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

The correct option is C

$\frac{1}{2}$

Explanation for the correct option:

STEP 1: Find the total number of three digits number possible

Total ways to form a three digit number using $1,2,3$ and $4$ is equal to:

$\begin{array}{rcl}{}^{4}P_3& =& \frac{4!}{\left(4-3\right)!}\\ & =& \frac{1\times 2\times 3\times 4}{1}\\ & =& 24\end{array}$

STEP 2: Finding the favorable cases

if a three digit number is divisible by $3$, then the sum of all the digits of that number should be $3,6,9,12,....$ or divisible by $3$

But, using the numbers $1,2,3$ and $4$ to form a three digit number, a sum of $3$ is not possible.

To obtain a sum of $6$, the three digits needed are $1,2,3$.

Hence, the number of ways to get a three digit number with sum of all digits as six is:

$\begin{array}{rcl}3!& =& 1\times 2\times 3\\ & =& 6\end{array}$

To obtain a sum of $9$, the three digits needed are $2,3,4$.

Hence, the number of ways to get a three digit number with sum of all digits as nine is:

$\begin{array}{rcl}3!& =& 1\times 2\times 3\\ & =& 6\end{array}$

Thus, the total number of favorable ways $=3!+3!$

STEP 3: Finding he required probability

$\begin{array}{rcl}P\left(E\right)& =& \frac{\mathbf{Number}\mathbf{}\mathbf{of}\mathbf{}\mathbf{favourable}\mathbf{}\mathbf{ways}}{\mathbf{Total}\mathbf{}\mathbf{number}\mathbf{}\mathbf{of}\mathbf{}\mathbf{ways}}\\ & =& \frac{3!+3!}{{}^{4}P_3}\\ & =& \frac{6+6}{24}\\ & =& \frac{12}{24}\\ & =& \frac{\mathbf{1}}{\mathbf{2}}\end{array}$

Hence, option C is correct.