In a class of 10 student, probability of exactly I students passing an examination is directly proportional to `i^(2).` Then answer the following questions:
The probability that exactly 5 students passing an examination is

Options

A. `1//11`
B. `5//77`
C. `25//77`
D. `10//77`

Explanation

To solve the problem step by step, we need to determine the probability that exactly 5 students pass the examination given that the probability of exactly i students passing is directly proportional to \( i^2 \).

### Step 1: Define the Probability Function
Let \( P(i) \) be the probability that exactly \( i \) students pass the examination. Since it is given that \( P(i) \) is directly proportional to \( i^2 \), we can express this as:
\[ P(i) = \lambda i^2 \]
where \( \lambda \) is a constant of proportionality.

### Step 2: Normalize the Probability
Since the total probability must sum to 1, we need to find \( \lambda \) such that:
\[ \sum_{i=0}^{10} P(i) = 1 \]
This gives us:
\[ \sum_{i=0}^{10} \lambda i^2 = 1 \]
Calculating the sum of squares from 0 to 10:
\[ \sum_{i=0}^{10} i^2 = 0^2 + 1^2 + 2^2 + \ldots + 10^2 = \frac{10(10 + 1)(2 \cdot 10 + 1)}{6} = \frac{10 \cdot 11 \cdot 21}{6} = 385 \]
Thus, we have:
\[ \lambda \cdot 385 = 1 \implies \lambda = \frac{1}{385} \]

### Step 3: Calculate the Probability for Exactly 5 Students
Now, we can find the probability that exactly 5 students pass the examination:
\[ P(5) = \lambda \cdot 5^2 = \frac{1}{385} \cdot 25 = \frac{25}{385} \]

### Step 4: Simplify the Probability
We can simplify \( \frac{25}{385} \):
\[ \frac{25}{385} = \frac{5}{77} \]

### Final Answer
The probability that exactly 5 students pass the examination is:
\[ \frac{5}{77} \]