In a test an examinee either guesses, copies or knows the answer to multiple choice question with five choices. The probability that he makes a guess is `1/4` and probability that he copies the answer is `1/8`. The probability that his answer is correct given that he copies is `1/10`. Find the probability that he knows the answer to the question given that he correctly answered it:

Options

A. `1`
B. `10/11`
C. `1/2`
D. `12/19`

Explanation

To solve the problem, we need to find the probability that the examinee knows the answer given that he answered correctly. We will use Bayes' theorem for this calculation.

### Step-by-Step Solution:

1. **Define Events**:
- Let \( E_1 \): the examinee guesses the answer.
- Let \( E_2 \): the examinee copies the answer.
- Let \( E_3 \): the examinee knows the answer.
- Let \( A \): the answer is correct.

2. **Given Probabilities**:
- \( P(E_1) = \frac{1}{4} \) (Probability of guessing)
- \( P(E_2) = \frac{1}{8} \) (Probability of copying)
- \( P(A | E_2) = \frac{1}{10} \) (Probability of correct answer given that he copies)

3. **Calculate Probability of Knowing the Answer**:
Since \( E_1, E_2, E_3 \) are mutually exclusive, we can find \( P(E_3) \):
\[ P(E_3) = 1 - P(E_1) - P(E_2) = 1 - \frac{1}{4} - \frac{1}{8} \]
To perform this calculation, we first convert \( \frac{1}{4} \) and \( \frac{1}{8} \) to a common denominator:
\[ P(E_1) = \frac{2}{8}, \quad P(E_2) = \frac{1}{8} \]
Thus,
\[ P(E_3) = 1 - \frac{2}{8} - \frac{1}{8} = 1 - \frac{3}{8} = \frac{5}{8} \]

4. **Calculate Probabilities of Correct Answers**:
- If the examinee guesses (\( E_1 \)):
\[ P(A | E_1) = \frac{1}{5} \quad \text{(only one correct answer out of five choices)} \]
- If the examinee copies (\( E_2 \)):
\[ P(A | E_2) = \frac{1}{10} \quad \text{(given)} \]
- If the examinee knows the answer (\( E_3 \)):
\[ P(A | E_3) = 1 \quad \text{(if he knows, he will definitely be correct)} \]

5. **Apply Bayes' Theorem**:
We want to find \( P(E_3 | A) \):
\[ P(E_3 | A) = \frac{P(E_3) \cdot P(A | E_3)}{P(E_1) \cdot P(A | E_1) + P(E_2) \cdot P(A | E_2) + P(E_3) \cdot P(A | E_3)} \]
Substituting the known values:
\[ P(E_3 | A) = \frac{\frac{5}{8} \cdot 1}{\frac{1}{4} \cdot \frac{1}{5} + \frac{1}{8} \cdot \frac{1}{10} + \frac{5}{8} \cdot 1} \]
Calculating the denominator:
\[ P(E_1) \cdot P(A | E_1) = \frac{1}{4} \cdot \frac{1}{5} = \frac{1}{20} \]
\[ P(E_2) \cdot P(A | E_2) = \frac{1}{8} \cdot \frac{1}{10} = \frac{1}{80} \]
\[ P(E_3) \cdot P(A | E_3) = \frac{5}{8} \cdot 1 = \frac{5}{8} \]
Now, finding a common denominator for the terms in the denominator:
\[ \text{LCM of } 20, 80, \text{ and } 8 = 80 \]
Converting:
\[ \frac{1}{20} = \frac{4}{80}, \quad \frac{1}{80} = \frac{1}{80}, \quad \frac{5}{8} = \frac{50}{80} \]
Therefore:
\[ \text{Denominator} = \frac{4}{80} + \frac{1}{80} + \frac{50}{80} = \frac{55}{80} \]

6. **Final Calculation**:
Now substituting back into Bayes' theorem:
\[ P(E_3 | A) = \frac{\frac{5}{8}}{\frac{55}{80}} = \frac{5}{8} \cdot \frac{80}{55} = \frac{5 \cdot 10}{11} = \frac{50}{11} \approx 0.9091 \]

Thus, the final answer is:
\[ P(E_3 | A) = \frac{10}{11} \]