There are n urns each containing (n+1) balls such that ith urn contains i white balls and (n+1-i) red balls. Let u_i be the event of selecting ith urn, i=1,2,3…, n and w denotes the event of getting a white ball. IfP(u_i)=c, where c is a constant then P(u_n/w) is equal to

Answer

To solve the problem, we need to find \( P(u_n | w) \), the probability of selecting the nth urn given that a white ball is drawn. We can use Bayes' theorem for this purpose.

### Step-by-step Solution:

1. **Understanding the Events**:
- Let \( u_i \) be the event of selecting the ith urn.
- Let \( w \) be the event of drawing a white ball.
- Each urn \( i \) contains \( i \) white balls and \( (n + 1 - i) \) red balls.

2. **Total Balls in Each Urn**:
- Each urn contains \( n + 1 \) balls (i.e., \( i \) white + \( (n + 1 - i) \) red).

3. **Calculating \( P(w | u_i) \)**:
- The probability of drawing a white ball from the ith urn is given by:
\[ P(w | u_i) = \frac{i}{n + 1} \]

4. **Calculating \( P(w) \)**:
- To find \( P(w) \), we need to consider all urns:
\[ P(w) = \sum_{i=1}^{n} P(w | u_i) P(u_i) \]
- Given \( P(u_i) = c \) (a constant), we have:
\[ P(w) = \sum_{i=1}^{n} \frac{i}{n + 1} \cdot c = \frac{c}{n + 1} \sum_{i=1}^{n} i \]
- The sum \( \sum_{i=1}^{n} i = \frac{n(n + 1)}{2} \), so:
\[ P(w) = \frac{c}{n + 1} \cdot \frac{n(n + 1)}{2} = \frac{cn}{2} \]

5. **Applying Bayes' Theorem**:
- By Bayes' theorem:
\[ P(u_n | w) = \frac{P(w | u_n) P(u_n)}{P(w)} \]
- We already know:
- \( P(w | u_n) = \frac{n}{n + 1} \)
- \( P(u_n) = c \)
- \( P(w) = \frac{cn}{2} \)
- Substituting these values:
\[ P(u_n | w) = \frac{\frac{n}{n + 1} \cdot c}{\frac{cn}{2}} = \frac{n \cdot c \cdot 2}{(n + 1) \cdot c \cdot n} \]
- The \( c \) and \( n \) cancel out:
\[ P(u_n | w) = \frac{2}{n + 1} \]

### Final Answer:
Thus, \( P(u_n | w) = \frac{2}{n + 1} \).