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

Options

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

Answer

To solve the problem, we need to find \( P(U_n | W) \), which is the conditional probability of selecting the \( n^{th} \) urn given that we have drawn a white ball. We will use the formula for conditional probability:

\[ P(U_n | W) = \frac{P(U_n \cap W)}{P(W)} \]

### Step 1: Calculate \( P(W) \)

First, we need to find \( P(W) \), the total probability of drawing a white ball from any urn.

1. **Probability of selecting the \( i^{th} \) urn**: Given \( P(U_i) = c \) for each urn \( i \).
2. **Probability of drawing a white ball from the \( i^{th} \) urn**: The \( i^{th} \) urn contains \( i \) white balls and \( (n + 1 - i) \) red balls, so the total number of balls in the \( i^{th} \) urn is \( n + 1 \). Therefore, the probability of drawing a white ball from the \( i^{th} \) urn is \( \frac{i}{n + 1} \).

Thus, the total probability \( P(W) \) can be calculated as follows:

\[ P(W) = \sum_{i=1}^{n} P(U_i) \cdot P(W | U_i) = \sum_{i=1}^{n} c \cdot \frac{i}{n + 1} \]

This simplifies to:

\[ P(W) = c \cdot \frac{1}{n + 1} \sum_{i=1}^{n} i \]

Using the formula for the sum of the first \( n \) natural numbers, we have:

\[ \sum_{i=1}^{n} i = \frac{n(n + 1)}{2} \]

Substituting this back, we get:

\[ P(W) = c \cdot \frac{1}{n + 1} \cdot \frac{n(n + 1)}{2} = \frac{cn}{2} \]

### Step 2: Calculate \( P(U_n \cap W) \)

Next, we find \( P(U_n \cap W) \), which is the probability of selecting the \( n^{th} \) urn and drawing a white ball.

1. **Probability of selecting the \( n^{th} \) urn**: \( P(U_n) = c \).
2. **Probability of drawing a white ball from the \( n^{th} \) urn**: The \( n^{th} \) urn contains \( n \) white balls, so the probability is \( \frac{n}{n + 1} \).

Thus, we have:

\[ P(U_n \cap W) = P(U_n) \cdot P(W | U_n) = c \cdot \frac{n}{n + 1} \]

### Step 3: Calculate \( P(U_n | W) \)

Now we can substitute \( P(U_n \cap W) \) and \( P(W) \) into the conditional probability formula:

\[ P(U_n | W) = \frac{P(U_n \cap W)}{P(W)} = \frac{c \cdot \frac{n}{n + 1}}{\frac{cn}{2}} \]

The \( c \) and \( n \) terms cancel out:

\[ P(U_n | W) = \frac{2}{n + 1} \]

### Final Answer

Thus, the required probability is:

\[ P(U_n | W) = \frac{2}{n + 1} \]