# Let A denote the event that a 6 -digit integer

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 :

1. (1) $\frac{9}{56}$

2. (2) $\frac{4}{9}$

3. (3) $\frac{3}{7}$

4. (4) $\frac{11}{27}$

Correct Option: , 2

Solution:

Total cases :

$\underline{6} \cdot \underline{6} \cdot \underline{5} \cdot \underline{4} \cdot \underline{3} \cdot \underline{2}$

$\mathrm{n}(\mathrm{s})=6 \cdot 6 !$

Favourable cases :

Number divisible by $3 \equiv$

Sum of digits must be divisible by 3

Case-I

$1,2,3,4,5,6$

Number of ways $=6 !$

Case-II

$0,1,2,4,5,6$

Number of ways $=5 \cdot 5 !$

Case-III

$0,1,2,3,4,5$

Number of ways $=5 \cdot 5 !$

$\mathrm{n}($ favourable $)=6 !+2 \cdot 5 \cdot 5 !$

$P=\frac{6 !+2 \cdot 5 \cdot 5 !}{6 \cdot 6 !}=\frac{4}{9}$