Let A and B be two independent events such that \(P(A) = 0.4\) and \(P(B) = 0.5\).
(A) \(P(A \cap B) = 0.2\)
(B) \(P(A \cup B) = 0.9\)
(C) \(P(A|B) = 0.4\)
(D) \(P(A' \cap B') = 0.3\)
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → (A), (C) and (D) only
For independent events, \(P(A \cap B) = P(A)P(B)\), and A’, B’ are also independent.
\( P(A \cap B) = 0.4 \times 0.5 = 0.2 \)
\( P(A \cup B) = 0.4 + 0.5 - 0.2 = 0.7 \)
\( \displaystyle P(A|B) = \frac{0.2}{0.5} = 0.4 \)
\( P(A' \cap B') = 0.6 \times 0.5 = 0.3 \)
So \(P(A \cap B) = 0.2\), \(P(A \cup B) = 0.7\), \(P(A|B) = 0.4\) and \(P(A' \cap B') = 0.3\).
Checking the statements
(A) \(P(A \cap B) = 0.2\) — Correct. \(0.4 \times 0.5 = 0.2\).
(B) \(P(A \cup B) = 0.9\) — Incorrect. It is \(0.9 - 0.2 = 0.7\); 0.9 forgets to subtract \(P(A \cap B)\).
(C) \(P(A|B) = 0.4\) — Correct. For independent events \(P(A|B) = P(A) = 0.4\).
(D) \(P(A' \cap B') = 0.3\) — Correct. \(0.6 \times 0.5 = 0.3\), which also equals \(1 - 0.7\).