Problem 11 · 2015 AMC 8
Medium
Counting & Probability
multiplication-principleprobability
In the small country of Mathland, all automobile license plates have four symbols. The first must be a vowel (A, E, I, O, or U), the second and third must be two different letters among the 21 non-vowels, and the fourth must be a digit (0 through 9). If the symbols are chosen at random subject to these conditions, what is the probability that the plate will read "AMC8"?
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Answer: B — 1/21,000.
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Hint 1 of 2
'AMC8' is one single plate. If every allowed plate is equally likely, you don't need to do anything with AMC8 itself — just count how many plates exist in all, and the probability is 1 over that.
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Hint 2 of 2
Multiply the choices slot by slot, but watch slot 3: it must differ from slot 2, so it has 21 − 1 = 20 options, not 21. (One slot's count depending on an earlier slot is common in plate/seating counts.)
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Approach: count all equally likely plates; the target is just one of them
- Count total plates by the rule of product, going slot by slot: 5 vowels, 21 non-vowels, then 20 (the second letter must differ from the third), then 10 digits.
- Total = 5 × 21 × 20 × 10 = 21,000 equally likely plates.
- 'AMC8' is exactly one of those, so its probability is 1/21,000.
- Why this transfers: the chance of one specific equally-likely outcome is simply 1 ÷ (number of outcomes) — the work is all in the careful count, especially the dependent slot.
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