If A and B are two independent events, the probability that both A and B occurs is `1/12` and probability that neither A nor B occurs is `1/2` then

Options

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

Answer

To solve the problem, we need to find the probabilities of events A and B given the conditions provided. Let's break it down step by step.

### Step 1: Define the Given Information
We know:
- The probability that both A and B occur is given as \( P(A \cap B) = \frac{1}{12} \).
- The probability that neither A nor B occurs is given as \( P(A' \cap B') = \frac{1}{2} \).

### Step 2: Use the Complement Rule
We can express the probability that neither A nor B occurs using the complement rule:
\[ P(A' \cap B') = 1 - P(A \cup B) \]
From the given information, we can substitute:
\[ \frac{1}{2} = 1 - P(A \cup B) \]
This implies:
\[ P(A \cup B) = 1 - \frac{1}{2} = \frac{1}{2} \]

### Step 3: Use the Formula for the Union of Two Events
The probability of the union of two events can be expressed as:
\[ P(A \cup B) = P(A) + P(B) - P(A \cap B) \]
Substituting the known values:
\[ \frac{1}{2} = P(A) + P(B) - \frac{1}{12} \]
Rearranging gives:
\[ P(A) + P(B) = \frac{1}{2} + \frac{1}{12} \]

### Step 4: Find a Common Denominator
To add \( \frac{1}{2} \) and \( \frac{1}{12} \), we need a common denominator. The least common multiple of 2 and 12 is 12:
\[ \frac{1}{2} = \frac{6}{12} \]
Thus:
\[ P(A) + P(B) = \frac{6}{12} + \frac{1}{12} = \frac{7}{12} \]

### Step 5: Set Up the Equations
Let \( P(A) = x \) and \( P(B) = y \). We now have two equations:
1. \( x + y = \frac{7}{12} \) (Equation 1)
2. \( xy = \frac{1}{12} \) (Equation 2)

### Step 6: Solve the System of Equations
From Equation 1, we can express \( x \) in terms of \( y \):
\[ x = \frac{7}{12} - y \]
Substituting this into Equation 2:
\[ \left(\frac{7}{12} - y\right)y = \frac{1}{12} \]
Expanding and rearranging gives:
\[ \frac{7y}{12} - y^2 = \frac{1}{12} \]
Multiplying through by 12 to eliminate the fraction:
\[ 7y - 12y^2 = 1 \]
Rearranging gives:
\[ 12y^2 - 7y + 1 = 0 \]

### Step 7: Use the Quadratic Formula
Now we can use the quadratic formula \( y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) where \( a = 12, b = -7, c = 1 \):
\[ D = b^2 - 4ac = (-7)^2 - 4 \cdot 12 \cdot 1 = 49 - 48 = 1 \]
Thus:
\[ y = \frac{7 \pm \sqrt{1}}{2 \cdot 12} = \frac{7 \pm 1}{24} \]
This gives us two possible values for \( y \):
1. \( y = \frac{8}{24} = \frac{1}{3} \)
2. \( y = \frac{6}{24} = \frac{1}{4} \)

### Step 8: Find Corresponding Values of \( x \)
Using \( x + y = \frac{7}{12} \):
- If \( y = \frac{1}{3} \), then \( x = \frac{7}{12} - \frac{1}{3} = \frac{7}{12} - \frac{4}{12} = \frac{3}{12} = \frac{1}{4} \).
- If \( y = \frac{1}{4} \), then \( x = \frac{7}{12} - \frac{1}{4} = \frac{7}{12} - \frac{3}{12} = \frac{4}{12} = \frac{1}{3} \).

### Final Result
Thus, the probabilities are:
- \( P(A) = \frac{1}{4} \) and \( P(B) = \frac{1}{3} \) or vice versa.