Question:

Step 1: Count total $2 \times 2$ matrices. Each of the 4 entries can be any of 5 elements, so total = $5^4 = 625$. Step 2: Count symmetric matrices. A $2 \times 2$ symmetric matrix has the form $\begin{pmatrix} a & b \\ b & d \end{pmatrix}$. Free entries: $a, b, d$ — each chosen from 5 elements. Count = $5^3 = 125$. Step 3: Count skew-symmetric matrices. A $2 \times 2$ skew-symmetric matrix has the form $\begin{pmatrix} 0 & b \\ -b & 0 \end{pmatrix}$. We need $b \in \{-2,-1,0,1,2\}$ and $-b \in A$. Since $A$ is symmetric about 0, all 5 values work. Count = 5. Step 4: Apply inclusion-exclusion. Symmetric $\cap$ skew-symmetric: matrix must satisfy $A = A^T$ and $A = -A^T$, so $A = -A$, meaning all entries are 0. Count = 1. Favorable = $125 + 5 - 1 = 129$. Step 5: Compute probability: $$P = \frac{129}{625} \approx 0.2064$$ Step 6: Compare with options. $\dfrac{1}{10} = 0.1$, $\dfrac{2}{10} = 0.2$, $\dfrac{3}{10} = 0.3$. Since $0.2064 > 0.1$ and $0.2064 > 0.2$ but $0.2064 < 0.3$, the probability is greater than $\dfrac{1}{10}$ and $\dfrac{2}{10}$.