If `A and B` are independent events of a random experiment such that `P(A nn B) = 1/6 and P( overlineA nn overlineB)=1/3` then `P(A)=`

Options

A. `(1)/(4)`
B. `(1)/(3)`
C. `(1)/(6)`
D. `(2)/(3)`

Explanation

To solve the problem step by step, we will use the information given about the independent events A and B.

### Step 1: Define the probabilities
Let \( P(A) = x \) and \( P(B) = y \).

### Step 2: Use the property of independent events
Since A and B are independent events, we have:
\[ P(A \cap B) = P(A) \cdot P(B) = x \cdot y \]
Given that \( P(A \cap B) = \frac{1}{6} \), we can write:
\[ xy = \frac{1}{6} \quad \text{(1)} \]

### Step 3: Use the probability of complements
We also know:
\[ P(\overline{A} \cap \overline{B}) = P(\overline{A}) \cdot P(\overline{B}) = (1 - x)(1 - y) \]
Given that \( P(\overline{A} \cap \overline{B}) = \frac{1}{3} \), we can write:
\[ (1 - x)(1 - y) = \frac{1}{3} \quad \text{(2)} \]

### Step 4: Expand equation (2)
Expanding equation (2):
\[ 1 - x - y + xy = \frac{1}{3} \]
Substituting \( xy = \frac{1}{6} \) from equation (1):
\[ 1 - x - y + \frac{1}{6} = \frac{1}{3} \]
Rearranging gives:
\[ 1 - x - y = \frac{1}{3} - \frac{1}{6} \]
Calculating the right side:
\[ \frac{1}{3} - \frac{1}{6} = \frac{2}{6} - \frac{1}{6} = \frac{1}{6} \]
Thus, we have:
\[ 1 - x - y = \frac{1}{6} \]

### Step 5: Rearranging to find \( x + y \)
Rearranging gives:
\[ x + y = 1 - \frac{1}{6} = \frac{5}{6} \quad \text{(3)} \]

### Step 6: Solve the system of equations
Now we have two equations:
1. \( xy = \frac{1}{6} \) (from equation (1))
2. \( x + y = \frac{5}{6} \) (from equation (3))

We can express \( y \) in terms of \( x \):
\[ y = \frac{5}{6} - x \]
Substituting into equation (1):
\[ x\left(\frac{5}{6} - x\right) = \frac{1}{6} \]
Expanding this gives:
\[ \frac{5}{6}x - x^2 = \frac{1}{6} \]
Multiplying through by 6 to eliminate the fraction:
\[ 5x - 6x^2 = 1 \]
Rearranging gives:
\[ 6x^2 - 5x + 1 = 0 \]

### Step 7: Solve the quadratic equation
Using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
Here, \( a = 6, b = -5, c = 1 \):
\[ x = \frac{-(-5) \pm \sqrt{(-5)^2 - 4 \cdot 6 \cdot 1}}{2 \cdot 6} \]
Calculating the discriminant:
\[ 25 - 24 = 1 \]
Thus:
\[ x = \frac{5 \pm 1}{12} \]
This gives us two possible values:
\[ x = \frac{6}{12} = \frac{1}{2} \quad \text{and} \quad x = \frac{4}{12} = \frac{1}{3} \]

### Step 8: Conclusion
Since \( P(A) \) must be a valid probability, we have:
\[ P(A) = \frac{1}{3} \quad \text{or} \quad P(A) = \frac{1}{2} \]
From the context of the problem, the valid value for \( P(A) \) is:
\[ P(A) = \frac{1}{3} \]