More or Less

As is often the case, it's easier to count the integers below $10000000$ which do not contain the digit $1$ .

This leaves $9$ possibilities for each digit: $0$ and $2$ to $9$ .

So, for the integers up to $9999999$ , we have $9^7-1=4782968$ integers which do not contain the digit $1$ . We had to subtract $1$ because we had counted $0$ as one of our possibilities.

Therefore, $9999999-4782968=5217031$ integers contain the digit $1$ , which is the majority.

I cheated & wrote a programme, 52.2 % to 1 dec. place.

the probability of any digit not being a 1 is 0.9. the probability of numbers with 7 digits or less not having 1 in them is 0.9^7=0.478(3sf). Thus the probability of such numbers containing 1 is (1-0.478)>0.5, thus the majority contain 1.