# probability – What’s more likely: \$7\$-digit number with no \$1\$’s or at least one \$1\$ among its digits?

A $$7$$-digit number is chosen at random. Which is more likely: the number has no $$1$$‘s among its digits or the number has at least one $$1$$ among its digits?

Here’s how I did it: The question is asking whether $$8(9)^6$$ (the number of those with no $$1$$‘s among its digits) or $$9(10)^6 – 8(9)^6$$ (the number of those with at least one $$1$$ among its digits). Some tedious multiplying shows that $$8(9)^6 = 4241528 < 4500000$$, which demonstrates that $$9(10^6) – 8(9)^6$$ i.e. the number having at least one $$1$$‘s among its digits is more likely.

However, I am wondering if there is a slicker way to get the answer without having to do any tedious multipication.