Let A denote the event that a 6-digit integer formed by

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. $\frac{9}{56}$

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

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

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


Correct Option: , 2

Solution:

Total cases :

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

$n(s)=6 \cdot 6 !$

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 $=5.5 !$

Case-III

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

Number of ways $=5.5 !$

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

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

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