If one number is to be chosen at random from the integers 0 through 9 and if a second number...

1. Translate the problem requirements: We need to find the probability that when two numbers are randomly chosen from \(\{0,1,2,3,4,5,6,7,8,9\}\), their product will be even. A product is even if at least one factor is even.
2. Identify the complementary approach: Since a product is even when at least one number is even, it's easier to find when the product is odd (both numbers must be odd) and subtract from 1.
3. Count favorable outcomes using the complement: Determine how many ways we can choose two odd numbers, then subtract from total possible outcomes.
4. Calculate the final probability: Convert the ratio to match the answer choices.

Execution of Strategic Approach

1. Translate the problem requirements

Let's break down what we're being asked to find. We have two separate random selections:

• First number: chosen from \(\{0, 1, 2, 3, 4, 5, 6, 7, 8, 9\}\)
• Second number: chosen from \(\{0, 1, 2, 3, 4, 5, 6, 7, 8, 9\}\)

We want to find the probability that when we multiply these two numbers together, the result is even.

Let's think about when a product is even: A product of two numbers is even if at least one of the numbers is even. This is because any number multiplied by an even number gives an even result.

For example: \(3 \times 4 = 12\) (even), \(6 \times 7 = 42\) (even), \(2 \times 8 = 16\) (even)

2. Identify the complementary approach

Since we need "at least one number to be even," it's often easier to think about the opposite situation first. When is a product odd?

A product is odd only when both numbers are odd. Think about it:

• Odd × Odd = Odd (like \(3 \times 5 = 15\))
• Even × Odd = Even (like \(4 \times 5 = 20\))
• Odd × Even = Even (like \(3 \times 6 = 18\))
• Even × Even = Even (like \(4 \times 6 = 24\))

So if we can find the probability that both numbers are odd, we can subtract that from 1 to get our answer.

This approach: \(\mathrm{P(product\,is\,even)} = 1 - \mathrm{P(product\,is\,odd)} = 1 - \mathrm{P(both\,numbers\,are\,odd)}\)

3. Count favorable outcomes using the complement

Let's identify the even and odd numbers in our set \(\{0, 1, 2, 3, 4, 5, 6, 7, 8, 9\}\):

• Even numbers: \(\{0, 2, 4, 6, 8\}\) → 5 numbers
• Odd numbers: \(\{1, 3, 5, 7, 9\}\) → 5 numbers

Now, what's the probability of choosing an odd number on the first draw?

\(\mathrm{P(first\,number\,is\,odd)} = \frac{5}{10} = \frac{1}{2}\)

What's the probability of choosing an odd number on the second draw?

\(\mathrm{P(second\,number\,is\,odd)} = \frac{5}{10} = \frac{1}{2}\)

Since the draws are independent (the second choice doesn't depend on the first), the probability that both are odd is:

\(\mathrm{P(both\,odd)} = \mathrm{P(first\,odd)} \times \mathrm{P(second\,odd)} = \frac{1}{2} \times \frac{1}{2} = \frac{1}{4}\)

4. Calculate the final probability

Now we can find what we actually want:

\(\mathrm{P(product\,is\,even)} = 1 - \mathrm{P(both\,numbers\,are\,odd)}\)

\(\mathrm{P(product\,is\,even)} = 1 - \frac{1}{4} = \frac{3}{4}\)

Let's verify this makes sense: Out of every 4 pairs of numbers we could choose, 3 of them will have an even product, and only 1 will have an odd product.

Final Answer: \(\frac{3}{4}\)

This matches answer choice D.

Common Faltering Points

Errors while devising the approach

Faltering Point 1: Misunderstanding when a product is even
Students often think a product is even only when both numbers are even, rather than understanding that a product is even when at least one number is even. This fundamental misunderstanding leads them to count only the cases where both numbers are even (\(5 \times 5 = 25\) cases) instead of recognizing the complementary approach.

Faltering Point 2: Overlooking the complementary probability strategy
Many students attempt to directly count all cases where the product is even (even×odd, odd×even, even×even), which is more complex and error-prone. They miss that using the complement (finding when the product is odd, then subtracting from 1) is much simpler since a product is odd only when both numbers are odd.

Errors while executing the approach

Faltering Point 1: Incorrectly categorizing 0 as odd
Some students mistakenly classify 0 as an odd number, leading them to count 6 odd numbers \(\{0,1,3,5,7,9\}\) instead of the correct 5 odd numbers \(\{1,3,5,7,9\}\). This changes their probability calculation to \(\frac{6}{10} \times \frac{6}{10} = \frac{36}{100} = \frac{9}{25}\) instead of the correct \(\frac{1}{2} \times \frac{1}{2} = \frac{1}{4}\).

Faltering Point 2: Arithmetic errors in fraction operations
Students may correctly identify that \(\mathrm{P(both\,odd)} = \frac{1}{2} \times \frac{1}{2} = \frac{1}{4}\), but then make errors when computing \(1 - \frac{1}{4}\). Common mistakes include getting \(\frac{4}{4} - \frac{1}{4} = \frac{4}{3}\) (incorrect subtraction) or confusing the final fraction conversion.

Errors while selecting the answer

Faltering Point 1: Selecting the complement probability instead of the desired probability
After correctly calculating \(\mathrm{P(both\,odd)} = \frac{1}{4}\), some students mistakenly select \(\frac{1}{4}\) (answer choice B) as their final answer, forgetting that they need to subtract this from 1 to get \(\mathrm{P(product\,is\,even)} = \frac{3}{4}\). They solve the complement correctly but forget the final step of the complementary approach.