Question

Let \(M\) be a randomly chosen non-empty subset of \(S = \{1,2,3,\ldots,2026\}\).
Which of the following is the probability that the product of all the elements of \(M\) is even?

• A. \(\dfrac{2^{1013}(2^{1013}-1)}{2^{2026}}\)
• B. \(\dfrac{2^{1013}}{2^{2026}}\)
• C. \(\dfrac{2^{1013}(2^{1013}-1)}{2^{2026}-1}\)
• D. \(\dfrac{1}{2^{2026}-1}\)

Hint:

The product is odd only when every chosen element is odd. Count non-empty all-odd subsets, subtract that fraction from 1, keeping the denominator as \(2^{2026}-1\) since \(M\) must be non-empty.

Answer 1

The Correct Option is C

Step 1: Count the total number of non-empty subsets.
The set \(S = \{1,2,\ldots,2026\}\) has \(2026\) elements, so it has \(2^{2026}\) subsets in total, including the empty set. Removing the empty set (since \(M\) is required to be non-empty) leaves \(2^{2026}-1\) equally likely choices for \(M\). This is the sample space size.

Step 2: Split the elements into odd and even.
Among \(1,2,\ldots,2026\), exactly half are odd and half are even, since \(2026\) is even. So there are \(1013\) odd numbers (\(1,3,5,\ldots,2025\)) and \(1013\) even numbers (\(2,4,\ldots,2026\)).

Step 3: Use the fact that a product is even unless every factor is odd.
The product of the elements of \(M\) is odd only when \(M\) contains no even number at all, that is, when \(M\) is a non-empty subset made entirely of odd numbers. If \(M\) contains even one even number, the whole product becomes even. So it is easier to first count the "product is odd" case and subtract from the total.

Step 4: Count subsets whose product is odd.
A subset with product odd must be a non-empty subset of the \(1013\) odd numbers only, chosen from any subset of these \(1013\) odd elements (with even numbers excluded entirely). There are \(2^{1013}\) subsets of the odd numbers, and removing the empty one leaves \(2^{1013}-1\) non-empty all-odd subsets.

Step 5: Compute the probability that the product is even.
\[ P(\text{even}) = 1 - P(\text{odd}) = 1 - \frac{2^{1013}-1}{2^{2026}-1} \]
Combine over the common denominator:
\[ P(\text{even}) = \frac{(2^{2026}-1) - (2^{1013}-1)}{2^{2026}-1} = \frac{2^{2026} - 2^{1013}}{2^{2026}-1} \]
Factor \(2^{1013}\) out of the numerator, since \(2^{2026} = 2^{1013}\cdot 2^{1013}\):
\[ P(\text{even}) = \frac{2^{1013}(2^{1013}-1)}{2^{2026}-1} \]

Step 6: Why the other options are wrong.
Option (A) and (B) use \(2^{2026}\) as the denominator, but the sample space excludes the empty set, so the correct denominator must be \(2^{2026}-1\), not \(2^{2026}\).
Option (D) is actually the probability of the complement event (product odd, with the empty-set case wrongly folded in), not the event asked for.

Final Answer:
\[ \boxed{P(\text{product is even}) = \dfrac{2^{1013}(2^{1013}-1)}{2^{2026}-1}} \]