Question

What proportion of the first 10,000 natural numbers contain a 3

Answer 1

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

$0.3439$

Answer 2

In the first 10 -> 10 - 9 = 1 (contain a 3).
In the first 100 -> 100 - 9 * 9 = 19 (contain a 3).
In the first 10^n -> 10^n - 9^n  (contain a 3).


$So for 10^4 -> 10^4 - 9^4 = 10.000 - 6.561 = 3.439$

The answer is 3.439 numbers contain a 3 in the first 10.000