Let A denote the event that a 6 -digit intege | vedclass

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Question

Total cases :

$\underline{6} \cdot \underline{6} \cdot \underline{\underline{5}} \cdot \underline{4} \cdot \underline{3} \cdot \underline{2}$

$n

Vedclass · Accepted Answer

$\frac{4}{9}$

Vedclass · Answer

Total cases :

$\underline{6} \cdot \underline{6} \cdot \underline{\underline{5}} \cdot \underline{4} \cdot \underline{3} \cdot \underline{2}$

$n(s)=6 \cdot 6 !$

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 $=5 \cdot 5 !$

Case$-III$

$0,1,2,3,4,5$

Number of ways $=5 \cdot 5 !$ $n ($ favourable $)=6 !+2 \cdot 5 \cdot 5 !$

$P=\frac{6 !+2 \cdot 5 \cdot 5 !}{6 \cdot 6 !}=\frac{4}{9}$

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 :

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