Understanding the Set and Subset Properties

The set is $S = \{1, 2, 3, \dots, 2026\}$. The total number of elements is $n = 2026$.

We need to find the probability that the product of elements in a randomly chosen non-empty subset $M$ of $S$ is even.

Analyzing Elements in Set S

The set $S$ contains both odd and even numbers.

• Number of odd elements = $1013$. (These are $1, 3, 5, \dots, 2025$)
• Number of even elements = $1013$. (These are $2, 4, 6, \dots, 2026$)

Calculating Total Possible Outcomes

The total number of subsets possible for a set with $n$ elements is $2^n$.

For set $S$, the total number of subsets is $2^{2026}$.

The question specifies a non-empty subset. So, we exclude the empty set. The total number of possible outcomes (non-empty subsets) is $2^{2026} - 1$.

Determining Favorable Outcomes (Even Product)

The product of elements in a subset $M$ is even if the subset $M$ contains at least one even number.

Consider the complementary event: the product of elements is odd.

The product of elements in $M$ is odd if and only if all elements chosen for $M$ are odd numbers.

Let $O$ be the set of odd numbers in $S$. We found $|O| = 1013$.

The subsets $M$ that result in an odd product must be subsets of $O$. The total number of subsets of $O$ is $2^{|O|} = 2^{1013}$.

Since $M$ must be non-empty, the number of non-empty subsets of $O$ (which result in an odd product) is $2^{1013} - 1$.

Calculating the Probability

Let $P(\text{Odd Product})$ be the probability that the product of elements of a randomly chosen non-empty subset is odd.

$ P(\text{Odd Product}) = \frac{\text{Number of non-empty subsets with odd product}}{\text{Total number of non-empty subsets}} $

$ P(\text{Odd Product}) = \frac{2^{1013} - 1}{2^{2026} - 1} $

The event that the product is even is the complement of the event that the product is odd.

Let $P(\text{Even Product})$ be the probability that the product of elements is even.

$ P(\text{Even Product}) = 1 - P(\text{Odd Product}) $

$ P(\text{Even Product}) = 1 - \frac{2^{1013} - 1}{2^{2026} - 1} $

To subtract, find a common denominator:

$ P(\text{Even Product}) = \frac{(2^{2026} - 1) - (2^{1013} - 1)}{2^{2026} - 1} $

Simplify the numerator:

$ P(\text{Even Product}) = \frac{2^{2026} - 1 - 2^{1013} + 1}{2^{2026} - 1} $

$ P(\text{Even Product}) = \frac{2^{2026} - 2^{1013}}{2^{2026} - 1} $

Factor out $2^{1013}$ from the numerator:

$ P(\text{Even Product}) = \frac{2^{1013}(2^{2026 - 1013} - 1)}{2^{2026} - 1} $

$ P(\text{Even Product}) = \frac{2^{1013}(2^{1013} - 1)}{2^{2026} - 1} $

This result matches one of the options.