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

First Case: Choose two non-zero digits ${ }^{9} \mathrm{C}_{2}$

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

Second Case: Choose one non-zero \& one zero as digit ${ }^{9} \mathrm{C}_{1}$

Number of 5 -digit numbers containg one

non zero and one zero both $=\left(2^{4}-1\right)$

Required prob.

$=\frac{\left({ }^{9} \mathrm{C}_{2} \times\left(2^{5}-2\right)+{ }^{9} \mathrm{C}_{1} \times\left(2^{4}-1\right)\right)}{9 \times 10^{4}}$

$=\frac{36 \times(32-2)+9 \times(16-1)}{9 \times 10^{4}}$

$=\frac{4 \times 30+15}{10^{4}}=\frac{135}{10^{4}}$