Question

Let X_n = {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 

Answer 1

Correct option  (C) p > q and p – q = 1/10 

Explanation:

when n = 10

Let A_r 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