KVPY SX SB Exam 2013 Question Paper

Question : 15 of 120

Marks: +1, -0

A box contains coupons labeled 1,2,3,... ,n. A coupon is picked at random and the number x is noted. The coupon is put back into the box and a new coupon is picked at random. The new number is y. Then the probability that one of the numbers x,y divides the other is (in the options below [r] denotes the largest integer less than or equal to r)

$1/2$
$1/n^2 ∑↖{n}↙{k=1}[n/k]$
$-1/n+1/n^2 ∑↖{n}↙{k=1}[n/k]$
$-1/n+2/n^2 ∑↖{n}↙{k=1}[n/k]$

Validate

Solution:

Let x = 1
favourable out comes (1, 1), (1, 2) .......(1, n) no. of favourable out comes when x = 1 =

[n/1]

∴ no. of favourable out comes when x = 1 or y = 1

=2[n/1]-1

Similarly
no. of favourable out comes when x = k or y = k but x, y ∉ {1, 2, ..... k - 1}

=2[n/k]-1

So probability

={∑↖{n}↙{k=1}[n/k]-(1+1....\n\times)}/\n^2

=2/n^2∑↖{n}↙{k=1}[n/k]-1/n

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