Question-7

Which of the following option(s) is(are) true?

• If events \(A\) and \(B\) are disjoint, then \(P(A \cup B) = P(A) + P(B)\).
• If events \(A\) and \(B\) are independent, then \(P(A \cup B) = P(A) + P(B)\).
• If events \(A\) and \(B\) are disjoint, then \(P(A \cap B) = P(A) \times P(B)\).
• If events \(A\) and \(B\) are independent, then \(P(A \cap B) = P(A) \times P(B)\).

Hint

Two events \(A\) and \(B\) are disjoint (mutually exclusive) if \(A \cap B = \phi\) and \(A\) and \(B\) are independent events if \(P(A | B ) = P(A | B^c) = P(A)\).

Answer

• If events \(A\) and \(B\) are disjoint, then \(P(A \cup B) = P(A) + P(B)\).
• If events \(A\) and \(B\) are independent, then \(P(A \cup B) = P(A) + P(B)\).
• If events \(A\) and \(B\) are disjoint, then \(P(A \cap B) = P(A) \times P(B)\).
• If events \(A\) and \(B\) are independent, then \(P(A \cap B) = P(A) \times P(B)\).

Solution

Two events \(A\) and \(B\) are mutually exclusive (disjoint) if \(A \cap B = \phi\). Then, \(P(A \cap B) = 0 \implies P(A \cup B) = P(A) + P(B).\)

\(P(A \cap B) = P(A | B) \times P(B) = P(A) \times P(B)\) because \(P(A | B ) = P(A | B^c) = P(A)\) if \(A\) and \(B\) are independent.