InterviewSolution
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InterviewSolution
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Let Xn = {1, 2, 3, ..., n} and let a subset A of Xn be chosen so that every pair of elements of A differ by at least 3. (For example, if n = 5, A can be ∅, {2} or {1,5} among others). When n = 10, let the probability that 1 ∈ A be p and let the probability that 2 ∈ A be q. Then -(A) p > q and p – q = 1/6(B) p < q and q – p = 1/6(C) p > q and p – q = 1/10(D) p < q and q – p = 1/10 |
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Answer» Correct option (C) p > q and p – q = 1/10 Explanation: when n = 10 Let Ar be no. of ways of selecting r numbers. No. of selection of A is = n (A0) + n(A1) + n (A2) + n(A3) + n(A4) = 1 + 10 + (7 + 6 + 5 +…+ 1) + (4 + 3 + 2 + 1) + (3 + 2 + 1) + (2 + 1) + 1 = 11 + 7.8/2 + 10 + 6 + 3 + 1 + 1 = 60 N(p) = n(no. of ways 1 is selected) = 1 + 7 + 4 + 3 + 2 + 1 + 1 = 19 N(q) = n(no. of ways 2 is selected) = 1 + 6 + 3 + 2 + 1 = 13 So p = 19/60 q = 13/60 p-q = 1/10 |
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