The probability that a randomly selected 2-digit number belongs

Total number of cases $={ }^{90} \mathrm{C}_{1}=90$

Now, $2^{\mathrm{n}}-2=(3-1)^{\mathrm{n}}-2$

${ }^{n} C_{0} 3^{n}-{ }^{n} C_{1} \cdot 3^{n-1}+\ldots+(-1)^{n-1} \cdot{ }^{n} C_{n-1} 3+(-1)^{n} \cdot{ }^{n} C_{n}-2$

$3\left(3^{\mathrm{n}-1}-\mathrm{n} 3^{\mathrm{n}-2}+\ldots+(-1)^{\mathrm{n}-1} \cdot \mathrm{n}\right)+(-1)^{\mathrm{n}}-2$

$\left(2^{\mathrm{n}}-2\right)$ is multiply of 3 only when $\mathrm{n}$ is odd

Req. Probability $=\frac{45}{90}=\frac{1}{2}$