If A and B are two events such that P(A) = 0.5, P(B) = 0.4 and \(P(A'\cup B') = 0.8\), then A and B are
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → independent events
By De Morgan’s law, \(A'\cup B'\) is the complement of \(A\cap B\):
\(P(A\cap B) = 1-P(A'\cup B')\)
\(P(A\cap B) = 1-0.8 = 0.2\)
Also
\(P(A)\,P(B) = 0.5\times 0.4 = 0.2\)
So \(P(A\cap B) = P(A)\,P(B)\), and A and B are independent.
Why the other options are incorrect
Option (2) mutually exclusive events — Incorrect. \(P(A\cap B) = 0.2\), not 0.
Option (3) events with \(A\subset B\) — Incorrect. Then \(P(A\cap B)\) would be P(A) = 0.5, not 0.2; also P(A) > P(B) here.
Option (4) exhaustive events — Incorrect. \(P(A\cup B) = 0.5+0.4-0.2 = 0.7\), not 1.