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.
Total numbers $= 4 \times 24 = 96$.

For the number to be divisible by $2$, the units digit must be $0, 2$, or $4$.

Case 1: Units digit is $0$.
The remaining $3$ places can be filled in $^{4}P_{3} = 24$ ways.

Case 2: Units digit is $2$ or $4$.
The units place can be filled in $2$ ways.
The thousands place can be filled in $3$ ways (excluding $0$ and the units digit).
The remaining $2$ places can be filled in $^{3}P_{2} = 6$ ways.
Number of such numbers $= 2 \times 3 \times 6 = 36$.

Total number of even 4-digit numbers $= 24 + 36 = 60$.

Required probability $= \dfrac{60}{96} = \dfrac{5}{8}$.

Answer: $\dfrac{5}{8}$