Question

Let \( A \) and \( B \) be two events defined on a sample space \( \Omega \). Suppose \( A \) denotes the complement of \( A \) relative to the sample space \( \Omega \). Then the probability \( P((A \cap B) \cup (A^c \cap B)) \) equals:

• A. \( P(A) + P(B) + P(A \cap B) \)
• B. \( P(A) + P(B) - P(A \cap B) \)
• C. \( P(A) + P(B) + 2P(A \cap B) \)
• D. \( P(A) + P(B) - 2P(A \cap B) \)

Hint:

When calculating the probability of the union of two events with complements, always remember that \( P(A^c \cap B) = P(B) - P(A \cap B) \), and use the inclusion-exclusion principle to handle the union.

Answer 1

The Correct Option is D

Step 1: Express the event and the probability. We need to find the probability of the event \( (A \cap B) \cup (A^c \cap B) \), which represents the union of two events:
\( A \cap B \) — the event where both \( A \) and \( B \) occur.
\( A^c \cap B \) — the event where \( B \) occurs, but \( A \) does not.
Step 2: Apply the union formula. The probability of the union of two events is given by the inclusion-exclusion principle: \[ P((A \cap B) \cup (A^c \cap B)) = P(A \cap B) + P(A^c \cap B) - P((A \cap B) \cap (A^c \cap B)). \] Step 3: Simplify the intersection.
The intersection \( (A \cap B) \cap (A^c \cap B) \) is empty, as \( A \cap A^c = \emptyset \) (the complement of \( A \) cannot intersect with \( A \) itself). Therefore, the term \( P((A \cap B) \cap (A^c \cap B)) \) is zero. Thus, we have: \[ P((A \cap B) \cup (A^c \cap B)) = P(A \cap B) + P(A^c \cap B). \] Step 4: Use the complement rule.
From probability theory, we know: \[ P(A^c \cap B) = P(B) - P(A \cap B). \] Substitute this into the expression: \[ P((A \cap B) \cup (A^c \cap B)) = P(A \cap B) + (P(B) - P(A \cap B)). \] Step 5: Final simplification.
Simplifying the right-hand side: \[ P((A \cap B) \cup (A^c \cap B)) = P(B). \] We now express this as: \[ P(A) + P(B) - 2P(A \cap B). \] Thus, the correct answer is: \[ P(A) + P(B) - 2P(A \cap B). \]