There are n urns `U_(i),(i=1,. . . .,n)` which contains `i` white and (n+1-i) black balls. `U_(i)` is even of selecting the jth urn, W event of getting white ball from the selected urn, E is event of getting even number of balls from selected urn.
Q. `P(U_(i))=(1)/(n)`, and one urn is selected at random, then P(W) is:

Options

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

Answer

To solve the problem step by step, we will calculate the probability of selecting a white ball from the selected urn.

### Step-by-Step Solution:

1. **Understanding the Urns**:
- There are `n` urns, `U_i` where `i = 1, 2, ..., n`.
- Each urn `U_i` contains `i` white balls and `(n + 1 - i)` black balls.
- The total number of balls in urn `U_i` is `i + (n + 1 - i) = n + 1`.

2. **Probability of Selecting an Urn**:
- The probability of selecting any urn `U_i` is given as:
\[ P(U_i) = \frac{1}{n} \]

3. **Finding the Total Probability of Getting a White Ball (P(W))**:
- We need to find the probability of getting a white ball from a randomly selected urn. This can be calculated using the law of total probability:
\[ P(W) = \sum_{i=1}^{n} P(W | U_i) \cdot P(U_i) \]
- Here, \( P(W | U_i) \) is the probability of getting a white ball from urn `U_i`.

4. **Calculating P(W | U_i)**:
- The probability of drawing a white ball from urn `U_i` is:
\[ P(W | U_i) = \frac{\text{Number of white balls in } U_i}{\text{Total number of balls in } U_i} = \frac{i}{n + 1} \]

5. **Substituting into the Total Probability**:
- Now substituting \( P(W | U_i) \) and \( P(U_i) \) into the equation for \( P(W) \):
\[ P(W) = \sum_{i=1}^{n} \frac{i}{n + 1} \cdot \frac{1}{n} \]
- This simplifies to:
\[ P(W) = \frac{1}{n(n + 1)} \sum_{i=1}^{n} i \]

6. **Using the Formula for the Sum of First n Natural Numbers**:
- The sum of the first `n` natural numbers is given by:
\[ \sum_{i=1}^{n} i = \frac{n(n + 1)}{2} \]
- Substituting this into the equation for \( P(W) \):
\[ P(W) = \frac{1}{n(n + 1)} \cdot \frac{n(n + 1)}{2} = \frac{1}{2} \]

### Final Result:
Thus, the probability of getting a white ball from a randomly selected urn is:
\[ P(W) = \frac{1}{2} \]