JEE Main 2023 — Mathematics Probability & Statistics

Given,

A fair $n(n>1)$n(n>1) faces die is rolled repeatedly until a number less than $n$n appears,

$${x}_{i}1234... {p}_{i}\frac{n-1}{n}\frac{1}{n}\cdot (\frac{n-1}{n})\frac{1}{{n}^{2}}\cdot \frac{n-1}{n}\frac{1}{{n}^{3}}\cdot (\frac{n-1}{n})...$$xi​1234...pi​nn−1​n1​⋅(nn−1​)n21​⋅nn−1​n31​⋅(nn−1​)...

Now we know that, mean is given by,

Mean $\displaystyle =\sum _{i=1}^{\infty }{p}_{i}{x}_{i}=1\cdot \frac{n-1}{n}+\frac{2}{n}\cdot (\frac{n-1}{n})+\frac{3}{{n}^{2}}(\frac{n-1}{n})+...$=i=1∑∞​pi​xi​=1⋅nn−1​+n2​⋅(nn−1​)+n23​(nn−1​)+...

$$\Rightarrow \frac{n}{9}=(1-\frac{1}{n})S........(1)$$⇒9n​=(1−n1​)S........(1)

Where, $S=1+\frac{2}{n}+\frac{3}{{n}^{2}}+\frac{4}{{n}^{3}}+.........(2)$S=1+n2​+n23​+n34​+.........(2)

$$\frac{1}{n}S=\frac{1}{n}+\frac{2}{{n}^{2}}+\frac{3}{{n}^{3}}+.........(3)$$n1​S=n1​+n22​+n33​+.........(3)

Now subtracting equation $(2)-(3)$(2)−(3) we get,

$$(1-\frac{1}{n})S=1+\frac{1}{n}+\frac{1}{{n}^{2}}+\frac{1}{{n}^{3}}+...$$(1−n1​)S=1+n1​+n21​+n31​+...

$$\Rightarrow (1-\frac{1}{n})S=\frac{1}{1-\frac{1}{n}}$$⇒(1−n1​)S=1−n1​1​

Now putting the value of $S$S in equation $(1)$(1) we get,

$$\Rightarrow \frac{n}{9}=(1-\frac{1}{n})\times \frac{1}{{(1-\frac{1}{n})}^{2}}=\frac{n}{n-1}$$⇒9n​=(1−n1​)×(1−n1​)21​=n−1n​

$$\Rightarrow n=10$$⇒n=10