The probability that a randomly selected 2 - digit number belongs...

Total two-digit numbers is $90$

So, total number of cases $= C1 90 = 90$

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

$= C0 n {3}^{n} - C1 n \cdot {3}^{n - 1} + \ldots + {\left(- 1\right)}^{n - 1} \cdot Cn - 1 n 3 + {\left(- 1\right)}^{n} \cdot Cn n - 2$

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

Here, the term ${\left(- 1\right)}^{n} - 2$ should be multiple of $3 .</p><p>So, \left({2}^{n} - 2\right) is multiple of$3$only when$n$is odd.</p><p>There are total$45$odd two-digit numbers.</p><p>Hence, the required probability =\dfrac{C1 45}{90} = \dfrac{45}{90} = \dfrac{1}{2}</p>$