Consider an event \(E\) such that a matrix of order \(2 \times 2\) is invertible with entries 0 or 1. Then, \(P(E)\) is …

\(\begin{aligned}& 2 \times 2 \rightarrow \text { Matrix } \\& |A| \neq 0 \\& n(s)=2 \times 2 \times 2 \times 2 \Rightarrow 16\end{aligned}\)

\(A\) can't be

\(\begin{aligned}& {\left[\begin{array}{ll}0 & 0 \\0 & 0\end{array}\right],\left[\begin{array}{ll}1 & 1 \\1 & 1\end{array}\right],\left[\begin{array}{ll}1 & 0 \\0 & 0\end{array}\right]} \\& {\left[\begin{array}{ll}1 & 1 \\0 & 0\end{array}\right],\left[\begin{array}{ll}1 & 0 \\1 & 0\end{array}\right],\left[\begin{array}{ll}0 & 0 \\1 & 1\end{array}\right]} \\ & {\left[\begin{array}{ll}0 & 1 \\0 & 1\end{array}\right],\left[\begin{array}{ll}0 & 0 \\1 & 0\end{array}\right],\left[\begin{array}{ll}0 & 0 \\0 & 1\end{array}\right]} \end{aligned}\)

\(\left[\begin{matrix}0 & 1 \\ 0 & 0\end{matrix}\right]\)
So, total \(\Rightarrow 10\) types of matrix we can't take

\(\begin{aligned}& \therefore P(\bar{A})=\frac{10}{16}=\frac{5}{8} \\& \therefore P(A)=1-P(\bar{A})=1-\frac{5}{8} \Rightarrow \frac{3}{8} \\\end{aligned}\)