Question

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 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)

• A. $\frac{1}{2}$
• B. $\frac{1}{n^2}{\sum }_{k=1}^n\left[\frac{n}{k}\right]$
• C. $-\frac{1}{n}+\frac{1}{n^2}{\sum }_{k=1}^n\left[\frac{n}{k}\right]$
• D. $-\frac{1}{n}+\frac{2}{n^2}{\sum }_{k=1}^n\left[\frac{n}{k}\right]$

Answer 1

Answer: (D)

$-\frac{1}{n}+\frac{2}{n^2}{\sum }_{k=1}^n\left[\frac{n}{k}\right]$

Let $x=1$

then favourable outcomes are $(1,1),(1,2)......(1,n)$.

$\therefore$ Number of favourable outcomes when $x=1$ is $\left[\frac{n}{1}\right]$.

$\therefore$ Number of favourable outcomes when $x=1$ or $y=1$ is $2\left[\frac{n}{1}\right]-1$

Number of favourable outcomes when $x=2$ or $y=2$ but $x\ne 1$, $y\ne 1$ is $2\left[\frac{n}{2}\right]-1$

Similarly,

Number of favourable outcomes when $x=k$ or $y=k$ but $x,y\notin \{1,2,....,k-1\}$ is $2\left[\frac{n}{k}\right]-1$

So probability = $\frac{\left(2\sum _{k=1}^n\left[\frac{n}{k}\right]\right)-(1+1+...(n\text{times}\left)\right)}{n^2}=\frac{2}{n^2}\sum _{k=1}^n\left[\frac{n}{k}\right]-\frac{1}{n}$