a,b,c$ each drawn independently from $\{1,\dots ,6\}$ ($216$ equally likely triples).
For each $b$, count pairs $(a,c)$ with $4ac<b^2$:
$b=1,2$: no pairs satisfy $4ac<b^2$ (minimum $4ac=4$).
$b=3$: need $ac\le 2$ — $(1,1),(1,2),(2,1)$ — 3 pairs.
$b=4$: need $ac\le 3$ — 5 pairs.
$b=5$: need $ac\le 6$ — 14 pairs.
$b=6$: need $ac\le 8$ — 16 pairs.
Total favourable $=0+0+3+5+14+16=38$.