Two independent events A and B are such that `P(A uu B) = 2//3`, and `P(A nn B) = 1//6` If `P(B) lt P(A)`, then what is P(B) equal to ?

Options

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

Answer

To solve the problem, we will follow these steps:

### Step 1: Understand the given information
We are given two independent events A and B. The probabilities provided are:
- \( P(A \cup B) = \frac{2}{3} \)
- \( P(A \cap B) = \frac{1}{6} \)

### Step 2: Use the properties of independent events
For independent events, the probability of their intersection is given by:
\[ P(A \cap B) = P(A) \cdot P(B) \]
From the information provided, we have:
\[ P(A) \cdot P(B) = \frac{1}{6} \] (Equation 1)

### Step 3: Use the formula for the union of two events
The probability of the union of two events is given by:
\[ P(A \cup B) = P(A) + P(B) - P(A \cap B) \]
Substituting the known values:
\[ \frac{2}{3} = P(A) + P(B) - \frac{1}{6} \]

### Step 4: Rearrange the equation
Rearranging the equation gives:
\[ P(A) + P(B) = \frac{2}{3} + \frac{1}{6} \]
To add these fractions, we need a common denominator. The least common multiple of 3 and 6 is 6.
\[ \frac{2}{3} = \frac{4}{6} \]
So,
\[ P(A) + P(B) = \frac{4}{6} + \frac{1}{6} = \frac{5}{6} \] (Equation 2)

### Step 5: Set up the system of equations
Now we have two equations:
1. \( P(A) \cdot P(B) = \frac{1}{6} \) (Equation 1)
2. \( P(A) + P(B) = \frac{5}{6} \) (Equation 2)

### Step 6: Substitute \( P(B) \) in terms of \( P(A) \)
From Equation 2, we can express \( P(B) \) in terms of \( P(A) \):
\[ P(B) = \frac{5}{6} - P(A) \]

### Step 7: Substitute into Equation 1
Now substitute \( P(B) \) into Equation 1:
\[ P(A) \left( \frac{5}{6} - P(A) \right) = \frac{1}{6} \]
Expanding this gives:
\[ \frac{5}{6} P(A) - P(A)^2 = \frac{1}{6} \]

### Step 8: Rearrange into a quadratic equation
Rearranging gives:
\[ P(A)^2 - \frac{5}{6} P(A) + \frac{1}{6} = 0 \]

### Step 9: Multiply through by 6 to eliminate fractions
Multiplying through by 6:
\[ 6P(A)^2 - 5P(A) + 1 = 0 \]

### Step 10: Solve the quadratic equation
Using the quadratic formula \( P(A) = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
Here, \( a = 6, b = -5, c = 1 \):
\[ P(A) = \frac{5 \pm \sqrt{(-5)^2 - 4 \cdot 6 \cdot 1}}{2 \cdot 6} \]
\[ P(A) = \frac{5 \pm \sqrt{25 - 24}}{12} \]
\[ P(A) = \frac{5 \pm 1}{12} \]

### Step 11: Calculate the two possible values for \( P(A) \)
1. \( P(A) = \frac{6}{12} = \frac{1}{2} \)
2. \( P(A) = \frac{4}{12} = \frac{1}{3} \)

### Step 12: Find \( P(B) \) using both values of \( P(A) \)
Using \( P(B) = \frac{5}{6} - P(A) \):
1. If \( P(A) = \frac{1}{2} \):
\[ P(B) = \frac{5}{6} - \frac{1}{2} = \frac{5}{6} - \frac{3}{6} = \frac{2}{6} = \frac{1}{3} \]
2. If \( P(A) = \frac{1}{3} \):
\[ P(B) = \frac{5}{6} - \frac{1}{3} = \frac{5}{6} - \frac{2}{6} = \frac{3}{6} = \frac{1}{2} \]

### Step 13: Check the condition \( P(B) < P(A) \)
Since we are given that \( P(B) < P(A) \):
- For \( P(A) = \frac{1}{2} \), \( P(B) = \frac{1}{3} \) (valid)
- For \( P(A) = \frac{1}{3} \), \( P(B) = \frac{1}{2} \) (not valid)

Thus, the only valid solution is:
\[ P(B) = \frac{1}{3} \]

### Final Answer
Therefore, \( P(B) = \frac{1}{3} \).

---