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$ways.<br><br>Total possible numbers = 4 \times 24 = 96$.

A number is divisible by $4$ if the number formed by its last two digits is divisible by $4$.

The possible last two digits from the given set are $04, 12, 20, 24, 32, 40$.

Case 1: Last two digits are $04, 20$, or $40$ ($3$ cases).

Since $0$ is used, the first two places can be filled from the remaining $3$ digits in ^{3}P_{2} = 6$ways.<br><br>Number of such numbers = 3 \times 6 = 18$.

Case 2: Last two digits are $12, 24$, or $32$ ($3$ cases).

Since $0$ is not used in the last two digits, the thousands place cannot be $0$. It can be filled in $2$ ways.

The hundreds place can be filled in $2$ ways.

Number of such numbers = $3 \times (2 \times 2) = 12$.

Total numbers divisible by $4 = 18 + 12 = 30$.

Required probability = $\dfrac{30}{96} = \dfrac{5}{16}$.

Answer: $\dfrac{5}{16}$