JEE Main 2024 — Probability Question with Solution

Equation is $$a{x}^2+bx+c=0$$

$$\mathrm{D}>0$$ [for roots to be real & distinct]

$$\Rightarrow b^2-4ac>0$$

For $$b<2$$ no value of $$a$$ & $$c$$ are possible

$$\begin{array}{rl}& \text{ For }b=3\Rightarrow ac<\frac{9}{4}\\ & (a,c)\in \{(1,1),(1,2),(2,1)\}\Rightarrow 3\text{ cases }\end{array}$$

For $$b=4\Rightarrow ac<4$$

$$(a,c)\in \{(1,1),(1,2),(2,1),(3,1),(1,3)\}\Rightarrow 5\text{ cases }$$

For $$b=5\Rightarrow ac<\frac{25}{4}$$

$$\begin{array}{rl}& (a,c)\in \{(1,1),(1,2),(2,1),(3,1),(1,3),(2,2),\\ & (4,1),(1,4),(3,2),(2,3),(5,1),(1,5),(1,6),\\ & (6,1)\}=14\text{ cases }\end{array}$$

For $$b=6\Rightarrow ac<9$$

$$\begin{array}{rl}& (a,c)\in \{(1,1),(1,2),(2,1),(3,1),(1,3),(2,2),\\ & (4,1),(1,4),(3,2),(2,3),(5,1),(1,5),(1,6),\\ & (6,1),(2,4),(4,2)\}=16\text{ cases }\end{array}$$

Total cases $$=3+5+14+16=38$$ cases

$$\Rightarrow$$ Probability, $$p=\frac{38}{216}$$

$$\Rightarrow 216p=38$$