Question

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. $\frac { 3 } { 4 }$

Answer 1

Answer: (C)

$\frac { 1 } { 2 }$

Answer 2

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