Total two-digit numbers is $90$

So, total number of cases $={}^{90}\mathrm{C}_1=90$

Now, $2^n-2={\left(3-1\right)}^n-2$

$={}^{n}C_0{3}^n-{}^{n}C_1·3^{n-1}+\dots +{\left(-1\right)}^{n-1}·{}^{n}C_{n-1}3+{\left(-1\right)}^n·{}^{n}C_n-2$

$=3\left(3^{n-1}-n{3}^{n-2}+\dots +{\left(-1\right)}^{n-1}·n\right)+{\left(-1\right)}^n-2$

Here, the term ${\left(-1\right)}^n-2$ should be multiple of $3.$

So, $\left(2^n-2\right)$ is multiple of $3$ only when $n$ is odd.

There are total $45$ odd two-digit numbers.

Hence, the required probability $=\frac{{}^{45}C_1}{90}=\frac{45}{90}=\frac{1}{2}$