Problem 21 · 2016 AMC 8
Hard
Counting & Probability
careful-countingcomplementary-counting
A top hat contains 3 red chips and 2 green chips. Chips are drawn randomly, one at a time without replacement, until all 3 of the reds are drawn or until both green chips are drawn. What is the probability that the 3 reds are drawn?
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Answer: B — 2/5.
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Hint 1 of 3
The real reframe: pretend you keep drawing until ALL 5 chips are out, ignoring the stop. Whoever's color finishes "first" (gets its full set out earliest) is decided entirely by the VERY LAST chip — because the last chip is the one color that didn't finish first.
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Hint 2 of 3
So "the 3 reds come out before both greens" is the same event as "the last chip in the full random order is green." Now you just need the chance a random one of the 5 positions — the last — is green.
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Hint 3 of 3
Each of the 5 chips is equally likely to be the last one drawn, so the probability the last is green is simply (number of greens) / (total chips).
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Approach: reframe as 'which color is last in a full shuffle'
- Imagine shuffling all 5 chips and revealing them one by one (ignore the early stop — it doesn't change the order). The reds are all out before the greens are all out exactly when the LAST chip revealed is green.
- Why: if the last chip is green, then both greens were NOT out before that point, so the reds must have completed first; if the last chip is red, the greens finished first.
- Every chip is equally likely to occupy the last position, so P(last is green) = 2 greens / 5 chips = 2/5.
- Why this transfers: "which group finishes drawing first" usually hinges only on the single LAST element of a full random order — a powerful shortcut that dodges casework entirely.
Another way — direct: reds win means the 5th-position chip is green:
- List which chip sits in the final (5th) position of a random arrangement: it's R, R, R, G, or G — 5 equally likely outcomes, 2 of them green.
- The reds-first event is exactly those 2 green-last outcomes, giving 2/5 = 2/5. (You can also enumerate all arrangements and count, landing on the same 2/5.)
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