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)`.
Proportional constant k is equal to

Options

A. 0.25
B. 0.5
C. 0.75
D. 0.35

Answer

To find the proportional constant \( k \) for the probability \( P(E_\lambda) \) that exactly \( \lambda \) out of \( n \) students pass the examination, we can follow these steps:

### Step-by-Step Solution

1. **Understanding the Proportionality**:
We know that \( P(E_\lambda) \) is directly proportional to \( \lambda^2 \). This can be expressed mathematically as:
\[ P(E_\lambda) = k \cdot \lambda^2 \]
where \( k \) is the proportional constant we need to determine.

2. **Considering All Possible Outcomes**:
The total probability of all possible outcomes (from 0 to \( n \) students passing) must equal 1. Therefore, we can write:
\[ P(E_0) + P(E_1) + P(E_2) + \ldots + P(E_n) = 1 \]

3. **Substituting the Proportional Expression**:
Substituting the expression for \( P(E_\lambda) \) into the total probability equation gives us:
\[ k \cdot 0^2 + k \cdot 1^2 + k \cdot 2^2 + \ldots + k \cdot n^2 = 1 \]
This simplifies to:
\[ k \cdot (0^2 + 1^2 + 2^2 + \ldots + n^2) = 1 \]

4. **Using the Formula for the Sum of Squares**:
The sum of the squares of the first \( n \) natural numbers is given by the formula:
\[ \sum_{i=0}^{n} i^2 = \frac{n(n+1)(2n+1)}{6} \]
Therefore, we can substitute this into our equation:
\[ k \cdot \frac{n(n+1)(2n+1)}{6} = 1 \]

5. **Solving for \( k \)**:
To find \( k \), we rearrange the equation:
\[ k = \frac{6}{n(n+1)(2n+1)} \]

### Final Result
Thus, the proportional constant \( k \) is:
\[ k = \frac{6}{n(n+1)(2n+1)} \]