What is the probability that a random 17-character identifier contains exactly five letters?

Instruction: Generate 17 independent characters uniformly from uppercase A–Z and digits 0–9, allowing repetition. Calculate the probability of exactly five letters. This is a hypothetical identifier, not a real VIN.

Context: Tests a binomial letter count under a precisely defined independent character generator.

Updated

Official answer

Under this simplified identifier model, each independent position is a letter with probability 26/36 and a digit with probability 10/36. Choose which five of the 17 positions contain letters:

P(exactly five letters) = C(17,5)(26/36)^5(10/36)^12

= C(17,5)26^5 10^12 / 36^17 ≈ 0.000256611, or 0.0256611%.

The numerator chooses five letter positions, fills each with one of 26 letters, and fills the other twelve with digits. Repetition is allowed. This is a hypothetical uniform string generator; it does not model the structural constraints of an actual vehicle identification number.

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