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 \( P(A|B) \), which is the probability that the student did his homework given that he answered correctly. We will use Bayes' theorem to find this probability.

### Step 1: Identify the probabilities
1. Let \( A \) be the event that the student does his homework.
2. Let \( B \) be the event that the student answers correctly.
3. We know that:
- \( P(A) = \frac{2}{3} \) (the probability that the student does his homework).
- \( P(A') = 1 - P(A) = 1 - \frac{2}{3} = \frac{1}{3} \) (the probability that the student does not do his homework).
- If the student does his homework, he is sure to identify the correct answer, so \( P(B|A) = 1 \).
- If the student does not do his homework, he guesses randomly among 5 options, so \( P(B|A') = \frac{1}{5} \).

### Step 2: Calculate \( P(B) \)
Using the law of total probability, we can find \( P(B) \):
\[ P(B) = P(B|A)P(A) + P(B|A')P(A') \]
Substituting the known values:
\[ P(B) = (1) \left(\frac{2}{3}\right) + \left(\frac{1}{5}\right) \left(\frac{1}{3}\right) \]
Calculating \( P(B) \):
\[ P(B) = \frac{2}{3} + \frac{1}{15} = \frac{10}{15} + \frac{1}{15} = \frac{11}{15} \]

### Step 3: Apply Bayes' Theorem
Now we can apply Bayes' theorem:
\[ P(A|B) = \frac{P(B|A)P(A)}{P(B)} \]
Substituting the known values:
\[ P(A|B) = \frac{(1) \left(\frac{2}{3}\right)}{\frac{11}{15}} \]
Calculating \( P(A|B) \):
\[ P(A|B) = \frac{\frac{2}{3}}{\frac{11}{15}} = \frac{2}{3} \times \frac{15}{11} = \frac{30}{33} = \frac{10}{11} \]

### Final Answer
Thus, the probability that the student did his homework given that he answered correctly is:
\[ \boxed{\frac{10}{11}} \]