In Example 94, if n is even and E denotes the event of choosing even numbered urn `(p(U_(i))=(1)/(n))`, then the value of `P(W//E)`, is

Options

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

Explanation

To solve the problem, we need to find the value of \( P(W \mid E) \), where \( E \) is the event of choosing an even-numbered urn and \( P(U_i) = \frac{1}{n} \) for \( i = 1, 2, \ldots, n \).

### Step-by-Step Solution:

1. **Understanding the Events**:
- Let \( E \) denote the event of choosing an even-numbered urn. Since \( n \) is even, the even-numbered urns are \( U_2, U_4, U_6, \ldots, U_n \).
- The total number of even-numbered urns is \( \frac{n}{2} \).

2. **Using the Conditional Probability Formula**:
- The conditional probability \( P(W \mid E) \) can be expressed using the formula:
\[ P(W \mid E) = \frac{P(W \cap E)}{P(E)} \]

3. **Calculating \( P(E) \)**:
- Since the probability of choosing any urn \( U_i \) is \( \frac{1}{n} \), the probability of choosing any of the even-numbered urns is:
\[ P(E) = P(U_2) + P(U_4) + \ldots + P(U_n) = \frac{1}{n} \times \frac{n}{2} = \frac{1}{2} \]

4. **Calculating \( P(W \cap E) \)**:
- We can express \( P(W \cap E) \) as the sum of the probabilities of \( W \) occurring in each of the even-numbered urns:
\[ P(W \cap E) = P(W \cap U_2) + P(W \cap U_4) + \ldots + P(W \cap U_n) \]
- Assuming independence, we can write:
\[ P(W \cap U_i) = P(U_i) \cdot P(W \mid U_i) \]
- Thus, we have:
\[ P(W \cap E) = \sum_{i=1}^{n/2} P(U_{2i}) \cdot P(W \mid U_{2i}) = \sum_{i=1}^{n/2} \left(\frac{1}{n} \cdot P(W \mid U_{2i})\right) \]

5. **Summing Up the Probabilities**:
- If we denote \( P(W \mid U_{2i}) \) as \( \frac{2i}{n+1} \) (as derived from the problem context), we can express:
\[ P(W \cap E) = \frac{1}{n} \left( \frac{2}{n+1} + \frac{4}{n+1} + \ldots + \frac{n}{n+1} \right) \]
- The sum \( 2 + 4 + \ldots + n = 2(1 + 2 + \ldots + \frac{n}{2}) = 2 \cdot \frac{n/2(n/2 + 1)}{2} = \frac{n(n + 2)}{4} \).

6. **Final Calculation**:
- Therefore, we can substitute back into our equation:
\[ P(W \cap E) = \frac{1}{n} \cdot \frac{n(n + 2)}{4(n + 1)} = \frac{n + 2}{4(n + 1)} \]

7. **Calculating \( P(W \mid E) \)**:
- Now substituting \( P(W \cap E) \) and \( P(E) \) into the conditional probability formula:
\[ P(W \mid E) = \frac{P(W \cap E)}{P(E)} = \frac{\frac{n + 2}{4(n + 1)}}{\frac{1}{2}} = \frac{n + 2}{2(n + 1)} \]

### Final Answer:
\[ P(W \mid E) = \frac{n + 2}{2(n + 1)} \]