If a four-digit PIN code does not allow repeating digits and must start with a non-zero digit, what is the probability that a randomly generated PIN satisfies these conditions?

Example Answer

If “randomly generated” means four independent uniform digits from 0 through 9, including leading zeroes, there are 10,000 possible PIN strings. A valid PIN has nine choices for the first digit. The next digit has nine choices, because it may be zero but cannot repeat the first. The remaining digits have eight and seven choices. That gives 9 times 9 times 8 times 7, or 4,536, valid PINs, and a probability of 45.36%.

The denominator changes if the generator already rules out a leading zero. Among those 9,000 strings, 4,536 are valid, giving 50.4%. If it already enforces both constraints, every generated PIN is valid. I would clarify the generator before choosing a denominator.

Worked example

P(valid) = 4536/10000 = 0.4536; P(no repeats | first digit nonzero) = 4536/9000 = 0.504.

Why this works

It separates an unconditional probability from a conditional one; both percentages are correct for their stated sample spaces.