InterviewSolution
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InterviewSolution
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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.A. `(5n-3)/(2(3n-1))`B. `(5n-3)/(3(3n-1))`C. `(3n-1)/((5n-3))`D. `(3n-1)/(2(5n-3))` |
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Answer» Correct Answer - B The number of ways of choosing two numbers from the given set is `.^(3n)C_(2)`. Let us divide given 3n numbers into three groups `G_(1),G_(2) " and " G_(3)` as follows : `G_(1) : 3,6,9,.., 3n` `G_(2) : 1,4,7,10, .., 3n-1` `G_(3) : 2,5,8,11,..,3n-2` We have, `a^(2)-b^(2)=(a-b)(a+b)` Therefore, `a^(2)-b^(2)` will be divisible by 3 if either a and b are chosen from the same group or one of them is chosen from group `G_(2)` and the other from group `G_(3)`. Therefore, the number of favourable elemnetary events is `(.^(n)C_(2)+ .^(n)C_(2)+ .^(n)C_(2))+ .^(n)C_(1) xx .^(n)C_(1)=3 .^(n)C_(2)+n^(2)` `therefore` Required probability `=(3xx .^(n)C_(2)+n^(2))/(.^(3n)C_(2))=(5n-3)/(3(3n-1))` |
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