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Q.
The probability that a randomly selected $2$ -digit number belongs to the set $\left\{n \in N:\left(2^{n}-2\right)\right.$. is a multiple of 3$\}$ is equal to
Total number of cases $={ }^{90} C_{1}=90$
Now, $2^{n}-2=(3-1)^{n}-2$
${ }^{n} C_{0} 3^{n}-{ }^{n} C_{1} \cdot 3^{n-1}+\ldots .+(-1)^{n-1} \cdot{ }^{n} C_{n-1} 3+(-1)^{n} \cdot{ }^{n} C_{n}-2$
$3\left(3^{n-1}-n 3^{n-2}+\ldots \ldots+(-1)^{n-1} \cdot n\right)+(-1)^{n}-2$
$\left(2^{n}-2\right)$ is multiply of $3$ only when $n$ is odd
Req. Probability $=\frac{45}{90}=\frac{1}{2}$