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Question

Out of 3n consecutive natural numbers, 3 natural numbers are chosen at random without replacement. The probability that the sum of the chosen numbers is divisible by 3 is

A
n(3n23n+2)2
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B
(3n23n+2)2(3n1)(3n2)
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C
(3n23n+2)(3n1)(3n2)
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D
n(3n1)(3n2)3(n1)
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Solution

The correct option is C (3n23n+2)(3n1)(3n2)
Let the sequence of 3n consecutive integers begin with m. Then

3n consecutive integers are
m,m+1,m+2,....m+(3n1)
3 integers from 3n can be selected in 3nC3 ways
Total no. of outcomes =3nC3

Now 3n integers can be divided into 3 groups

G1:n numbers of form 3p
G2:n numbers of form 3p+1
G3:n numbers of form 3p+2

The sum of 3 integers chosen from 3n integers will be divisible by 3 if either all the three are from same group or one integer from each group. The no. of ways that the three integers are from same group is nC3+nC3+nC3nC1 and no. of ways that the integers are from different group is nC1×nC1×nC1

favourable cases =(nC3+nC3+nC3)+(nC1×nC1×nC1)

Required probability =3.nC3+(nC1)33nC3=3n23n+2(3n1)(3n2)

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