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 \( P\left(\frac{A}{A^c \cup B^c}\right) \) given:

- \( P(A) = 0.5 \)
- \( P(B) = 0.69 \)
- \( P(A | B) = 0.5 \)

### Step 1: Find \( P(A \cap B) \)

Using the definition of conditional probability, we have:

\[ P(A | B) = \frac{P(A \cap B)}{P(B)} \]

Substituting the known values:

\[ 0.5 = \frac{P(A \cap B)}{0.69} \]

Now, we can solve for \( P(A \cap B) \):

\[ P(A \cap B) = 0.5 \times 0.69 = 0.345 \]

### Step 2: Find \( P(A \cap B^c) \)

We know that:

\[ P(A) = P(A \cap B) + P(A \cap B^c) \]

Substituting the known values:

\[ 0.5 = 0.345 + P(A \cap B^c) \]

Now, we can solve for \( P(A \cap B^c) \):

\[ P(A \cap B^c) = 0.5 - 0.345 = 0.155 \]

### Step 3: Find \( P(A^c \cup B^c) \)

Using the formula for the probability of the union of two events:

\[ P(A^c \cup B^c) = 1 - P(A \cap B) \]

We already know \( P(A \cap B) = 0.345 \):

\[ P(A^c \cup B^c) = 1 - 0.345 = 0.655 \]

### Step 4: Find \( P(A \cap (A^c \cup B^c)) \)

Using the property of intersections:

\[ P(A \cap (A^c \cup B^c)) = P(A \cap A^c) + P(A \cap B^c) \]

Since \( P(A \cap A^c) = 0 \):

\[ P(A \cap (A^c \cup B^c)) = 0 + P(A \cap B^c) = P(A \cap B^c) = 0.155 \]

### Step 5: Find \( P(A^c \cup B^c) \)

We already calculated \( P(A^c \cup B^c) = 0.655 \).

### Step 6: Calculate \( P\left(\frac{A}{A^c \cup B^c}\right) \)

Now we can find the desired probability:

\[ P\left(\frac{A}{A^c \cup B^c}\right) = \frac{P(A \cap (A^c \cup B^c))}{P(A^c \cup B^c)} = \frac{0.155}{0.655} \]

Calculating this gives:

\[ P\left(\frac{A}{A^c \cup B^c}\right) \approx 0.236 \]

### Final Answer

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

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