Question 701448
A more general question would be:
Are there more n-digit numbers with a 1 in them than without a 1 in them?
The fraction of n-digit numbers without a 1 in them
(the probability that 1 would not be any of the n digits) is
{{{b[n]=(8/9)(9/10)^(n-1)}}}
because the first digit cannot be zero,
and it would not be 1 for {{{8/9}}} of the possible digits,
while the other {{{(n-1)}}} digits have no restriction,
and would not be 1 in {{{9/10}}} of the cases.
If you were taught the name you would recognize that as a
"geometric sequence" or "geometric progression".
Even if you did not know of a name for them,
it is clear that the fraction of n-digit numbers without a 1 in them
(or the probability of an n-digit number not having a 1)
decreases as towards zero as {{{n}}} increases.
It is {{{8/9}}} for 1-digit numbers,
{{{(8/9)(9/10)=8/10}}} for 2-digit numbers,
and keeps getting smaller as we keep including more {{{(9/10)<1}}} factors.
For 7-digit numbers the fraction of them without a 1 is
{{{(8/9)(9/10)^6=0.472392<1/2}}} ,
so there are {{{highlight(more)}}} 7-digit numbers with a 1 in them than without a 1 in them.
 
EXTRAS:
Calculating {{{b[6]}}}, I would find that {{{b[6]>1/2}}},
but how would I find the value of {{{n}}} where {{{b[n]}}} becomes smaller than {{{1/2}}} without calculating and tabulating {{{b[n]}}} values for n=1, 2, 3, ...?
That's where you use logarithms.
Sorry, I'm getting carried away.