The probability that a randomly chosen 5-digit number is made from exactly two digits is:

First we note that a 5-digit number runs from 10000 to 99999, so its left-most digit can be 1,2,…,9. Hence the total number of 5-digit numbers is

$$9\times10^4=90000.$$

Our task is to count how many of these 90000 numbers use exactly two distinct digits. Both chosen digits must appear at least once inside the 5 positions.

We begin by choosing the two digits. Choosing two different digits out of the ten possible digits 0-9 is done by the combination formula

$$\,^nC_r=\frac{n!}{r!(n-r)!},$$

so

$$\binom{10}{2}=45$$

unordered pairs of digits are possible. Now we split these 45 pairs into two convenient classes, depending on whether 0 is one of the chosen digits.

Class 1: Neither digit is 0. There are 9 non-zero digits, so the number of such unordered pairs is

$$\binom{9}{2}=36.$$

Fix any one of these 36 pairs, say {a,b} with a≠b and both a,b∈{1,…,9}. We form 5-length strings from {a,b} such that both a and b occur and the first digit is non-zero (already guaranteed here). Without any restriction each of the 5 positions could be chosen in 2 ways, giving

$$2^5=32$$

strings. From these 32 we must discard the two monochromatic strings aaaaa and bbbbb in which both digits do not appear. Hence

$$32-2=30$$

valid numbers arise from any fixed pair {a,b}. Multiplying by the 36 choices of the pair gives

$$36\times30=1080$$

valid 5-digit numbers in Class 1.

Class 2: Exactly one digit is 0. Here the pair is {0,b} with b∈{1,…,9}. There are

$$9$$

such pairs. For each pair the first digit of the number cannot be 0 (otherwise we would not have a 5-digit number), so the first digit is forced to be b. The remaining four positions can be filled with either 0 or b, but 0 must appear at least once among them so that both digits occur. For the last four places we have

$$2^4=16$$

total strings, of which exactly one, namely bbbb, contains no 0. Therefore the admissible strings for the last four places are

$$16-1=15.$$\

Thus each pair {0,b} contributes 15 numbers, and altogether Class 2 supplies

$$9\times15=135$$

valid 5-digit numbers.

Adding the two classes together, the number of favourable 5-digit numbers is

$$1080+135=1215.$$

The required probability is therefore

$$\frac{\text{favourable}}{\text{total}}=\frac{1215}{90000}.$$

We now simplify this fraction. Both numerator and denominator are divisible by 45, because

$$1215=45\times27,\qquad 90000=45\times2000.$$

So we get

$$\frac{1215}{90000}=\frac{27}{2000}.$$

To match the form given in the options we multiply numerator and denominator by 5:

$$\frac{27}{2000}=\frac{27\times5}{2000\times5}=\frac{135}{10000}=\frac{135}{10^4}.$$

This expression appears as Option A.

Hence, the correct answer is Option A.