A) \[\frac{2}{3}\]
B) \[\frac{2}{7}\]
C) \[\frac{1}{2}\]
D) \[\frac{3}{4}\]
Correct Answer: C
Solution :
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 favorable ways \[=3\,!+3\,!\] Hence required probability \[=\frac{3\,!\,+\,3\,!}{4\,!}=\frac{12}{24}=\frac{1}{2}.\]
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