Suppose n students appear in an examination. Let X = the number of students who pass the examination. Suppose `P(X=k)=lambdak^2`, then value of `lambda` is

Options

A. `2/(n(n+1))`
B. `6/(n(n+1)(2n+1))`
C. `4/(n^2(n+1)^2)`
D. `(2n+1)/(n^2(n+1)^2)`

Answer

To solve the problem, we need to find the value of \( \lambda \) given the probability distribution of the number of students passing the examination. The probability that \( X = k \) is given as \( P(X = k) = \lambda k^2 \).

### Step-by-Step Solution:

1. **Understanding the Probability Distribution**:
We know that \( P(X = k) = \lambda k^2 \) for \( k = 0, 1, 2, \ldots, n \). The total probability must sum to 1:
\[ \sum_{k=0}^{n} P(X = k) = 1 \]

2. **Writing the Total Probability**:
We can express the total probability as:
\[ \sum_{k=0}^{n} P(X = k) = \sum_{k=0}^{n} \lambda k^2 = \lambda \sum_{k=0}^{n} k^2 \]

3. **Calculating the Sum of Squares**:
The sum of squares of the first \( n \) natural numbers is given by the formula:
\[ \sum_{k=1}^{n} k^2 = \frac{n(n + 1)(2n + 1)}{6} \]
Note that \( \sum_{k=0}^{n} k^2 = 0^2 + \sum_{k=1}^{n} k^2 = 0 + \frac{n(n + 1)(2n + 1)}{6} \).

4. **Setting Up the Equation**:
We can now set up the equation:
\[ \lambda \cdot \frac{n(n + 1)(2n + 1)}{6} = 1 \]

5. **Solving for \( \lambda \)**:
Rearranging the equation gives:
\[ \lambda = \frac{6}{n(n + 1)(2n + 1)} \]

### Final Answer:
Thus, the value of \( \lambda \) is:
\[ \lambda = \frac{6}{n(n + 1)(2n + 1)} \]