JEE Main 2022 — Probability Question with Solution

For $x^2+\alpha x+\beta >0\forall x\in R$ to hold, we should have ${\alpha }^2-4\beta <0$

If $\alpha =1,\beta$ can be 1, 2, 3, 4, 5, 6 i.e., 6 choices

If $\alpha =2,\beta$ can be 2, 3, 4, 5, 6 i.e., 5 choices

If $\alpha =3,\beta$ can be $3,4,5,6$ i.e., 4 choices

If $\alpha =4,\beta$ can be 5 or 6 i.e., 2 choices

If $\alpha =6$, No possible value for $\beta$ i.e., 0 choices

Hence total favourable outcomes

$$\begin{array}{rl}& =6+5+4+2+0+0\\ & \\ & =17\end{array}$$

Total possible choices for $\alpha$ and $\beta =6\times 6=36$

Required probability $=\frac{17}{36}$