we know that
It is much easier to compute the number of numbers that do not contain a \(3\).
In the
first \(10\) numbers \(9\) do not contain a \(3\).
In the first \(100\) numbers \(81\) do not contain a \(3\). (This
is because these numbers are formed by selecting a digit out of
\([0, 1, 2, 4, 5, 6, 7, 8, 9]\) for
the first position and another digit out of the same set for the last position and this may
be done in
\(9*9 = 81 ways.\)
In the first \(1000\) numbers, reasoning as above, there are
\(9^{3}\) numbers that do not contain a \(3\).
Now the pattern should be becoming obvious.
In the first
\(10^{n}\) numbers there are \(9^{n}\)
numbers that do not contain a \(3\).
So
In the first
\(10^{4}\) numbers there are \(9^{4}\)
numbers that do not contain a \(3\).
\(9^{4}= 6,561\)
\(10,000- 6,561= 3,439\)
Thus
the ratio of those numbers
that do contain a 3 over the total is
equal to
\(Ratio=\frac{3,439}{10,000} \\ \\ Ratio=0.3439\)
therefore
the answer is
\(0.3439\)
