If A and B are two events defined on a sample space with the probabilities `P(A)=0.5, P(B)=0.69 and P((A)/(B))=0.5`, thent the value of `P((A)/(A^(c )uuB^(c )))` is equal to

Options

A. `(2)/(5)`
B. `(3)/(13)`
C. `(31)/(131)`
D. `(100)/(131)`

Answer

To solve the problem, we need to find the value of \( P\left(\frac{A}{A^c \cup B^c}\right) \).

### Step 1: Understand the Given Information
We have:
- \( P(A) = 0.5 \)
- \( P(B) = 0.69 \)
- \( P(A|B) = 0.5 \)

From the definition of conditional probability, we know:
\[ P(A|B) = \frac{P(A \cap B)}{P(B)} \]
Thus, we can find \( P(A \cap B) \):
\[ P(A \cap B) = P(A|B) \cdot P(B) = 0.5 \cdot 0.69 = 0.345 \]

### Step 2: Calculate \( P(A^c \cup B^c) \)
Using the formula for the union of two events:
\[ P(A^c \cup B^c) = 1 - P(A \cap B) \]
We know \( P(A \cap B) = 0.345 \), so:
\[ P(A^c \cup B^c) = 1 - P(A \cap B) = 1 - 0.345 = 0.655 \]

### Step 3: Calculate \( P(A \cap (A^c \cup B^c)) \)
Using the distributive property of probability:
\[ P(A \cap (A^c \cup B^c)) = P((A \cap A^c) \cup (A \cap B^c)) \]
Since \( A \cap A^c = \emptyset \) (the intersection of an event and its complement is empty), we have:
\[ P(A \cap (A^c \cup B^c)) = P(A \cap B^c) \]

### Step 4: Calculate \( P(A \cap B^c) \)
We can find \( P(A \cap B^c) \) using the formula:
\[ P(A) = P(A \cap B) + P(A \cap B^c) \]
Thus,
\[ P(A \cap B^c) = P(A) - P(A \cap B) = 0.5 - 0.345 = 0.155 \]

### Step 5: Calculate \( P\left(\frac{A}{A^c \cup B^c}\right) \)
Now we can find:
\[ P\left(\frac{A}{A^c \cup B^c}\right) = \frac{P(A \cap (A^c \cup B^c))}{P(A^c \cup B^c)} = \frac{P(A \cap B^c)}{P(A^c \cup B^c)} \]
Substituting the values we calculated:
\[ P\left(\frac{A}{A^c \cup B^c}\right) = \frac{0.155}{0.655} \]

### Step 6: Final Calculation
Calculating the final value:
\[ P\left(\frac{A}{A^c \cup B^c}\right) \approx 0.236 \]

### Conclusion
Thus, the value of \( P\left(\frac{A}{A^c \cup B^c}\right) \) is approximately \( 0.236 \).

---