JEE Mathematics 2023 Question with Solution

Given: A fair $n$ faces die is rolled until a number less than $n$ appears.

Find: The value of $n$ if the mean number of tosses is $\frac{n}{9}$.

Let the first success be getting a number less than $n$. Then

$$P(\text{success}) = \frac{n-1}{n}, \qquad P(\text{failure}) = \frac{1}{n}$$

So the mean is written as

$$1 \cdot \frac{n-1}{n} + 2 \cdot \frac{1}{n} \cdot \frac{n-1}{n} + 3 \cdot \left( \frac{1}{n} \right)^2 \cdot \left( \frac{n-1}{n} \right) + \dots$$

Hence,

$$\frac{n}{9} = \frac{n-1}{n} \left( 1 + 2 \cdot \frac{1}{n} + 3 \cdot \left( \frac{1}{n} \right)^2 + \dots \right)$$

Using

$$1 + 2r + 3r^2 + \dots = \frac{1}{(1-r)^2}$$

with $r = \frac{1}{n}$,

$$\frac{n}{9} = \frac{n-1}{n} \cdot \left( 1 - \frac{1}{n} \right)^{-2}$$

Now,

$$\frac{n}{9} = \frac{n-1}{n} \cdot \frac{n^2}{(n-1)^2}$$

So,

$$\frac{n}{9} = \frac{n}{n-1}$$

Therefore,

$$n-1 = 9$$

which gives

$$n = 10$$

Therefore, the value of $n$ is $10$.

Given: Success occurs when the die shows any of $1,2,3,\dots,n-1$.

Find: $n$ from the given mean $\frac{n}{9}$.

The probability that the process stops on the first toss is

$$\frac{n-1}{n}$$

The probability that it stops on the second toss is

$$\frac{1}{n} \cdot \frac{n-1}{n}$$

The probability that it stops on the third toss is

$$\left( \frac{1}{n} \right)^2 \cdot \frac{n-1}{n}$$

Thus the expectation becomes

$$E(X) = \sum_{k=1}^{\infty} k \left( \frac{1}{n} \right)^{k-1} \frac{n-1}{n}$$

This matches the expanded series used above. Evaluating it gives

$$E(X) = \frac{n-1}{n} \cdot \frac{1}{\left(1-\frac{1}{n}\right)^2} = \frac{n}{n-1}$$

Since this is equal to $\frac{n}{9}$,

$$\frac{n}{9} = \frac{n}{n-1}$$

Cancelling $n \neq 0$,

$$\frac{1}{9} = \frac{1}{n-1}$$

Hence,

$$n-1 = 9 \Rightarrow n = 10$$

So the required numerical value is $10$.