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:
If a students selected at random is found to have passed the examination, then the probability that he was the only student who has passed the examination is

Options

A. `1//3025`
B. `1//605`
C. `1//275`
D. `1//121`

Explanation

To solve the problem, we need to find the probability that a student who has passed the examination is the only student who passed, given that the probability of exactly \( i \) students passing is proportional to \( i^2 \).

### Step-by-Step Solution:

1. **Define the Probability Function**:
Let \( P(E_i) \) be the probability of exactly \( i \) students passing the examination. According to the problem, this probability is directly proportional to \( i^2 \). Therefore, we can express this as:
\[ P(E_i) = k \cdot i^2 \]
where \( k \) is a constant of proportionality.

2. **Calculate Total Probability**:
The total probability for all possible outcomes (from 0 to 10 students passing) must equal 1. Thus, we have:
\[ P(E_0) + P(E_1) + P(E_2) + \ldots + P(E_{10}) = 1 \]
Substituting the expression for \( P(E_i) \):
\[ k \cdot (0^2 + 1^2 + 2^2 + \ldots + 10^2) = 1 \]
The sum of squares from 0 to 10 can be calculated as:
\[ 0^2 + 1^2 + 2^2 + \ldots + 10^2 = \frac{10(10 + 1)(2 \cdot 10 + 1)}{6} = \frac{10 \cdot 11 \cdot 21}{6} = 385 \]
Thus, we have:
\[ k \cdot 385 = 1 \implies k = \frac{1}{385} \]

3. **Find Individual Probabilities**:
Now we can find the probabilities for each case:
\[ P(E_0) = k \cdot 0^2 = 0 \]
\[ P(E_1) = k \cdot 1^2 = \frac{1}{385} \]
\[ P(E_2) = k \cdot 2^2 = \frac{4}{385} \]
\[ P(E_3) = k \cdot 3^2 = \frac{9}{385} \]
\[ P(E_4) = k \cdot 4^2 = \frac{16}{385} \]
\[ P(E_5) = k \cdot 5^2 = \frac{25}{385} \]
\[ P(E_6) = k \cdot 6^2 = \frac{36}{385} \]
\[ P(E_7) = k \cdot 7^2 = \frac{49}{385} \]
\[ P(E_8) = k \cdot 8^2 = \frac{64}{385} \]
\[ P(E_9) = k \cdot 9^2 = \frac{81}{385} \]
\[ P(E_{10}) = k \cdot 10^2 = \frac{100}{385} \]

4. **Calculate \( P(A) \)**:
Let \( A \) be the event that a student passes the examination. The probability of event \( A \) is given by:
\[ P(A) = P(E_1) + P(E_2) + P(E_3) + \ldots + P(E_{10}) \]
\[ P(A) = \frac{1}{385} + \frac{4}{385} + \frac{9}{385} + \frac{16}{385} + \frac{25}{385} + \frac{36}{385} + \frac{49}{385} + \frac{64}{385} + \frac{81}{385} + \frac{100}{385} \]
\[ P(A) = \frac{1 + 4 + 9 + 16 + 25 + 36 + 49 + 64 + 81 + 100}{385} = \frac{385}{385} = 1 \]

5. **Find \( P(E_1 | A) \)**:
We need to find the conditional probability \( P(E_1 | A) \):
\[ P(E_1 | A) = \frac{P(E_1)}{P(A)} = \frac{\frac{1}{385}}{1} = \frac{1}{385} \]

### Final Answer:
Thus, the probability that a student who has passed the examination is the only student who passed is:
\[ \boxed{\frac{1}{385}} \]