In a class of 10 student, probability of exactly `i` students passing an examination is directly proportional to `i^(2).` If a student is selected at random, then the probability that he has passed the examination is

Options

A. (a) `1//7`
B. (b) `11//35`
C. (c) `11//14`
D. (d) None of these

Explanation

To solve the problem step by step, we will follow the logical flow presented in the video transcript.

### Step 1: Define the Events
Let \( E_i \) be the event that exactly \( i \) out of 10 students pass the examination. The probability of \( E_i \) is directly proportional to \( i^2 \).

### Step 2: Write the Probability Expression
We can express the probability of \( E_i \) as:
\[ P(E_i) = k \cdot i^2 \]
where \( k \) is a constant of proportionality.

### Step 3: Find the Value of \( k \)
Since the events \( E_0, E_1, \ldots, E_{10} \) are mutually exclusive and exhaustive, we have:
\[ P(E_0) + P(E_1) + \ldots + P(E_{10}) = 1 \]
Substituting the expression for \( P(E_i) \):
\[ k \cdot 0^2 + k \cdot 1^2 + k \cdot 2^2 + \ldots + k \cdot 10^2 = 1 \]
This simplifies to:
\[ k \cdot (0 + 1^2 + 2^2 + \ldots + 10^2) = 1 \]
Using the formula for the sum of squares:
\[ \sum_{i=1}^{n} i^2 = \frac{n(n+1)(2n+1)}{6} \]
For \( n = 10 \):
\[ \sum_{i=1}^{10} i^2 = \frac{10 \cdot 11 \cdot 21}{6} = 385 \]
Thus, we have:
\[ k \cdot 385 = 1 \implies k = \frac{1}{385} \]

### Step 4: Write the Probability of Each Event
Now we can write:
\[ P(E_i) = \frac{1}{385} \cdot i^2 \]

### Step 5: Calculate the Probability of Passing
Let \( A \) be the event that a randomly selected student has passed the examination. We need to find \( P(A) \):
\[ P(A) = \sum_{i=0}^{10} P(E_i) \cdot P(A | E_i) \]
Where \( P(A | E_i) = \frac{i}{10} \) (since if \( i \) students pass, the probability of selecting one of them is \( \frac{i}{10} \)).

### Step 6: Substitute and Simplify
Substituting the values:
\[ P(A) = \sum_{i=0}^{10} P(E_i) \cdot \frac{i}{10} = \sum_{i=0}^{10} \left(\frac{1}{385} \cdot i^2 \cdot \frac{i}{10}\right) \]
This simplifies to:
\[ P(A) = \frac{1}{385 \cdot 10} \sum_{i=0}^{10} i^3 \]

### Step 7: Calculate the Sum of Cubes
Using the formula for the sum of cubes:
\[ \sum_{i=1}^{n} i^3 = \left(\frac{n(n+1)}{2}\right)^2 \]
For \( n = 10 \):
\[ \sum_{i=1}^{10} i^3 = \left(\frac{10 \cdot 11}{2}\right)^2 = 55^2 = 3025 \]
Thus:
\[ P(A) = \frac{1}{3850} \cdot 3025 \]

### Step 8: Final Calculation
Calculating \( P(A) \):
\[ P(A) = \frac{3025}{3850} = \frac{11}{14} \]

### Conclusion
The probability that a randomly selected student has passed the examination is:
\[ \boxed{\frac{11}{14}} \]