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Two natural numbers x and y are chosen at random from the set `{1,2,3,4,...3n}`. find the probability that `x^2-y^2` is divisible by 3.

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Two natural numbers x and y are chosen at random from the set {1,2,3,4,...3n}. find the probability that x2-y2 is divisible by 3. A. 5n-3 2(3n-1) B. 5n-3 3(3n-1) C. 3n-1 (5n-3) D. 3n-1 2(5n-3) Select the correct answer from above options

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Correct Answer - B 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, a2-b2=(a-b)(a+b) Therefore, a2-b2 will be divisible by 3 if either a and b are chosen from the same group or one of them is chosen from group G2 and the other from group G3 . Therefore, the number of favourable elemnetary events is (.nC2+.nC2+.nC2)+.nC1×.nC1=3.nC2+n2 ∴ Required probability = 3×.nC2+n2 .3nC2 = 5n-3 3(3n-1)

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