Passage: A 4-digit number is selected at random formed by using the...

The total number of $4$-digit numbers that can be formed using the digits $0, 1, 2, 3, 4$ without repetition is calculated by filling the thousands place in $4$ ways (excluding $0$) and the remaining $3$ places in ^{4}P_{3} = 24$ways.<br><br>Total numbers$= 4 \times 24 = 96$<br><br>For a number to be divisible by 6$, it must be divisible by both $2$ and $3$.

For divisibility by $3$, the sum of the digits must be a multiple of $3$. The sum of all five digits is $0 + 1 + 2 + 3 + 4 = 10$. To choose $4$ digits whose sum is a multiple of $3$, the excluded digit must be $1$ or $4$.

Case 1: The excluded digit is $1$. The selected digits are $0, 2, 3, 4$.
For divisibility by $2$, the unit digit must be $0, 2$, or $4$.
If the unit digit is $0$, the remaining places can be filled in $3! = 6$ ways.
If the unit digit is $2$, the thousands place can be filled in $2$ ways (excluding $0$), and the remaining places in $2! = 2$ ways, giving $2 \times 2 = 4$ ways.
If the unit digit is $4$, similarly, there are $2 \times 2 = 4$ ways.
Total numbers in Case 1 $= 6 + 4 + 4 = 14$

Case 2: The excluded digit is $4$. The selected digits are $0, 1, 2, 3$.
For divisibility by $2$, the unit digit must be $0$ or $2$.
If the unit digit is $0$, the remaining places can be filled in $3! = 6$ ways.
If the unit digit is $2$, the thousands place can be filled in $2$ ways (excluding $0$), and the remaining places in $2! = 2$ ways, giving $2 \times 2 = 4$ ways.
Total numbers in Case 2 $= 6 + 4 = 10$

Total favorable outcomes $= 14 + 10 = 24$

Required probability $= \dfrac{24}{96} = \dfrac{1}{4}$

Answer: $\dfrac{1}{4}$