Problem 13 · AMC 8 Stretch
Core
Counting & Probability
asking-key-questionsconsidering-extreme-cases
A drawer holds \(6\) red, \(7\) green, \(4\) blue, and \(9\) yellow ribbons, all the same shape so you can't tell them apart by feel. In the dark, how many ribbons must you pull out to be sure you have at least one of EVERY color?
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Answer: 23 ribbons
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Hint 1 of 3
Ask the key question: with the worst luck, how many ribbons could you pull and still be missing one color (have only \(3\) of the \(4\) colors)?
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Hint 2 of 3
To stay missing a color as long as possible, imagine grabbing every ribbon of the three biggest colors first, and none of the smallest.
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Hint 3 of 3
The smallest color is blue (\(4\)). Add up the other three counts: \(6 + 7 + 9\). After all those, you still have no blue. What does the next ribbon have to be?
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Approach: Asking the key question — the worst-case grab
- Ask the key question: what is the most ribbons you can hold while still missing a color?
- The unluckiest run grabs every ribbon of the three most common colors and skips the rarest (blue): \(6\) (red) \(+\ 7\) (green) \(+\ 9\) (yellow) \(= 22\) ribbons, and you still have zero blue.
- The \(23\)rd ribbon has no choice but to be blue, completing all four colors. So you must take \(23\) ribbons to be certain. (You might get lucky much sooner, but 'certain' means even in the worst case.)
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