Let  $$A=[a_{ij}]$$  be a square matrix of order 2 with entries either 0 or 1. Let $$E$$  be the event that  $$A$$  is an invertible matrix.  Then the probability  $$P(E)$$ is:

Let $$A = [a_{ij}]$$ be a $$2 \times 2$$ matrix with entries either 0 or 1, and let us find the probability that $$A$$ is invertible.

A $$2 \times 2$$ matrix has 4 entries, each of which can be 0 or 1, so the total number of such matrices is $$2^4 = 16$$.

Since $$A = \begin{bmatrix} a_{11} & a_{12} \\ a_{21} & a_{22} \end{bmatrix}$$ is invertible exactly when $$\det(A) = a_{11}a_{22} - a_{12}a_{21} \neq 0,$$ and because each product of two entries in \{0,1\} can be either 0 or 1, the determinant vanishes precisely when $$a_{11}a_{22} = a_{12}a_{21}.$$

Next, the probability that a product of two bits is 0 is $$\tfrac{3}{4}$$ (at least one factor is 0) and that it is 1 is $$\tfrac{1}{4}$$ (both are 1). Therefore, the probability that both products are 0 is $$\tfrac{3}{4}\cdot\tfrac{3}{4}=\tfrac{9}{16}$$, and that both are 1 is $$\tfrac{1}{4}\cdot\tfrac{1}{4}=\tfrac{1}{16}$$. Adding these gives $$P(\det(A)=0)=\tfrac{9}{16}+\tfrac{1}{16}=\tfrac{10}{16}.$$

Hence the probability that the matrix is invertible is $$1 - \tfrac{10}{16} = \tfrac{6}{16} = \tfrac{3}{8}.$$

The correct answer is Option (3): $$\frac{3}{8}$$.