MCQ
A three digit number is formed by using numbers $1, 2, 3$ and $4$. The probability that the number is divisible by $3$, is
- A$\frac{2}{3}$
- B$\frac{2}{7}$
- ✓$\frac{1}{2}$
- D$\frac{3}{4}$
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 required probability $ = \frac{{3\,!\, + \,3\,!}}{{4\,!}} = \frac{{12}}{{24}} = \frac{1}{2}.$
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