Q50 / 90 · KVPY 2013 (SB/SX) · Mathematics
A box contains coupons labeled $1,2,3$....n. A coupon is picked at random and the number $\mathrm{x}$ 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 $\mathrm{x}$, y divides the other is (in the options below [r] denotes the largest integer less than or equal to $\mathrm{r}$ )
- A$\frac{1}{2}$
- B$\frac{1}{n^{2}} \sum_{k=1}^{n}\left[\frac{n}{k}\right]$
- C$-\frac{1}{n}+\frac{1}{n^{2}} \sum_{k=1}^{n}\left[\frac{n}{k}\right]$
- Correct answerD$-\frac{1}{n}+\frac{2}{n^{2}} \sum_{k=1}^{n}\left[\frac{n}{k}\right]$
Solution
Let $x=1$
favourable out comes $(1,1),(1,2) \ldots \ldots(1, n)$ no. of favourable out comes when $x=1$
$$
=\left[\frac{\mathrm{n}}{1}\right]
$$
$\therefore$ no. of favourable out comes when $x=1$ or $y=1$
$$
=2\left[\frac{\mathrm{n}}{1}\right]-1
$$
no. of favourable out comes when $x=2$ or $y=2$ but $x \neq 1, y \neq 1$
$$
2\left[\frac{\mathrm{n}}{2}\right]-1
$$
Similarly
no. of favourable out comes when $x=k$ or $y=k$ but $x, y \notin\{1,2, \ldots . . k-1\}$
$$
2\left[\frac{\mathrm{n}}{\mathrm{k}}\right]-1
$$
So probability $=\frac{\sum_{k=1}^{n}\left[\frac{n}{k}\right]-(1+1 \ldots . . n \text { times })}{n^{2}}$
$$
=\frac{2}{\mathrm{n}^{2}} \sum_{\mathrm{k}=1}^{n}\left[\frac{\mathrm{n}}{\mathrm{k}}\right]-\frac{1}{\mathrm{n}}
$$
Hence, the correct answer is option: 'D'.