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

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14.Probability

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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

$\frac{2}{7}$

C

$\frac{1}{2}$

D

$\frac{3}{4}$

Solution

(c) 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 required probability $ = \frac{{3\,!\, + \,3\,!}}{{4\,!}} = \frac{{12}}{{24}} = \frac{1}{2}.$

Standard 11

Mathematics

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