Problem 8 · 2013 AMC 8
Easy
Counting & Probability
enumerate-outcomes
A fair coin is tossed 3 times. What is the probability of at least two consecutive heads?
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Answer: C — 3/8.
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Hint 1 of 2
Only 8 outcomes exist for 3 flips — small enough to just list them all and circle the winners. "Consecutive" means the two heads must be neighbors (positions 1-2 or 2-3), not scattered.
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Hint 2 of 2
When the sample space is tiny (23 = 8), full enumeration beats any clever formula. Probability = (favorable outcomes) ÷ (total outcomes).
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Approach: enumerate all 8 equally-likely outcomes
- All 8 outcomes: HHH, HHT, HTH, HTT, THH, THT, TTH, TTT. Each is equally likely, so we just count favorables.
- "At least two consecutive heads" needs HH side by side: HHH, HHT, THH — that's 3. (HTH fails: its heads aren't neighbors.)
- Probability = 3 ÷ 8 = 3/8.
- Watch the trap: HTH has two heads but they're not consecutive, so it doesn't count — reading "consecutive" carefully is the whole problem.
Another way — complement — subtract the 'no HH' outcomes:
- Sometimes it's easier to count the unwanted outcomes. Strings of 3 flips with no two heads adjacent: TTT, TTH, THT, HTT, HTH — 5 of them.
- So favorable = 8 − 5 = 3, giving probability 3/8 = 3/8. (When 'at least' makes direct counting fiddly, count the opposite.)
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