Problem 21 · 2002 AMC 8
Stretch
Counting & Probability
symmetrycomplementary-counting
Harold tosses a nickel four times. The probability that he gets at least as many heads as tails is
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Answer: E — 11/16.
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Hint 1 of 2
A coin has no preference for heads over tails, so "heads ≥ tails" must be exactly as likely as "tails ≥ heads." Those two cases together cover *everything* — but they double-count something.
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Hint 2 of 2
The only outcome that fits both is a 2–2 tie. So the two equal halves overlap precisely on the tie: 2·(what you want) = 1 + P(tie). Find the tie, and you're done.
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Approach: use heads–tails symmetry
- Heads and tails are interchangeable, so P(heads ≥ tails) = P(tails ≥ heads). Together these events cover all outcomes, overlapping only on the 2–2 tie.
- By inclusion-exclusion, P(H≥T) + P(T≥H) = 1 + P(tie), i.e. 2·P(H≥T) = 1 + P(tie). The tie has C(4,2) = 6 of the 16 equally likely outcomes, so P(tie) = 6/16 = 3/8.
- Then P(heads ≥ tails) = (1 + 3/8) ÷ 2 = 11/16.
- *Why this transfers:* when two symmetric events together fill the whole sample space, you don't count both — you write 2·P = 1 + P(overlap) and only the small overlap needs counting.
Another way — just count the winning outcomes:
- "At least as many heads as tails" over 4 flips means 2, 3, or 4 heads.
- Ways: C(4,2) + C(4,3) + C(4,4) = 6 + 4 + 1 = 11, out of 2⁴ = 16 outcomes.
- Probability = 11/16.
Mark:
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