A three digit number is formed by using numbers 1, 2, 3 and 4. The probability that the number is divisible by 3, is:

Total number of ways to form the numbers of three digit with 1, 2, 3 and 4 are ${ }^4{P}_3=4!=24$.

If the numbers are divisible by three then their sum of digits must be 3, 6 or 9 But sum 3 is impossible. Then for sum 6, digits are 1, 2, 3

Number of ways = 3 !

Similarly for sum 9, digits are 2, 3, 4. Number of ways= 3! Thus number of favourable ways $=3!+3!$

Hence probability $=\frac{3!+3!}{4!}=\frac{12}{24}=\frac{1}{2}$.