1. Key Concepts and Formulas

• Geometric Distribution: This distribution models the number of independent Bernoulli trials required to achieve the first success. If $X$ is the random variable representing the number of trials, and $p$ is the probability of success in a single trial, then the probability mass function (PMF) is $P(X=k) = q^{k-1}p$, where $q = 1-p$ is the probability of failure, and $k = 1, 2, 3, \ldots$.
• Mean of Geometric Distribution: The expected value (mean) of a Geometric Distribution is given by $E[X] = \frac{1}{p}$. This formula is crucial for efficiency in solving such problems.
• Sum of Infinite Geometric Progression (GP): For a GP with first term $a$ and common ratio $r$ where $|r|<1$, the sum to infinity is $S_{\infty} = \frac{a}{1-r}$. This will be used in deriving the mean from first principles.

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2. Step-by-Step Solution

Step 1: Identify the Geometric Distribution and Define Probabilities

• What we are doing: We need to recognize that the problem describes a sequence of independent trials (rolling a die repeatedly) until a specific event (getting a number less than $n$) occurs for the first time. This fits the definition of a Geometric Distribution. We will then define "success" and "failure" and calculate their probabilities, $p$ and $q$.

• Why we are doing it: Correctly identifying the distribution and its parameters ($p$ and $q$) is the foundational step for solving any probability problem of this type.

• Defining Success: The problem states the process stops when "a number less than $n$ appears."

  • Favorable outcomes (numbers less than $n$): $\{1, 2, \ldots, n-1\}$. There are $(n-1)$ such outcomes.
  • Total possible outcomes when rolling an $n$-faced die: $\{1, 2, \ldots, n\}$. There are $n$ such outcomes.

• Calculating Probability of Success ($p$): $p = P(\text{getting a number less than } n) = \frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \frac{n-1}{n}$

• Calculating Probability of Failure ($q$): A failure means getting a number that is not less than $n$. The only number on the die that satisfies this is $n$. Alternatively, we use $q=1-p$: $q = 1 - p = 1 - \frac{n-1}{n} = \frac{n-(n-1)}{n} = \frac{1}{n}$ Since $n > 1$ (given), $p$ is always positive and $q$ is always between 0 and 1, ensuring a valid probability distribution.

Step 2: State the Mean of the Geometric Distribution

• What we are doing: We recall the standard formula for the mean (expected value) of a Geometric Distribution.
• Why we are doing it: This formula provides a direct way to calculate the mean once $p$ is known, significantly simplifying the problem.

Let $X$ be the number of tosses required. $X$ follows a Geometric Distribution. The mean of $X$ is given by: $E[X] = \frac{1}{p}$

Step 3: (Optional Derivation) Derive the Mean of the Geometric Distribution

• What we are doing: Although we know the formula $E[X] = 1/p$, the original solution included its derivation using series summation. This demonstrates a deeper understanding of expected values and series.
• Why we are doing it: This derivation is a valuable skill for JEE, as some problems might require summing such series from first principles or when the formula isn't immediately obvious.

The expected value $E[X]$ is defined as $E[X] = \sum_{k=1}^{\infty} k \cdot P(X=k)$. Using $P(X=k) = q^{k-1}p$: $E[X] = \sum_{k=1}^{\infty} k \cdot (q^{k-1}p)$ Factor out $p$: $E[X] = p \sum_{k=1}^{\infty} k q^{k-1} = p (1 \cdot q^0 + 2 \cdot q^1 + 3 \cdot q^2 + \ldots)$ Let $S' = 1 + 2q + 3q^2 + 4q^3 + \ldots \quad \ldots(A)$ This is an Arithmetico-Geometric Progression (AGP). To sum it, we multiply by $q$ and subtract: $qS' = \quad q + 2q^2 + 3q^3 + \ldots \quad \ldots(B)$ Subtract (B) from (A): $S' - qS' = 1 + (2q-q) + (3q^2-2q^2) + (4q^3-3q^3) + \ldots$ $S'(1-q) = 1 + q + q^2 + q^3 + \ldots$ The right-hand side is an infinite Geometric Progression with first term $a=1$ and common ratio $q$. Since $q = \frac{1}{n}$ and $n>1$, we have $0 < q < 1$, so the sum converges to $\frac{1}{1-q}$. $S'(1-q) = \frac{1}{1-q}$ Solving for $S'$: $S' = \frac{1}{(1-q)^2}$ Now substitute $S'$ back into the expression for $E[X]$: $E[X] = p \cdot S' = p \cdot \frac{1}{(1-q)^2}$ Since $p = 1-q$, we replace $(1-q)$ with $p$: $E[X] = p \cdot \frac{1}{p^2} = \frac{1}{p}$ This confirms the formula for the mean of a Geometric Distribution.

Step 4: Express the Mean in terms of $n$

• What we are doing: We substitute the specific probability of success, $p = \frac{n-1}{n}$, into the mean formula $E[X] = \frac{1}{p}$.
• Why we are doing it: This allows us to have an expression for the mean that depends only on $n$, which we can then use to solve for $n$.

Using $E[X] = \frac{1}{p}$ and $p = \frac{n-1}{n}$: $E[X] = \frac{1}{\frac{n-1}{n}}$ $E[X] = \frac{n}{n-1}$

Step 5: Use the Given Mean to Solve for $n$

• What we are doing: The problem states that the mean number of tosses is $\frac{n}{9}$. We will equate our derived expression for $E[X]$ with this given value and solve for $n$.
• Why we are doing it: This is the final algebraic step to find the value of $n$ as required by the question.

We have $E[X] = \frac{n}{n-1}$ and the problem states $E[X] = \frac{n}{9}$. Equating these two expressions: $\frac{n}{n-1} = \frac{n}{9}$ Since $n > 1$ (given in the problem), we know $n \neq 0$. Therefore, we can safely divide both sides by $n$: $\frac{1}{n-1} = \frac{1}{9}$ Taking the reciprocal of both sides: $n-1 = 9$ Solving for $n$: $n = 9 + 1$ $n = 10$ This value $n=10$ satisfies the condition $n>1$.

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3. Common Mistakes & Tips

• Incorrectly Defining $p$: The most frequent error is miscalculating the probability of "success." Always clearly define what constitutes a success and a failure based on the problem statement.
• Forgetting the Mean Formula: While deriving the mean ($E[X] = 1/p$) is good practice, knowing it directly can save significant time in JEE, especially when time is critical.
• Algebraic Errors: Be careful with fractions and algebraic manipulations, particularly when solving the final equation. Double-check your steps.

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4. Summary

This problem effectively tests the understanding and application of the Geometric Distribution. We first identified the problem as a Geometric Distribution scenario and accurately determined the probability of success, $p = \frac{n-1}{n}$, and failure, $q = \frac{1}{n}$. We then utilized the formula for the mean of a Geometric Distribution, $E[X] = \frac{1}{p}$, substituting the value of $p$ to express the mean in terms of $n$ as $E[X] = \frac{n}{n-1}$. Finally, by equating this derived mean with the given mean of $\frac{n}{9}$, we solved the resulting algebraic equation to find $n=10$.

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5. Final Answer

The final answer is $\boxed{10}$.