KVPY SX SB Exam 2015 Question Paper

Question : 89 of 120

Marks: +1, -0

Let

X_n = {1, 2, 3, ..., n}

and let a subset A of

X_n

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 -

$p > q \and p- q = 1/6$
$p < q \and q – p= 1/6$
$p > q \and p - q =1/ 10$
$p < q \and q - p=1 /10$

Validate

Solution:

when n = 10
Let A

_r

be no. of ways of selecting r numbers.
No. of selection of A is

= n (A_0) + n(A_1) + n (A_2) + n(A_3) + n(A_4)

= 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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