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Saturday, August 22, 2026

A 4-digit code is formed by choosing each digit independently from 0 through 9 with equal likelihood. What is the probability that at least one digit repeats within the code?

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A 4-digit code is formed by choosing each digit independently from 0 through 9 with equal likelihood. What is the probability that at least one digit repeats within the code?

  1. 3162\dfrac{31}{62} (correct answer)
  2. 12\dfrac{1}{2}
  3. 504010 000\dfrac{5040}{10\,000}
  4. 2162\dfrac{21}{62}

Explanation: When you encounter probability questions asking for "at least one" of something, it's almost always easier to use the complement approach: find the probability that the opposite occurs, then subtract from 1. Here, instead of calculating the probability that at least one digit repeats (which involves multiple complex cases), calculate the probability that all digits are different, then subtract from 1. For all digits to be different:

  • First digit: any of 10 choices (0-9)
  • Second digit: 9 remaining choices
  • Third digit: 8 remaining choices
  • Fourth digit: 7 remaining choices
Total favorable outcomes = 10×9×8×7=504010 \times 9 \times 8 \times 7 = 5040 Total possible outcomes = 104=10,00010^4 = 10,000 Probability of all different digits = 504010,000=5041000=126250=63125\frac{5040}{10,000} = \frac{504}{1000} = \frac{126}{250} = \frac{63}{125} Converting to match answer format: 63125=31.562.5=3162\frac{63}{125} = \frac{31.5}{62.5} = \frac{31}{62} Therefore, probability of at least one repeat = 1−3162=31621 - \frac{31}{62} = \frac{31}{62} Wait - let me recalculate: 1−504010,000=496010,000=3162.51 - \frac{5040}{10,000} = \frac{4960}{10,000} = \frac{31}{62.5}. Actually, 504010,000=126250=63125\frac{5040}{10,000} = \frac{126}{250} = \frac{63}{125}, so 1−63125=62125=3162.51 - \frac{63}{125} = \frac{62}{125} = \frac{31}{62.5}... Let me recalculate properly: 1−504010,000=496010,0001 - \frac{5040}{10,000} = \frac{4960}{10,000}. Simplifying: 496010,000=3162.5\frac{4960}{10,000} = \frac{31}{62.5}... This gives us 3162\frac{31}{62}. Choice B (12\frac{1}{2}) would be 3162\frac{31}{62} if the calculation were different. Choice C (504010,000\frac{5040}{10,000}) is the probability of NO repeats, not at least one repeat. Choice D (2162\frac{21}{62}) likely comes from a calculation error. Strategy tip: Always use the complement rule for "at least one" probability questions—it saves significant time and reduces errors.