Problem 14 · 2013 AMC 8
Easy
Counting & Probability
independent-events
Abe holds 1 green and 1 red jelly bean in his hand. Bob holds 1 green, 1 yellow, and 2 red jelly beans in his hand. Each randomly picks a jelly bean to show the other. What is the probability that the colors match?
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Answer: C — 3/8.
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Hint 1 of 2
A "match" can only happen on a color both people own. Yellow is Bob-only, so it can never match — the only matchable colors are green and red. Handle those two cases separately.
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Hint 2 of 2
Two independent picks: within a case multiply (AND), across the separate cases add (OR). First filter to colors both hands share.
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Approach: split into the only two matchable colors
- Only green and red exist in both hands (Bob's yellow can never match), so a match is "both green" OR "both red."
- Both green = P(Abe green) × P(Bob green) = (1/2)(1/4) = 1/8 — multiply because both must happen.
- Both red = P(Abe red) × P(Bob red) = (1/2)(2/4) = 1/4.
- Add the mutually exclusive cases: 1/8 + 1/4 = 1/8 + 2/8 = 3/8.
- Sanity check via counting: 2 × 4 = 8 equally-likely pairs; matches are GG (1 way) and RR (1 × 2 = 2 ways) = 3, so 3/8. Same answer.
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