A code consists of 3 letters followed by 3 digits. If the letters must be vowels (A, E, I, O, U) and digits cannot repeat, what is the probability that a randomly generated code contains exactly one 'A'?

Example Answer

Assuming each of the three letters is chosen independently and uniformly from the five vowels, the probability of exactly one A is 48/125, or 38.4%. I choose which of the three positions contains A. That position has probability 1/5, and each other position has probability 4/5 of containing a different vowel. So the result is 3 times 1/5 times 4/5 squared.

The no-repeat rule applies to the digits in the question, not to the letters. Provided digit selection is independent of letter selection, the digit count cancels from numerator and denominator. If the interviewer also forbids repeated vowels, the answer changes: three distinct vowels contain A with probability 3/5.

Worked example

There are 3 × 4 × 4 = 48 favorable vowel strings out of 5³ = 125. Each can be paired with the same 10 × 9 × 8 = 720 digit strings.

Why this works

It follows the stated constraint and does not silently impose distinct vowels while using a repeated-vowel denominator.