- Home
- Standard 11
- Mathematics
14.Probability
hard
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
$\frac{9}{56}$
B
$\frac{4}{9}$
C
$\frac{3}{7}$
D
$\frac{11}{27}$
(JEE MAIN-2021)
Solution
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 $)=6 !+2 \cdot 5 \cdot 5 !$
$P=\frac{6 !+2 \cdot 5 \cdot 5 !}{6 \cdot 6 !}=\frac{4}{9}$
Standard 11
Mathematics
