Let `P(A) = 1/3` , `P(B) = 1/6` where A and B are independent events then

Options

A. `P(A//B)=(2)/(3)`
B. `P(A//B')=(1)/(3)`
C. `P(A'//B')=(1)/(3)`
D. `P(A//A cup B)=(1)/(4)`

Answer

To solve the problem, we need to find the probabilities related to the independent events A and B, given that \( P(A) = \frac{1}{3} \) and \( P(B) = \frac{1}{6} \).

### Step-by-Step Solution:

1. **Understanding Independence**:
Since A and B are independent events, we know that:
\[ P(A \cap B) = P(A) \cdot P(B) \]

2. **Calculating \( P(A \cap B) \)**:
Substitute the values of \( P(A) \) and \( P(B) \):
\[ P(A \cap B) = \frac{1}{3} \cdot \frac{1}{6} = \frac{1}{18} \]

3. **Finding \( P(A | B) \)**:
The conditional probability \( P(A | B) \) can be calculated using the formula:
\[ P(A | B) = \frac{P(A \cap B)}{P(B)} \]
Substitute the known values:
\[ P(A | B) = \frac{\frac{1}{18}}{\frac{1}{6}} = \frac{1}{18} \cdot \frac{6}{1} = \frac{1}{3} \]

4. **Finding \( P(A | B') \)**:
Since A and B are independent, A and \( B' \) (the complement of B) are also independent. Thus:
\[ P(A | B') = P(A) \]
Therefore:
\[ P(A | B') = \frac{1}{3} \]

5. **Finding \( P(A' | B') \)**:
First, calculate \( P(A') \):
\[ P(A') = 1 - P(A) = 1 - \frac{1}{3} = \frac{2}{3} \]
Since A and B are independent, A' and B' are also independent:
\[ P(A' | B') = P(A') \]
Thus:
\[ P(A' | B') = \frac{2}{3} \]

6. **Finding \( P(A | A \cup B) \)**:
We know:
\[ P(A \cup B) = P(A) + P(B) - P(A \cap B) \]
Substitute the values:
\[ P(A \cup B) = \frac{1}{3} + \frac{1}{6} - \frac{1}{18} \]
To add these fractions, find a common denominator (which is 18):
\[ P(A \cup B) = \frac{6}{18} + \frac{3}{18} - \frac{1}{18} = \frac{8}{18} = \frac{4}{9} \]
Now, calculate \( P(A | A \cup B) \):
\[ P(A | A \cup B) = \frac{P(A)}{P(A \cup B)} = \frac{\frac{1}{3}}{\frac{4}{9}} = \frac{1}{3} \cdot \frac{9}{4} = \frac{3}{4} \]

### Summary of Results:
- \( P(A | B) = \frac{1}{3} \)
- \( P(A | B') = \frac{1}{3} \)
- \( P(A' | B') = \frac{2}{3} \)
- \( P(A | A \cup B) = \frac{3}{4} \)