Question

Let $A$denote the event that a $6$-digit integer formed by $0,1,2,3,4,5,6$ without repetitions, be divisible by $3$. Then the probability of event $A$ is equal to

• A. $\frac{4}{9}$
• B. $\frac{9}{56}$
• C. $\frac{3}{7}$
• D. $\frac{11}{27}$

Answer 1

The correct option is A

$\frac{4}{9}$

Determining the probability

W have, $0,1,2,3,4,5,6$ using which $6$-digit number without repetition is to be formed.

We know that $0$cannot be used in the first place, So the possibility of filling the first place is by $6$ remaining digits, now we have five more numbers and zero, so

there is $6$ possible ways of filling the second place, similarly

for third place, its is $5$, for fourth place it is $4$, for fifth place it is $3$, for sixth place it is $2$.

Therefore sample space is $6\times 6\times 5\times 4\times 3\times 2=4320$

$A=$ $6$-digit integers divisible by three without repetition.

Condition for a number divisible by $3$ is a sum of the digits in the number is divisible by $3$

$\left(0,1,2,3,4,5\right),\left(0,1,2,4,5,6\right),\left(1,2,3,4,5,6\right)$ these are the sets with the possibility of forming a number divisible by $3$.

Therefore the total such number formed are:

$$\begin{gathered} =\left(5\times 5\times 4\times 3\times 2\times 1\right)+\left(5\times 5\times 4\times 3\times 2\times 1\right)+\left(6\times 5\times 4\times 3\times 2\times 1\right)\left[\text{Refer the above explanation}\right] \\ =600+600+720 \\ =1920 \end{gathered}$$

Therefore, $P\left(A\right)=\frac{1920}{4320}=\frac{4}{9}$

Hence, option (A) is the correct answer