If two numbers $x$ and $y$ are chosen one after the other at random with replacement from the set of number…
If two numbers $x$ and $y$ are chosen one after the other at random with replacement from the set of number $\{1,2,3$ ..., 10$\}$, then the probability that $\left|x^2-y^2\right|$ is divisible by 6 is
$\frac{8}{25}$
$\frac{6}{25}$
$\frac{3}{10}$
$\frac{13}{50}$
Solution
Total number of ways in which two numbers can be selected from $\{1,2,3, \ldots ., 10\}=10^2$
$\left|x^2-y^2\right|$ is divisible by 6
$\Rightarrow|(x-y)(x+y)|$ is divisible by 6
$\Rightarrow$ either $(x-y)$ is divisible by 6 or $(x+y)$ is divisible by 6 .
So, possible outcomes are \begin{array}{|c|c|c|}\hline Values of \boldsymbol{x} & Values of \boldsymbol{y} & No. of cases \\\hline 1 & 1,5,7 & 3 \\2 & 2,4,8,10 & 4 \\3 & 3,9 & 2 \\4 & 4,2,8,10 & 4 \\5 & 5,1,7 & 3 \\6 & 6 & 1 \\7 & 7,1,5 & 3 \\8 & 8,2,4,10 & 4 \\9 & 9,3 & 2 \\10 & 10,4,2,8 & 4 \\\hline & & \mathbf{3 0} \\\hline\end{array} Required probability $=\frac{30}{10^2}=\frac{3}{10}$.