Question

If \( 0 < P(A) < 1 \) and \( 0 < P(B) < 1 \), and \( P(A \cap B) = P(A)P(B) \), then

• A. \( P(B|A) = P(B) - P(A) \)
• B. \( P(A^c \cup B^c) = P(A^c) + P(B^c) \)
• C. \( P(A \cup B) = P(A^c)P(B^c) \)
• D. \( P(A|B) = P(A) - P(B) \)

Hint:

A key property of independent events A and B is that their complements (\( A^c, B^c \)) are also independent. This leads to the useful identity: \( P(A^c \cap B^c) = P(A^c)P(B^c) \). Using De Morgan's Law, this is equivalent to \( P((A \cup B)^c) = P(A^c)P(B^c) \), which simplifies to \( 1 - P(A \cup B) = (1-P(A))(1-P(B)) \).

Answer 1

The Correct Option is C

The given condition \( P(A \cap B) = P(A)P(B) \) means that events A and B are independent. Let's analyze each option based on this fact. \[\begin{array}{rl} \bullet & \text{(A) \( P(B|A) = P(B) - P(A) \): The definition of conditional probability is \( P(B|A) = \frac{P(A \cap B)}{P(A)} \). Since A and B are independent, \( P(A \cap B) = P(A)P(B) \). So, \( P(B|A) = \frac{P(A)P(B)}{P(A)} = P(B) \). The statement \( P(B|A) = P(B) - P(A) \) is incorrect.} \\ \bullet & \text{(B) \( P(A^c \cup B^c) = P(A^c) + P(B^c) \): This would imply that \( A^c \) and \( B^c \) are mutually exclusive, which is generally not true for independent events. This statement is incorrect.} \\ \bullet & \text{(C) \( P(A \cup B) = P(A^c)P(B^c) \): Let's evaluate both sides.} \\ \bullet & \text{(D) \( P(A|B) = P(A) - P(B) \): Similar to (A), for independent events, \( P(A|B) = P(A) \). The statement is incorrect.} \\ \end{array}\]