If A and B are two independent events such that `P(overline(A) cap B)=2//15 " and " P(A cap overline(B))=1//6`, then P(B), is

Options

A. `(1)/(5) " or ", (4)/(5)`
B. `(1)/(6) " or", (5)/(6)`
C. `(4)/(5) " or", (1)/(6)`
D. `(5)/(6) " or", (1)/(5)`

Explanation

To solve the problem step by step, we need to find the probability of event B, given the probabilities of the intersections of events A and B.

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

Since A and B are independent events, we have:
- \( P(A \cap B) = P(A) \cdot P(B) = x \cdot y \)

### Step 2: Use the given information
We are given:
1. \( P(\overline{A} \cap B) = \frac{2}{15} \)
2. \( P(A \cap \overline{B}) = \frac{1}{6} \)

Using the properties of probabilities, we can express these in terms of \( x \) and \( y \):
- \( P(\overline{A}) = 1 - P(A) = 1 - x \)
- \( P(\overline{B}) = 1 - P(B) = 1 - y \)

Thus, we can write:
1. \( P(\overline{A} \cap B) = P(\overline{A}) \cdot P(B) = (1 - x) \cdot y = \frac{2}{15} \)
2. \( P(A \cap \overline{B}) = P(A) \cdot P(\overline{B}) = x \cdot (1 - y) = \frac{1}{6} \)

### Step 3: Set up the equations
From the above expressions, we can set up the following equations:
1. \( (1 - x) \cdot y = \frac{2}{15} \) (Equation 1)
2. \( x \cdot (1 - y) = \frac{1}{6} \) (Equation 2)

### Step 4: Solve Equation 1 for y
From Equation 1:
\[ y - xy = \frac{2}{15} \]
Rearranging gives:
\[ y(1 - x) = \frac{2}{15} \]
Thus,
\[ y = \frac{2}{15(1 - x)} \tag{1} \]

### Step 5: Substitute y in Equation 2
Now substitute \( y \) from Equation (1) into Equation 2:
\[ x \cdot \left(1 - \frac{2}{15(1 - x)}\right) = \frac{1}{6} \]
Simplifying the left side:
\[ x \cdot \left(\frac{15(1 - x) - 2}{15(1 - x)}\right) = \frac{1}{6} \]
This simplifies to:
\[ x \cdot \frac{15 - 15x - 2}{15(1 - x)} = \frac{1}{6} \]
\[ x \cdot \frac{13 - 15x}{15(1 - x)} = \frac{1}{6} \]

### Step 6: Cross-multiply and simplify
Cross-multiplying gives:
\[ 6x(13 - 15x) = 15(1 - x) \]
Expanding both sides:
\[ 78x - 90x^2 = 15 - 15x \]
Rearranging gives:
\[ 90x^2 - 93x + 15 = 0 \]

### Step 7: Solve the quadratic equation
Using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
Here, \( a = 90, b = -93, c = 15 \):
\[ x = \frac{93 \pm \sqrt{(-93)^2 - 4 \cdot 90 \cdot 15}}{2 \cdot 90} \]
Calculating the discriminant:
\[ = 8649 - 5400 = 3249 \]
\[ x = \frac{93 \pm 57}{180} \]
Calculating the two possible values for \( x \):
1. \( x = \frac{150}{180} = \frac{5}{6} \)
2. \( x = \frac{36}{180} = \frac{1}{5} \)

### Step 8: Find y for both values of x
Using \( y = \frac{2}{15(1 - x)} \):
1. For \( x = \frac{5}{6} \):
\[ y = \frac{2}{15(1 - \frac{5}{6})} = \frac{2}{15 \cdot \frac{1}{6}} = \frac{2 \cdot 6}{15} = \frac{12}{15} = \frac{4}{5} \]
2. For \( x = \frac{1}{5} \):
\[ y = \frac{2}{15(1 - \frac{1}{5})} = \frac{2}{15 \cdot \frac{4}{5}} = \frac{2 \cdot 5}{15 \cdot 4} = \frac{10}{60} = \frac{1}{6} \]

### Conclusion
Thus, the possible values for \( P(B) \) are \( \frac{4}{5} \) and \( \frac{1}{6} \).