Given two $3$-distinct-digit-natural numbers. Prove that the probability that at least one of both is a multiple of $10$ is $16/81\simeq 0.197530..$ .

Question

Given two $3$-distinct-digit-natural numbers. Prove that the probability that at least one of both is a multiple of $10$ is $$16/81\simeq 0.197530..$$

My observation is https://www.wolframalpha.com/input/?i=9%21%2810%21-9%21-9%21%29%2Fbinomial%2810%21-9%21%2C2%29 , but why ?

Answer 1

Consider one such number.

There are $9\cdot 9 \cdot 8$ three-digit numbers with distinct digits. (Pick first, then pick second, then pick third)

$9\cdot 8$ of them are multiples of 10. (Pick first, then pick second)

Hence $\frac19$ chance of getting it.

Therefore the chance to get exactly one (among two) is $\binom21 \cdot \frac19 \frac89 = \frac{16}{81}.$

Hence your question might be wrong, because we didn’t add the probability that the two are both multiples of 10.