The total number of ways of forming four-digit numbers from the set \( \{1, 2, 3, 4, 5\} \) is: \[ { }^{5}C_{4} \times 4! = 120. \] We know that if the number is divisible by 3, then the sum of its digits should be a multiple of 3. So, only one case (\( \{1, 2, 4, 5\} \)) satisfies this condition. The total number of favorable cases is: \[ 4! = 24. \] The required probability is: \[ \frac{24}{120} = \frac{1}{5}. \]