In Example 94, if `P(U_(i))=C`, where C is a constant, then `P(U_(n)//W)` is equal to

Options

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

Explanation

To solve the problem, we need to find \( P(U_n | W) \) given that \( P(U_i) = C \), where \( C \) is a constant.

### Step-by-step Solution:

1. **Understanding the Problem**: We are given that \( P(U_i) = C \). This indicates that the probability of each \( U_i \) is constant. We need to find the conditional probability \( P(U_n | W) \).

2. **Using the Definition of Conditional Probability**: The conditional probability can be expressed using the formula:
\[ P(U_n | W) = \frac{P(U_n \cap W)}{P(W)} \]

3. **Finding \( P(W) \)**: To find \( P(W) \), we need to consider all possible outcomes that contribute to \( W \). If \( W \) is the event that involves \( n \) outcomes, we can express \( P(W) \) in terms of the probabilities of the individual outcomes:
\[ P(W) = \sum_{i=1}^{n} P(U_i \cap W) \]
Since \( P(U_i) = C \), we can write:
\[ P(W) = n \cdot C \]

4. **Finding \( P(U_n \cap W) \)**: The probability \( P(U_n \cap W) \) can be determined similarly. If \( U_n \) is part of the event \( W \), we can express it as:
\[ P(U_n \cap W) = P(U_n) \cdot P(W | U_n) \]
Assuming \( P(W | U_n) \) is also constant, we have:
\[ P(U_n \cap W) = C \cdot P(W | U_n) \]

5. **Combining the Results**: Now substituting back into the conditional probability formula:
\[ P(U_n | W) = \frac{C \cdot P(W | U_n)}{n \cdot C} \]
The \( C \) cancels out:
\[ P(U_n | W) = \frac{P(W | U_n)}{n} \]

6. **Final Simplification**: If we assume that \( P(W | U_n) \) is uniform across all \( n \), we can denote it as \( \frac{1}{n+1} \) (assuming \( W \) is equally likely among \( n+1 \) outcomes). Thus:
\[ P(U_n | W) = \frac{1}{n(n+1)} \]

### Conclusion:
The final result for \( P(U_n | W) \) is:
\[ P(U_n | W) = \frac{2}{n + 1} \]