Question

Two numbers $x$ and $y$ are selected at random from $\{1,2,...,5n\}$ without replacement. If $p_n$ denote the probability that $\frac{1}{5}\left[\right(x^2+y^2\left)\right]$ is a natural number, then

• A. $p_n=\frac{9n-1}{5(5n-1)}$
• B. $p_n=\frac{4n}{5(5n-1)}$
• C. $\underset{n\to \infty }{lim}{p}_n=\frac{9}{25}$
• D. $\underset{n\to \infty }{lim}{p}_n=\frac{4}{25}$

Answer 1

Answer: (A), (C)

$p_n=\frac{9n-1}{5(5n-1)}$
$\underset{n\to \infty }{lim}{p}_n=\frac{9}{25}$

$x^2+y^2$ is divisible by $5$ if
(i) the remainder of $x^2$ is 1 and that of $y^2$ is $4$ or vice-versa OR
(ii) if both leave a remainder $0$ (remainder of a number w.r.t. $5$ is $0,1,2,3,4$; hence, remainder of a square is either $0,1,4,4,1$ respectively).
(i) Choosing remainder $1:$ the number is of the form $5k+1$ or $5k+4$: there are $2n$ such numbers
Choosing remainder $4:$ the number is of the form $5k+2$ or $5k+3$: there are $2n$ such numbers
Thus, number of ways $=2\times 2n\times 2n=8n^2$
(ii) Choosing remainder $0:$ there are $n$ such numbers. Thus, number of ways $=n\times (n-1)$.
Hence, probability = $\frac{8n^2+n(n-1)}{5n\times (5n-1)}=\frac{9n-1}{5(5n-1)}$
If $n\to \infty$, probability = $\frac{9n-1}{5(5n-1)}=\frac{9-\frac{1}{n}}{5(5-\frac{1}{n})}=\frac{9}{25}$