The probability that a randomly chosen 5 - digit number is made...

Total number of five-digit numbers are $9\left({10}^{4}\right)$.

Case I : Choosing two non-zero digits i.e. in $C2 9$ ways.

So, number of five-digit numbers containing both digits $= {2}^{5} - 2$.

Case II : Choose one non-zero and one zero as digit i.e. in $C1 9$ ways.

So, number of five-digit numbers containing one non-zero and one zero both $= {2}^{4} - 1$.

So, favourable number of ways $= C2 9 \left({2}^{5} - 2\right) + C1 9 \left({2}^{4} - 1\right) = 36\left(30\right) + 9\left(15\right) = 1080 + 135 = 1215$

Thus, required probability $= \dfrac{1215}{9 \times {10}^{4}} = \dfrac{135}{{10}^{4}}$.