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Two numbers a and b are chosen at random from the set {1,2,3,..,3n}. The probability that `a^(3)+b^(3)` is divisible by 3, is

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Two numbers a and b are chosen at random from the set {1,2,3,..,3n}. The probability that a3+b3 is divisible by 3, is A. 1 2 B. (1)(4) C. 1 6 D. 1 3 Select the correct answer from above options

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Correct Answer - D The number of ways of choosing two numbers from the given set is .3nC2. Let us divide given 3n numbers into three groups G1,G2 and G3 as follows : G1:3,6,9,..,3n G2:1,4,7,10,..,3n-1 G3:2,5,8,11,..,3n-2 We have, a3+b3=(a+b)3-3ab(a+b) Therefore, a3+b3 will be divisible by 3, if a+b is divisible by 3. Now, a+b will be divisible by 3 in the following cases : (i) Both the numbers belong to the first group. (ii) One of the two numbers belongs to second group and one of them belongs to the third group. ∴ Favourable number of elementry events =.nC2+.nC1×.nC1 ) .3nC2 = 1 3

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