$A$ and $B$ each select one number at random from the distinct numbers $1,2,3, \ldots, n$ and the…
$A$ and $B$ each select one number at random from the distinct numbers $1,2,3, \ldots, n$ and the probability that the number selected by $A$ is less than the number selected by $B$ is $\frac{1009}{2019}$. Now, the probability that the number selected by $B$ is the number immediately next to the number selected by $A$ is
$\frac{2018}{2019}$
$\frac{2018}{(2019)^2}$
$\frac{2000}{(2019)}$
$\frac{2000}{(2019)^2}$
Solution
It is given that, $A$ and $B$ each select one number at random from the distinct numbers 1,2 , $3, \ldots, n$, then the probability that the number
selected by $A$ is less than the number selected by $B$ is
$
\begin{array}{rlrl}
& & \frac{{ }^n C_2}{n \times n} & =\frac{n(n-1)}{2(n \times n)}=\frac{1009}{2019} \text { (given) } \\
\Rightarrow \quad \frac{n-1}{2 n} & =\frac{1009}{2019} \Rightarrow 2019 n-2019=2018 n \\
\Rightarrow \quad n & =2019
\end{array}
$
Now, number of ways selecting numbers from the distinct numbers $1,2,3, \ldots, 2019$, by $B$ is the number immediately next to the number selected by $A$ is 2018 , because there are 2018 pairs of consecutive numbers.
$
\text { So, required probability }=\frac{2018}{2019 \times 2019}=\frac{2018}{(2019)^2}
$