Question

If \( P(A) = \frac{1}{4} \), \( P(B) = \frac{1}{3} \) and \( P(A \cup B) = \frac{1}{2} \), then \( P(A \cap B) = \)

• A. \( \frac{1}{4} \)
• B. \( \frac{1}{3} \)
• C. \( \frac{1}{12} \)
• D. \( \frac{1}{2} \)

Hint:

The formula \( P(A \cup B) = P(A) + P(B) - P(A \cap B) \) is fundamental in probability and allows you to find the probability of the intersection of two events if you know the probabilities of the individual events and their union.

Answer 1

The Correct Option is C

Step 1: Use the formula for the probability of the union of two events.
The formula is \( P(A \cup B) = P(A) + P(B) - P(A \cap B) \). Step 2: Substitute the given probabilities into the formula.
We have \( P(A) = \frac{1}{4} \), \( P(B) = \frac{1}{3} \), and \( P(A \cup B) = \frac{1}{2} \). Substituting these values:
$$\frac{1}{2} = \frac{1}{4} + \frac{1}{3} - P(A \cap B)$$ Step 3: Solve for \( P(A \cap B) \).
First, find a common denominator for \( \frac{1}{4} \) and \( \frac{1}{3} \), which is 12:
$$\frac{1}{4} = \frac{3}{12}$$
$$\frac{1}{3} = \frac{4}{12}$$
So, the equation becomes:
$$\frac{1}{2} = \frac{3}{12} + \frac{4}{12} - P(A \cap B)$$
$$\frac{1}{2} = \frac{7}{12} - P(A \cap B)$$
Now, isolate \( P(A \cap B) \):
$$P(A \cap B) = \frac{7}{12} - \frac{1}{2}$$
To subtract these fractions, find a common denominator, which is 12: $$\frac{1}{2} = \frac{6}{12}$$
So,
$$P(A \cap B) = \frac{7}{12} - \frac{6}{12} = \frac{7 - 6}{12} = \frac{1}{12}$$