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STD 11 - 14. probability — Maths STD 11 Science — Question

CBSE BoardEnglish MediumSTD 11 ScienceMathsSTD 11 - 14. probability1 Mark
MCQ
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
    $\frac{9}{56}$
  • ✓
    $\frac{4}{9}$
  • C
    $\frac{3}{7}$
  • D
    $\frac{11}{27}$
✓

Answer

Correct option: B.
$\frac{4}{9}$
b
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 $)=6 !+2 \cdot 5 \cdot 5 !$

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

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