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

Options

A. `(1)/(6)`
B. `(2)/(3)`
C. `(1)/(2)`
D. `(1)/(3)`

Answer

To solve the problem, we need to determine the probability that a randomly selected two-digit number belongs to the set defined by the condition that \(2^n - 2\) is a multiple of 3, where \(n\) is a natural number.

### Step-by-Step Solution:

1. **Identify the Range of Two-Digit Numbers:**
The two-digit numbers range from 10 to 99. Therefore, the total count of two-digit numbers is:
\[ 99 - 10 + 1 = 90 \]

2. **Understanding the Condition:**
We need to find when \(2^n - 2\) is a multiple of 3. This can be rewritten as:
\[ 2^n - 2 \equiv 0 \mod 3 \]
This simplifies to:
\[ 2^n \equiv 2 \mod 3 \]

3. **Finding the Values of \(n\):**
We can evaluate \(2^n \mod 3\):
- For \(n = 1\): \(2^1 \equiv 2 \mod 3\)
- For \(n = 2\): \(2^2 \equiv 1 \mod 3\)
- For \(n = 3\): \(2^3 \equiv 2 \mod 3\)
- For \(n = 4\): \(2^4 \equiv 1 \mod 3\)
- Continuing this pattern, we see that \(2^n \equiv 2 \mod 3\) when \(n\) is odd.

4. **Counting the Odd Natural Numbers:**
The odd natural numbers starting from 1 up to some limit can be counted. The odd numbers in the range of natural numbers are:
\[ 1, 3, 5, 7, 9, \ldots \]
Since we are interested in the two-digit numbers, we need to find how many odd numbers fall within the range of two-digit numbers.

5. **Finding Odd Numbers from 10 to 99:**
The odd two-digit numbers start from 11 and end at 99. They can be listed as:
\[ 11, 13, 15, \ldots, 99 \]
This forms an arithmetic sequence where:
- First term \(a = 11\)
- Common difference \(d = 2\)
- Last term \(l = 99\)

To find the number of terms \(n\) in this sequence, we can use the formula for the \(n\)-th term of an arithmetic sequence:
\[ l = a + (n-1)d \]
Plugging in the values:
\[ 99 = 11 + (n-1) \cdot 2 \]
Simplifying this:
\[ 99 - 11 = (n-1) \cdot 2 \implies 88 = (n-1) \cdot 2 \implies n-1 = 44 \implies n = 45 \]

6. **Calculating the Probability:**
The probability \(P\) that a randomly selected two-digit number belongs to the set where \(2^n - 2\) is a multiple of 3 (i.e., \(n\) is odd) is given by:
\[ P = \frac{\text{Number of favorable outcomes}}{\text{Total outcomes}} = \frac{45}{90} = \frac{1}{2} \]

### Final Answer:
The probability that a randomly selected two-digit number belongs to the set where \(2^n - 2\) is a multiple of 3 is \(\frac{1}{2}\).