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

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