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 :

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 : (A) (9/56) (B) (4/9) (C) (3/7) (D) (11/27)

Tardigrade Admin · Accepted Answer

Total cases: underline6 ⋅ underline6 ⋅ underline5 ⋅ underline4 ⋅ underline3 ⋅ underline2 n ( s )=6 ⋅ 6 ! Favourable cases: Number divisible by 3 ≡ 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 ! n ( favourable )=6 !+2 ⋅ 5 ⋅ 5 ! P =(6 !+2 ⋅ 5 ⋅ 5 !/6 ⋅ 6 !)=(4/9)

Q. 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 :

$\frac{9}{56}$

$\frac{4}{9}$

$\frac{3}{7}$

$\frac{11}{27}$

Q. 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 :

569​

94​

73​

2711​

Q. 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 :

$\frac{9}{56}$

$\frac{4}{9}$

$\frac{3}{7}$

$\frac{11}{27}$