For any events `A` and `B`. Given `P(AuuB)=0.6`, `P(A)=P(B)`, `P(B//A)=0.8`. Then the value of `P[AnnbarB)uu(barAnnB)]` is

Options

A. `1//3`
B. `1//2`
C. `1//4`
D. `1//5`

Explanation

To solve the problem step by step, let's break down the information given and apply the relevant probability formulas.

### Given:
1. \( P(A \cup B) = 0.6 \)
2. \( P(A) = P(B) \)
3. \( P(B|A) = 0.8 \)

### Step 1: Define \( P(A) \) and \( P(B) \)
Since \( P(A) = P(B) \), we can denote \( P(A) = P(B) = p \).

### Step 2: Use the conditional probability
From the conditional probability formula:
\[ P(B|A) = \frac{P(A \cap B)}{P(A)} \]
Substituting the known values:
\[ 0.8 = \frac{P(A \cap B)}{p} \]
This gives us:
\[ P(A \cap B) = 0.8p \tag{1} \]

### Step 3: Use the formula for the union of two events
Using the formula for the probability of the union of two events:
\[ P(A \cup B) = P(A) + P(B) - P(A \cap B) \]
Substituting the known values:
\[ 0.6 = p + p - P(A \cap B) \]
Substituting \( P(A \cap B) \) from equation (1):
\[ 0.6 = p + p - 0.8p \]
This simplifies to:
\[ 0.6 = 2p - 0.8p \]
\[ 0.6 = 1.2p \]
Now, solving for \( p \):
\[ p = \frac{0.6}{1.2} = 0.5 \]

### Step 4: Find \( P(A \cap B) \)
Now substituting \( p \) back into equation (1):
\[ P(A \cap B) = 0.8 \times 0.5 = 0.4 \]

### Step 5: Calculate \( P(A \cap B^c \cup A^c \cap B) \)
We want to find:
\[ P(A \cap B^c \cup A^c \cap B) \]
Using the formula:
\[ P(A \cap B^c \cup A^c \cap B) = P(A) + P(B) - 2P(A \cap B) \]
Substituting the values we found:
\[ P(A \cap B^c \cup A^c \cap B) = 0.5 + 0.5 - 2 \times 0.4 \]
Calculating this gives:
\[ P(A \cap B^c \cup A^c \cap B) = 1 - 0.8 = 0.2 \]

### Final Answer
Thus, the value of \( P(A \cap B^c \cup A^c \cap B) \) is:
\[ \boxed{0.2} \]