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 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 $\frac{4}{9}$

Number of elements in sample space $=6\times 6!$
Favourable case $=(0,1,2,3,4,5)$ or $(0,1,2,4,5,6)$ or $(1,2,3,4,5,6)$
Number of favourable case $=5\times 5!+5\times 5!+6!$
$=1920$
Required probability $=\frac{1920}{6\times 6!}=\frac{4}{9}$