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 ball. If `P(U_(i)) prop` i, where i=1,2,3,..,n, then `lim_(n to oo) P(W)` is equal to

Options

A. `1`
B. `(2)/(3)`
C. `(1)/(4)`
D. `(3)/(4)`

Explanation

To solve the problem step by step, we will analyze the given information and apply the concepts of probability.

### Step 1: Define the Events
Let \( U_i \) be the event of selecting the \( i^{th} \) urn, where \( i = 1, 2, \ldots, n \). The \( i^{th} \) urn contains \( i \) white balls and \( (n + 1 - i) \) red balls.

### Step 2: Probability of Selecting an Urn
The problem states that the probability of selecting the \( i^{th} \) urn, \( P(U_i) \), is proportional to \( i \). Thus, we can write:
\[ P(U_i) = \lambda i \]
for some constant \( \lambda \).

### Step 3: Normalize the Probabilities
Since the sum of the probabilities must equal 1, we have:
\[ \sum_{i=1}^{n} P(U_i) = 1 \]
Substituting our expression for \( P(U_i) \):
\[ \sum_{i=1}^{n} \lambda i = 1 \]
The sum of the first \( n \) natural numbers is given by:
\[ \sum_{i=1}^{n} i = \frac{n(n + 1)}{2} \]
Thus, we can write:
\[ \lambda \cdot \frac{n(n + 1)}{2} = 1 \]
Solving for \( \lambda \):
\[ \lambda = \frac{2}{n(n + 1)} \]

### Step 4: Calculate the Probability of Getting a White Ball
Now, we need to find \( P(W) \), the probability of drawing a white ball. We use the law of total probability:
\[ P(W) = \sum_{i=1}^{n} P(W | U_i) P(U_i) \]
The probability of drawing a white ball from the \( i^{th} \) urn is:
\[ P(W | U_i) = \frac{i}{n + 1} \]
Substituting \( P(U_i) \) and \( P(W | U_i) \):
\[ P(W) = \sum_{i=1}^{n} \frac{i}{n + 1} \cdot \lambda i \]
This simplifies to:
\[ P(W) = \frac{1}{n + 1} \sum_{i=1}^{n} \lambda i^2 \]
Substituting \( \lambda = \frac{2}{n(n + 1)} \):
\[ P(W) = \frac{2}{n(n + 1)(n + 1)} \sum_{i=1}^{n} i^2 \]
The sum of the squares of the first \( n \) natural numbers is given by:
\[ \sum_{i=1}^{n} i^2 = \frac{n(n + 1)(2n + 1)}{6} \]
Substituting this into our expression for \( P(W) \):
\[ P(W) = \frac{2}{n(n + 1)(n + 1)} \cdot \frac{n(n + 1)(2n + 1)}{6} \]
This simplifies to:
\[ P(W) = \frac{2(2n + 1)}{6(n + 1)} = \frac{2n + 1}{3(n + 1)} \]

### Step 5: Take the Limit as \( n \to \infty \)
Now we need to find:
\[ \lim_{n \to \infty} P(W) = \lim_{n \to \infty} \frac{2n + 1}{3(n + 1)} \]
Dividing the numerator and the denominator by \( n \):
\[ = \lim_{n \to \infty} \frac{2 + \frac{1}{n}}{3 + \frac{3}{n}} = \frac{2 + 0}{3 + 0} = \frac{2}{3} \]

### Final Answer
Thus, the limit of \( P(W) \) as \( n \) approaches infinity is:
\[ \lim_{n \to \infty} P(W) = \frac{2}{3} \]