If A and B are two events such that `P (A) = (1)/(2) P (B) = (1)/(3), P ((A)/(B))= (1)/(4), ` then `P (A' nn B')` is equal to

Options

A. `(1)/(12)`
B. `(3)/(4)`
C. `(1)/(4)`
D. `(3)/(16)`

Answer

To solve the problem, we need to find \( P(A' \cap B') \), which is the probability that neither event A nor event B occurs. We can use De Morgan's theorem to express this in terms of the union of A and B:

### Step 1: Use De Morgan's Theorem
According to De Morgan's theorem:
\[ P(A' \cap B') = P((A \cup B)') \]
This means we need to find the probability of the complement of the union of A and B.

### Step 2: Find \( P(A \cup B) \)
We can calculate \( P(A \cup B) \) using the formula:
\[ P(A \cup B) = P(A) + P(B) - P(A \cap B) \]
From the problem, we know:
- \( P(A) = \frac{1}{2} \)
- \( P(B) = \frac{1}{3} \)
- \( P(A | B) = \frac{1}{4} \)

### Step 3: Calculate \( P(A \cap B) \)
We can find \( P(A \cap B) \) using the definition of conditional probability:
\[ P(A | B) = \frac{P(A \cap B)}{P(B)} \]
Substituting the known values:
\[ \frac{1}{4} = \frac{P(A \cap B)}{\frac{1}{3}} \]
Multiplying both sides by \( \frac{1}{3} \):
\[ P(A \cap B) = \frac{1}{4} \times \frac{1}{3} = \frac{1}{12} \]

### Step 4: Substitute Values into the Union Formula
Now we can substitute the values into the union formula:
\[ P(A \cup B) = P(A) + P(B) - P(A \cap B) \]
\[ P(A \cup B) = \frac{1}{2} + \frac{1}{3} - \frac{1}{12} \]

### Step 5: Find a Common Denominator
The common denominator for 2, 3, and 12 is 12. We convert each fraction:
\[ P(A \cup B) = \frac{6}{12} + \frac{4}{12} - \frac{1}{12} \]
\[ P(A \cup B) = \frac{6 + 4 - 1}{12} = \frac{9}{12} = \frac{3}{4} \]

### Step 6: Find \( P(A' \cap B') \)
Now we can find \( P(A' \cap B') \):
\[ P(A' \cap B') = P((A \cup B)') = 1 - P(A \cup B) \]
\[ P(A' \cap B') = 1 - \frac{3}{4} = \frac{1}{4} \]

### Final Answer
Thus, the probability \( P(A' \cap B') \) is:
\[ \boxed{\frac{1}{4}} \]

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