Let A, B and F be events with P(F) = 0.5, \(P(A\cap F) = 0.2\), \(P(B\cap F) = 0.25\) and \(P(A\cap B\cap F) = 0.1\). Consider the following statements:
(A) P(A|F) = 0.4
(B) \(P(B'|F) = 0.75\)
(C) \(P(A\cup B|F) = 0.7\)
(D) \(P(A\cap B|F) = 0.2\)
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
Each conditional probability given F is the probability of the part inside F divided by P(F) = 0.5:
\( \displaystyle P(A|F) = \frac{0.2}{0.5} = 0.4 \)
\( \displaystyle P(B|F) = \frac{0.25}{0.5} = 0.5 \)
\( \displaystyle P(A\cap B|F) = \frac{0.1}{0.5} = 0.2 \)
Conditional probabilities given F follow the same rules as ordinary probabilities, including the addition theorem and the complement rule:
\(P(A\cup B|F) = P(A|F)+P(B|F)-P(A\cap B|F)\)
\(P(A\cup B|F) = 0.4+0.5-0.2 = 0.7\)
So P(A|F) = 0.4, P(B|F) = 0.5, \(P(B'|F) = 1-0.5 = 0.5\), \(P(A\cup B|F) = 0.7\) and \(P(A\cap B|F) = 0.2\).
Checking the statements
(A) P(A|F) = 0.4 — Correct. \(\frac{0.2}{0.5} = 0.4\).
(B) \(P(B'|F) = 0.75\) — Incorrect. \(P(B'|F) = 1-P(B|F) = 1-0.5 = 0.5\).
(C) \(P(A\cup B|F) = 0.7\) — Correct. \(0.4+0.5-0.2 = 0.7\).
(D) \(P(A\cap B|F) = 0.2\) — Correct. \(\frac{0.1}{0.5} = 0.2\).