A purse contain n coins of unknown values .a coin is drawn from it at random and is found to be a rupee .Then the chance that it is the only rupee coin in the purse is

Options

A. `(1)/(n)`
B. `(2)/(n+1)`
C. `(2)/((n(n+1)))`
D. `(2)/((n+1))`

Answer

To solve the problem step by step, we need to find the probability that there is only one rupee coin in the purse after drawing a coin and finding it to be a rupee coin.

### Step-by-Step Solution:

1. **Understanding the Problem**:
We have a purse containing \( n \) coins of unknown values. We draw one coin at random, and it turns out to be a rupee coin. We need to find the probability that this is the only rupee coin in the purse.

2. **Defining Events**:
- Let \( A \) be the event that there is only one rupee coin in the purse.
- Let \( B \) be the event that a rupee coin is drawn.

3. **Using Conditional Probability**:
We want to find \( P(A|B) \), the probability that there is only one rupee coin given that we have drawn a rupee coin. By Bayes' theorem, we have:
\[ P(A|B) = \frac{P(B|A) \cdot P(A)}{P(B)} \]

4. **Calculating \( P(B|A) \)**:
If there is only one rupee coin in the purse, the probability of drawing it (event \( B \)) is:
\[ P(B|A) = 1 \]
(since if there is only one rupee coin, drawing it will always yield a rupee).

5. **Calculating \( P(A) \)**:
The probability of having exactly one rupee coin among \( n \) coins can be considered as follows:
- If there is one rupee coin, the other \( n-1 \) coins can be any non-rupee coins.
- The probability of having exactly one rupee coin among \( n \) coins is \( \frac{1}{n} \).

6. **Calculating \( P(B) \)**:
To find \( P(B) \), we need to consider the cases where there could be \( k \) rupee coins (where \( k \) can be from 1 to \( n \)):
- If there is 1 rupee coin, the probability of drawing it is \( \frac{1}{n} \).
- If there are 2 rupee coins, the probability of drawing a rupee coin is \( \frac{2}{n} \).
- If there are 3 rupee coins, the probability is \( \frac{3}{n} \), and so on.

Thus, the total probability \( P(B) \) can be computed as:
\[ P(B) = \frac{1}{n} + \frac{2}{n} + \frac{3}{n} + \ldots + \frac{n}{n} = \frac{1 + 2 + 3 + \ldots + n}{n} = \frac{\frac{n(n+1)}{2}}{n} = \frac{n+1}{2} \]

7. **Putting It All Together**:
Now substituting back into Bayes' theorem:
\[ P(A|B) = \frac{P(B|A) \cdot P(A)}{P(B)} = \frac{1 \cdot \frac{1}{n}}{\frac{n+1}{2}} = \frac{2}{n(n+1)} \]

### Final Answer:
The probability that there is only one rupee coin in the purse given that a rupee coin was drawn is:
\[ P(A|B) = \frac{2}{n(n+1)} \]