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 :
Show Hint
Problems involving divisibility by a small number 'k' can often be solved by analyzing the remainders (modulo k). Look for repeating patterns, which simplifies the condition on 'n'.
We need to find the condition on n for which $2^n - 2$ is a multiple of 3.
$2^n - 2 \equiv 0 \pmod{3}$.
$2^n \equiv 2 \pmod{3}$.
Let's check the pattern of powers of 2 modulo 3.
$2^1 = 2 \equiv 2 \pmod{3}$.
$2^2 = 4 \equiv 1 \pmod{3}$.
$2^3 = 8 \equiv 2 \pmod{3}$.
$2^4 = 16 \equiv 1 \pmod{3}$.
The pattern of $2^n \pmod{3}$ is 2, 1, 2, 1, ...
We can see that $2^n \equiv 2 \pmod{3}$ when n is an odd number.
So, the condition is that n must be an odd natural number.
The problem asks for the probability that a randomly selected 2-digit number is odd.
The set of 2-digit numbers is $\{10, 11, 12, ..., 99\}$.
Total number of 2-digit numbers = $99 - 10 + 1 = 90$.
Now, we need to find the number of favorable outcomes, which is the number of odd 2-digit numbers.
The odd 2-digit numbers are $\{11, 13, 15, ..., 99\}$.
This is an arithmetic progression with first term $a=11$, last term $l=99$, and common difference $d=2$.
Number of terms = $\frac{l-a}{d} + 1 = \frac{99-11}{2} + 1 = \frac{88}{2} + 1 = 44 + 1 = 45$.
The probability is the ratio of favorable outcomes to total outcomes.
Probability = $\frac{\text{Number of odd 2-digit numbers}}{\text{Total number of 2-digit numbers}} = \frac{45}{90} = \frac{1}{2}$.
