Problem 5 · AMC 8 Stretch
Core
Counting & Probability
accounting-for-all-possibilitiescomplementary-countingorganizing-data
A license plate starts with three digits, like \(N_1 N_2 N_3\), where each digit is chosen at random from 0 through 9 (so 007 and 000 are allowed). (1) What is the probability that the three digits are all different? (2) What is the probability that they are NOT all different (some repeat)?
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Answer: All different: 0.72. Not all different: 0.28
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Hint 1 of 4
First count ALL possible three-digit strings. Each of the 3 positions can be any of 10 digits, so use the multiplication principle.
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Hint 2 of 4
Now count the strings where all three digits are different. The first digit can be anything; each later digit must avoid the digits already used.
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Hint 3 of 4
All-different count: \(10\times 9\times 8\). Total: \(10\times 10\times 10\). Probability = all-different / total.
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Approach: Multiplication principle plus complementary counting
- Total strings: each of the 3 positions has 10 choices, so by the multiplication principle there are \(10\times 10\times 10=1000\) possible strings (000 through 999).
- All three different: the first digit can be any of 10, the second must differ from the first (9 left), the third must differ from both (8 left): \(10\times 9\times 8=720\). So \(P(\text{all different})=\frac{720}{1000}=0.72\).
- Not all different is the opposite event, so \(P(\text{not all different})=1-0.72=0.28\).
- Check: exactly-two-equal gives \(3\times(10\times 9)=270\) and all-three-equal gives \(10\), total \(280\), which is \(0.28\). It matches!
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