JEE Main 2025 — Probability Question with Solution

$A={\left[a_y\right]}_{2\times 2}$ and entries are 0 or 1. $\begin{aligned} & \therefore\left|\begin{array}{ll} a & b \\ c & d \end{array}\right|=0 \\ & \Rightarrow a d-b c=0 \end{aligned}$ Case I: $ad=bc=1$ $\therefore a=b=c=d=1$ Case II: $ad=bc=0$ $\begin{array}{ll} a=0, d=0 & b=0, c=0 \\ a=0, d=1 & b=0, c=1 \\ a=1, d=0 & b=1, c=0 \end{array}$ $\therefore$ Total 10 cases when matrix is non invertible Total possible matrix $=2^4=16$ Required probability of invertible $=\frac{16-10}{16}=\frac{6}{16}=\frac{3}{8}$