If A, B, C are independent events such that `P(Acap B) = 1/2, P(Bcap C) = 1/3 and P(C cap A) = 1/6,` then respective values of `P(A), P(B) and P(C)` are

Options

A. `1/2, 1/3, 1`
B. `1/2,1, 1/3`
C. `1/3, 1, 1/2`
D. `1, 1/2, 1/3`

Answer

To solve the problem, we need to find the probabilities of independent events A, B, and C given the intersections of these events. The information provided is:

1. \( P(A \cap B) = \frac{1}{2} \)
2. \( P(B \cap C) = \frac{1}{3} \)
3. \( P(C \cap A) = \frac{1}{6} \)

Since A, B, and C are independent events, we can express the probabilities of their intersections in terms of their individual probabilities:

\[ P(A \cap B) = P(A) \cdot P(B) \]
\[ P(B \cap C) = P(B) \cdot P(C) \]
\[ P(C \cap A) = P(C) \cdot P(A) \]

Let's denote:

- \( P(A) = x \)
- \( P(B) = y \)
- \( P(C) = z \)

Now we can rewrite the equations based on the given probabilities:

1. \( x \cdot y = \frac{1}{2} \) (Equation 1)
2. \( y \cdot z = \frac{1}{3} \) (Equation 2)
3. \( z \cdot x = \frac{1}{6} \) (Equation 3)

### Step 1: Solve for one variable in terms of another

From Equation 1, we can express \( y \) in terms of \( x \):

\[ y = \frac{1}{2x} \]

### Step 2: Substitute into the second equation

Now substitute \( y \) into Equation 2:

\[ \left(\frac{1}{2x}\right) \cdot z = \frac{1}{3} \]

Multiply both sides by \( 2x \):

\[ z = \frac{2x}{3} \]

### Step 3: Substitute into the third equation

Now substitute \( z \) into Equation 3:

\[ \left(\frac{2x}{3}\right) \cdot x = \frac{1}{6} \]

This simplifies to:

\[ \frac{2x^2}{3} = \frac{1}{6} \]

### Step 4: Solve for \( x \)

Multiply both sides by 6 to eliminate the fraction:

\[ 4x^2 = 1 \]

Now divide by 4:

\[ x^2 = \frac{1}{4} \]

Taking the square root gives:

\[ x = \frac{1}{2} \]

### Step 5: Find \( y \) and \( z \)

Now that we have \( x \), substitute back to find \( y \):

\[ y = \frac{1}{2x} = \frac{1}{2 \cdot \frac{1}{2}} = 1 \]

Next, substitute \( x \) into the equation for \( z \):

\[ z = \frac{2x}{3} = \frac{2 \cdot \frac{1}{2}}{3} = \frac{1}{3} \]

### Final Values

Thus, we have:

- \( P(A) = x = \frac{1}{2} \)
- \( P(B) = y = 1 \)
- \( P(C) = z = \frac{1}{3} \)

### Conclusion

The respective values of \( P(A), P(B), \) and \( P(C) \) are:

- \( P(A) = \frac{1}{2} \)
- \( P(B) = 1 \)
- \( P(C) = \frac{1}{3} \)