alleryThree fair coins are tossed. Let E = "at least two heads" and F = "first coin shows tail". Then P(E|F) is
A1/4
B1/8
C1/2
D3/8
Answer & Solution
Correct answer: A. 1/4
1. Sample space of 3 coins has 8 equally likely outcomes.
2. E = {HHH, HHT, HTH, THH}, F = {THH, THT, TTH, TTT}.
3. E ∩ F = {THH}, so P(E ∩ F) = 1/8; P(F) = 4/8 = 1/2.
4. P(E|F) = P(E ∩ F) / P(F) = (1/8) / (1/2) = 1/4.
_Source: syllabus Mathematics, Ch 13 "Probability", §13.2_
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