QJEEM23Maaaeb: JEE Main 2023 Mathematics - Class 12 JEEM23Maaaeb Solution: A fair faces die is rolled repeatedly until a number less than…

In this case, the success is defined as getting a number less than $n$ when rolling an $n$ -sided die. The probability of success, $p$ , in each roll is then $(n-1)/n$ , and the probability of failure, $q = 1 - p$ , is $1/n$ .

$\text { Mean }=\sum\limits_{i=1}^{\infty} p_i x_i=1 \cdot \frac{n-1}{n}+\frac{2}{n} \cdot\left(\frac{n-1}{n}\right)+\frac{3}{n^2}\left(\frac{n-1}{n}\right)+\ldots$

$\frac{n}{9}=\left(1-\frac{1}{n}\right) S$ ......(1)

where

$\begin{aligned}& S=1+\frac{2}{n}+\frac{3}{n^2}+\frac{4}{n^3}+\ldots \\\\& \frac{1}{n} S=\frac{1}{n}+\frac{2}{n^2}+\frac{3}{n^3}+\ldots \\\\& ----------------\\\\& \left(1-\frac{1}{n}\right) S=1+\frac{1}{n}+\frac{1}{n^2}+\frac{1}{n^3}+\ldots\end{aligned}$

$\begin{aligned}& \Rightarrow \left(1-\frac{1}{n}\right) S=\frac{1}{1-\frac{1}{n}} \\\\& \Rightarrow \frac{n}{9}=\left(1-\frac{1}{n}\right) \times \frac{1}{\left(1-\frac{1}{n}\right)^2}=\frac{n}{n-1}\end{aligned}$

$\Rightarrow$ n = 10