If X and Y are two events such that `PX//Y)=(1)/(2), P(Y//X)=(1)/(3)` and `P(XcapY)(1)/(6)`. Then, which of the following is/are correct ?

Options

A. `P(XcupY)=2//3`
B. X and Y are independent
C. X and Y are not independent
D. `P(X^(c )capY)=1//3`

Explanation

To solve the problem, we need to use the given probabilities and the relationships between them. Let's denote the probabilities as follows:

- \( P(X|Y) = \frac{1}{2} \)
- \( P(Y|X) = \frac{1}{3} \)
- \( P(X \cap Y) = \frac{1}{6} \)

### Step 1: Use the definition of conditional probability

The conditional probability \( P(X|Y) \) is defined as:

\[ P(X|Y) = \frac{P(X \cap Y)}{P(Y)} \]

Substituting the known values:

\[ \frac{1}{2} = \frac{\frac{1}{6}}{P(Y)} \]

### Step 2: Solve for \( P(Y) \)

To find \( P(Y) \), we can rearrange the equation:

\[ P(Y) = \frac{1}{6} \cdot 2 = \frac{1}{3} \]

### Step 3: Use the definition of the other conditional probability

Now, we use the definition of \( P(Y|X) \):

\[ P(Y|X) = \frac{P(X \cap Y)}{P(X)} \]

Substituting the known values:

\[ \frac{1}{3} = \frac{\frac{1}{6}}{P(X)} \]

### Step 4: Solve for \( P(X) \)

Rearranging gives us:

\[ P(X) = \frac{1}{6} \cdot 3 = \frac{1}{2} \]

### Step 5: Verify the values

Now we have:

- \( P(X) = \frac{1}{2} \)
- \( P(Y) = \frac{1}{3} \)
- \( P(X \cap Y) = \frac{1}{6} \)

### Step 6: Check the relationships

We can check if the calculated probabilities satisfy the conditional probabilities:

1. **For \( P(X|Y) \)**:
\[ P(X|Y) = \frac{P(X \cap Y)}{P(Y)} = \frac{\frac{1}{6}}{\frac{1}{3}} = \frac{1}{2} \quad \text{(Correct)} \]

2. **For \( P(Y|X) \)**:
\[ P(Y|X) = \frac{P(X \cap Y)}{P(X)} = \frac{\frac{1}{6}}{\frac{1}{2}} = \frac{1}{3} \quad \text{(Correct)} \]

### Conclusion

The calculated probabilities are consistent with the given conditional probabilities. Therefore, the values of \( P(X) \) and \( P(Y) \) are correct.