There are n students in a class. Ler `P(E_lambda)` be the probability that exactly `lambda` out of n pass the examination. If `P(E_lambda)` is directly proportional to `lambda^2(0lelambdalen)`.
If a selected student has been found to pass the examination, then the probability that he is the only student to have passed the examination, is

Options

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

Answer

To solve the problem, we need to find the probability that a selected student who has passed the examination is the only student to have passed, given that there are \( n \) students in total and the probability of exactly \( \lambda \) students passing is proportional to \( \lambda^2 \).

### Step-by-Step Solution:

1. **Define the Probability Function**:
Given that \( P(E_\lambda) \) is directly proportional to \( \lambda^2 \), we can express this as:
\[ P(E_\lambda) = k \lambda^2 \]
where \( k \) is a constant.

2. **Normalization Condition**:
The sum of probabilities for all possible outcomes (from \( \lambda = 0 \) to \( \lambda = n \)) must equal 1:
\[ \sum_{\lambda=0}^{n} P(E_\lambda) = 1 \]
This gives us:
\[ P(E_0) + P(E_1) + P(E_2) + \ldots + P(E_n) = k(0^2 + 1^2 + 2^2 + \ldots + n^2) = 1 \]

3. **Sum of Squares Formula**:
The sum of the squares of the first \( n \) natural numbers is given by:
\[ 1^2 + 2^2 + \ldots + n^2 = \frac{n(n+1)(2n+1)}{6} \]
Therefore, we have:
\[ k \cdot \frac{n(n+1)(2n+1)}{6} = 1 \]
Solving for \( k \):
\[ k = \frac{6}{n(n+1)(2n+1)} \]

4. **Calculate \( P(E_\lambda) \)**:
Substitute \( k \) back into the probability function:
\[ P(E_\lambda) = \frac{6}{n(n+1)(2n+1)} \lambda^2 \]

5. **Find the Probability of Passing**:
The probability that a randomly selected student has passed the examination, \( P(A) \), can be calculated using the law of total probability:
\[ P(A) = \sum_{\lambda=1}^{n} P(E_\lambda) \cdot P(A | E_\lambda) \]
Given that if \( \lambda \) students pass, the probability that a specific student is among them is \( \frac{\lambda}{n} \):
\[ P(A) = \sum_{\lambda=1}^{n} P(E_\lambda) \cdot \frac{\lambda}{n} \]
Substitute \( P(E_\lambda) \):
\[ P(A) = \sum_{\lambda=1}^{n} \frac{6}{n(n+1)(2n+1)} \lambda^2 \cdot \frac{\lambda}{n} = \frac{6}{n^2(n+1)(2n+1)} \sum_{\lambda=1}^{n} \lambda^3 \]
The sum of cubes is:
\[ \sum_{\lambda=1}^{n} \lambda^3 = \left( \frac{n(n+1)}{2} \right)^2 \]
Therefore:
\[ P(A) = \frac{6}{n^2(n+1)(2n+1)} \cdot \left( \frac{n(n+1)}{2} \right)^2 = \frac{6n(n+1)}{4(2n+1)} = \frac{3n(n+1)}{2(2n+1)} \]

6. **Find the Conditional Probability**:
We need the probability that exactly one student passed given that the selected student passed:
\[ P(E_1 | A) = \frac{P(A | E_1) P(E_1)}{P(A)} \]
Where \( P(A | E_1) = 1 \) (if exactly one student passed, the selected student must be that one) and \( P(E_1) = \frac{6}{n(n+1)(2n+1)} \cdot 1^2 = \frac{6}{n(n+1)(2n+1)} \):
\[ P(E_1 | A) = \frac{1 \cdot \frac{6}{n(n+1)(2n+1)}}{\frac{3n(n+1)}{2(2n+1)}} = \frac{6}{n(n+1)(2n+1)} \cdot \frac{2(2n+1)}{3n(n+1)} = \frac{12}{3n^2} = \frac{4}{n^2} \]

### Final Answer:
The probability that a selected student who has passed the examination is the only student to have passed is:
\[ \frac{4}{n^2} \]