If x is to be chosen at random from the set {1, 2, 3, 4} and y is to be...

1. Translate the problem requirements: We need to find the probability that the product xy is even when x is randomly selected from \({1, 2, 3, 4}\) and y is randomly selected from \({5, 6, 7}\). A product is even if at least one factor is even.
2. Identify the nature of numbers in each set: Categorize each number as even or odd to understand which combinations will produce even products.
3. Count favorable outcomes using the multiplication principle: Determine how many of the total possible xy combinations result in an even product.
4. Calculate the probability as a fraction: Express the ratio of favorable outcomes to total outcomes and match with answer choices.

Execution of Strategic Approach

1. Translate the problem requirements

Let's understand what we're looking for in everyday terms. We have two bags of numbers: one bag contains \({1, 2, 3, 4}\) and another contains \({5, 6, 7}\). We're going to pick one number from each bag and multiply them together. We want to know: what's the chance that this product will be an even number?

A product is even when at least one of the numbers being multiplied is even. Think of it this way: if you multiply any even number by anything else, you'll always get an even result. It's like having an "even contamination" - once even is involved, the whole product becomes even.

Process Skill: TRANSLATE - Converting the probability question into a clear understanding of when products are even

2. Identify the nature of numbers in each set

Let's look at each number and identify whether it's even or odd:

From set \({1, 2, 3, 4}\):

• 1 is odd
• 2 is even
• 3 is odd
• 4 is even

So we have 2 odd numbers and 2 even numbers in the first set.

From set \({5, 6, 7}\):

• 5 is odd
• 6 is even
• 7 is odd

So we have 2 odd numbers and 1 even number in the second set.

This means: if we pick an even number from the first set (2 or 4), our product will definitely be even no matter what we pick from the second set. If we pick an odd number from the first set (1 or 3), our product will be even only if we pick the even number (6) from the second set.

3. Count favorable outcomes using the multiplication principle

Let's list all possible combinations and see which ones give us even products:

When x = 1 (odd): \(1 \times 5 = 5\) (odd), \(1 \times 6 = 6\) (even), \(1 \times 7 = 7\) (odd)
When x = 2 (even): \(2 \times 5 = 10\) (even), \(2 \times 6 = 12\) (even), \(2 \times 7 = 14\) (even)
When x = 3 (odd): \(3 \times 5 = 15\) (odd), \(3 \times 6 = 18\) (even), \(3 \times 7 = 21\) (odd)
When x = 4 (even): \(4 \times 5 = 20\) (even), \(4 \times 6 = 24\) (even), \(4 \times 7 = 28\) (even)

Total possible outcomes: \(4 \times 3 = 12\) combinations

Even products: \(1 \times 6\), \(2 \times 5\), \(2 \times 6\), \(2 \times 7\), \(3 \times 6\), \(4 \times 5\), \(4 \times 6\), \(4 \times 7\)
Counting these up: 8 favorable outcomes

Alternatively, we can count the unfavorable outcomes (odd products): \(1 \times 5\), \(1 \times 7\), \(3 \times 5\), \(3 \times 7\)
That's 4 unfavorable outcomes, so \(12 - 4 = 8\) favorable outcomes.

Process Skill: CONSIDER ALL CASES - Systematically examining every possible combination to ensure we don't miss any outcomes

4. Calculate the probability as a fraction

Now we can calculate our probability:

Probability = Number of favorable outcomes / Total number of outcomes
Probability = \(\frac{8}{12} = \frac{2}{3}\)

Let's verify this makes sense: We found that 8 out of 12 possible combinations result in even products. This means there's a \(\frac{2}{3}\) chance of getting an even product.

Looking at our answer choices, \(\frac{2}{3}\) corresponds to choice D.

5. Final Answer

The probability that xy will be even is \(\frac{2}{3}\).

Answer: D

Common Faltering Points

Errors while devising the approach

1. Misunderstanding when a product is even

Students often think that BOTH numbers need to be even for the product to be even, when actually a product is even if AT LEAST ONE of the factors is even. This fundamental misunderstanding leads them to count only cases where both x and y are even, drastically undercounting favorable outcomes.

2. Confusing "probability that xy is even" with "probability that xy is odd"

Some students might inadvertently solve for when the product is odd instead of even, especially if they're rushing through the problem setup. Since we want \(\mathrm{P(even)}\), but they calculate \(\mathrm{P(odd)}\), they would get \(\frac{1}{3}\) instead of \(\frac{2}{3}\).

3. Attempting to use complex probability formulas unnecessarily

Students might try to apply conditional probability or other advanced concepts when a simple enumeration approach works perfectly fine. This can lead to overcomplication and errors in a straightforward counting problem.

Errors while executing the approach

1. Incorrectly identifying odd vs even numbers

Basic arithmetic errors can occur when students rush through identifying which numbers are odd or even. For example, they might mistakenly classify numbers, leading to wrong categorizations in their systematic listing.

2. Missing combinations when listing all possibilities

When systematically listing the 12 possible combinations (\(4 \times 3\)), students might accidentally skip some pairs or double-count others. This is especially likely when working quickly under time pressure, leading to an incorrect count of favorable outcomes.

3. Arithmetic errors when calculating products

Students might make simple multiplication mistakes when computing products like \(3 \times 6\) or \(4 \times 7\), which could lead them to incorrectly classify whether a particular outcome is even or odd.

Errors while selecting the answer

1. Forgetting to reduce the fraction 8/12

After correctly finding 8 favorable outcomes out of 12 total, students might select an answer choice that represents \(\frac{8}{12}\) in unreduced form, not realizing they need to simplify it to \(\frac{2}{3}\) to match the given answer choices.

2. Calculating 1 minus their result when they shouldn't

If students have been working with complementary probability concepts recently, they might incorrectly calculate \(1 - \frac{8}{12} = \frac{4}{12} = \frac{1}{3}\), thinking they need the complement of their calculated probability.