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

Total number of $4$-digit numbers formed using $0, 1, 2, 3, 4$ without repetition:

The thousands place can be filled in $4$ ways (excluding $0$).

The remaining $3$ places can be filled in ^{4}P_{3} = 24$ways.<br><br>Total possible numbers$= 4 \times 24 = 96$.<br><br>A number is divisible by 3$ if the sum of its digits is a multiple of $3$.

The sum of all given digits is $0 + 1 + 2 + 3 + 4 = 10$.

To choose $4$ digits whose sum is a multiple of $3$, the excluded digit must leave a remainder of $1$ when divided by $3$. Thus, the excluded digit can be $1$ or $4$.

Case 1: Excluding $1$, the digits are $0, 2, 3, 4$.

Number of $4$-digit numbers $= 3 \times 3! = 18$.

Case 2: Excluding $4$, the digits are $0, 1, 2, 3$.

Number of $4$-digit numbers $= 3 \times 3! = 18$.

Total favorable outcomes $= 18 + 18 = 36$.

Required probability $= \dfrac{36}{96} = \dfrac{3}{8}$.

Answer: $\dfrac{3}{8}$