A multiple - choice question has 5 options of which only one is correct. If a student does home work, then he is sure to identify the correct answer, otherwise the answer is chosen at random. Let A ne the event that the student does his home work and B be the event that the student answers correctly. If `P(A)=(2)/(3)`, then `P((A)/(B))` is euqal to

Options

A. `(10)/(11)`
B. `(4)/(5)`
C. `(3)/(7)`
D. `(6)/(7)`

Answer

To solve the problem, we need to find the conditional probability \( P(A|B) \), where:

- \( A \) is the event that the student does his homework.
- \( B \) is the event that the student answers correctly.

Given:
- \( P(A) = \frac{2}{3} \)
- If the student does homework, he is sure to identify the correct answer, so \( P(B|A) = 1 \).
- If the student does not do homework (event \( A' \)), he guesses the answer randomly from 5 options, so \( P(B|A') = \frac{1}{5} \).

### Step-by-step Solution:

1. **Calculate \( P(A') \)**:
\[ P(A') = 1 - P(A) = 1 - \frac{2}{3} = \frac{1}{3} \]

2. **Use the Law of Total Probability to find \( P(B) \)**:
\[ P(B) = P(B|A) \cdot P(A) + P(B|A') \cdot P(A') \]
Substituting the known values:
\[ P(B) = (1) \cdot \left(\frac{2}{3}\right) + \left(\frac{1}{5}\right) \cdot \left(\frac{1}{3}\right) \]
\[ P(B) = \frac{2}{3} + \frac{1}{15} \]

3. **Finding a common denominator to add the fractions**:
The least common multiple of 3 and 15 is 15.
\[ P(B) = \frac{10}{15} + \frac{1}{15} = \frac{11}{15} \]

4. **Use Bayes' Theorem to find \( P(A|B) \)**:
\[ P(A|B) = \frac{P(B|A) \cdot P(A)}{P(B)} \]
Substituting the values:
\[ P(A|B) = \frac{(1) \cdot \left(\frac{2}{3}\right)}{\frac{11}{15}} \]

5. **Simplifying the expression**:
\[ P(A|B) = \frac{\frac{2}{3}}{\frac{11}{15}} = \frac{2}{3} \cdot \frac{15}{11} = \frac{30}{33} = \frac{10}{11} \]

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