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

Answer

To solve the problem, we need to find the probabilities of events X and Y based on the given conditions. Let's break down the steps systematically.

### Step 1: Write down the given probabilities
We have the following probabilities:
1. \( P(X|Y) = \frac{1}{2} \)
2. \( P(Y|X) = \frac{1}{3} \)
3. \( P(X \cap Y) = \frac{1}{6} \)

### Step 2: Use the definition of conditional probability
The conditional probability \( P(X|Y) \) can be expressed 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 3: Solve for \( P(Y) \)
Cross-multiplying gives:
\[ P(Y) = \frac{1}{6} \times 2 = \frac{1}{3} \]

### Step 4: Use the definition of the other conditional probability
Now we can use \( P(Y|X) \):
\[ P(Y|X) = \frac{P(X \cap Y)}{P(X)} \]
Substituting the known values:
\[ \frac{1}{3} = \frac{\frac{1}{6}}{P(X)} \]

### Step 5: Solve for \( P(X) \)
Cross-multiplying gives:
\[ P(X) = \frac{1}{6} \times 3 = \frac{1}{2} \]

### Step 6: Verify the union of probabilities
We can use the formula for the union of two events:
\[ P(X \cup Y) = P(X) + P(Y) - P(X \cap Y) \]
Substituting the values we found:
\[ P(X \cup Y) = \frac{1}{2} + \frac{1}{3} - \frac{1}{6} \]

### Step 7: Find a common denominator and calculate
The common denominator for 2, 3, and 6 is 6:
\[ P(X \cup Y) = \frac{3}{6} + \frac{2}{6} - \frac{1}{6} = \frac{4}{6} = \frac{2}{3} \]

### Step 8: Conclusion
We have determined:
- \( P(X) = \frac{1}{2} \)
- \( P(Y) = \frac{1}{3} \)
- \( P(X \cap Y) = \frac{1}{6} \)
- \( P(X \cup Y) = \frac{2}{3} \)

Based on these calculations, we can conclude that options A and B are correct.