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(A) The total number of $2$-digit numbers is $90$ (from $10$ to $99$).We need to check when $(2^{n}-2)$ is a multiple of $3$.Consider the expression m

Vedclass · Accepted Answer

$\frac{1}{2}$

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(A) The total number of $2$-digit numbers is $90$ (from $10$ to $99$).We need to check when $(2^{n}-2)$ is a multiple of $3$.Consider the expression modulo $3$:$2 \equiv -1 \pmod{3}$So,$2^{n}-2 \equiv (-1)^{n}-2 \pmod{3}$.If $n$ is even,$(-1)^{n}-2 = 1-2 = -1 \equiv 2 \pmod{3}$.If $n$ is odd,$(-1)^{n}-2 = -1-2 = -3 \equiv 0 \pmod{3}$.Thus,$(2^{n}-2)$ is a multiple of $3$ if and only if $n$ is an odd number.In the set of $2$-digit numbers ${10, 11, 12, \dots, 99}$,the odd numbers are ${11, 13, 15, \dots, 99}$.The number of odd integers in this range is $\frac{99-11}{2} + 1 = 44 + 1 = 45$.The probability is $\frac{	ext{Number of favorable outcomes}}{	ext{Total number of outcomes}} = \frac{45}{90} = \frac{1}{2}$.

The probability that a randomly selected $2$-digit number belongs to the set $\{n \in N : (2^{n}-2) \text{ is a multiple of } 3\}$ is equal to:

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