A box contains 100 balls, numbered from 1 to 100. If three balls are selected at random and with replacement...

1. Translate the problem requirements: We need to find the probability that when three balls are drawn with replacement from balls numbered 1-100, the sum of their numbers is odd. This requires understanding when a sum of three numbers is odd.
2. Identify the fundamental rule for odd sums: Determine the conditions under which the sum of three numbers is odd by analyzing the relationship between odd/even numbers.
3. Calculate individual probabilities: Find the probability of drawing an odd or even numbered ball, since balls are numbered 1-100.
4. Enumerate favorable scenarios: List all combinations of odd/even draws that result in an odd sum, since there are only a few possibilities.
5. Apply probability multiplication and addition: Calculate the probability of each favorable scenario and sum them to get the final answer.

Execution of Strategic Approach

1. Translate the problem requirements

Let's start by understanding what we're being asked. We have a box with 100 balls numbered 1 to 100. We're drawing three balls with replacement (meaning we put each ball back before drawing the next one). We want to find the probability that when we add up the three numbers we drew, we get an odd sum.

The key insight here is understanding when a sum becomes odd. Let's think about this with simple examples first:

• \(3 + 5 + 7 = 15\) (odd)
• \(2 + 4 + 6 = 12\) (even)
• \(1 + 2 + 3 = 6\) (even)
• \(1 + 3 + 5 = 9\) (odd)

Process Skill: TRANSLATE - Converting the probability question into understanding when sums are odd

2. Identify the fundamental rule for odd sums

Let's figure out the pattern for when three numbers add up to an odd sum. Remember these basic rules:

• Odd + Odd = Even
• Even + Even = Even
• Odd + Even = Odd

For three numbers, let's consider all possibilities:

• Odd + Odd + Odd = (Odd + Odd) + Odd = Even + Odd = Odd ✓
• Odd + Odd + Even = (Odd + Odd) + Even = Even + Even = Even ✗
• Odd + Even + Even = Odd + (Even + Even) = Odd + Even = Odd ✓
• Even + Even + Even = (Even + Even) + Even = Even + Even = Even ✗

So the sum is odd in exactly these cases:

• All three numbers are odd (OOO)
• Exactly one number is odd and two are even (OEE, EOE, EEO)

Process Skill: INFER - Recognizing the non-obvious pattern that determines when sums are odd

3. Calculate individual probabilities

Now let's find the probability of drawing an odd or even numbered ball.

Among numbers 1 to 100:

• Odd numbers: 1, 3, 5, 7, ..., 99 (there are 50 odd numbers)
• Even numbers: 2, 4, 6, 8, ..., 100 (there are 50 even numbers)

So:

• Probability of drawing an odd number = \(\frac{50}{100} = \frac{1}{2}\)
• Probability of drawing an even number = \(\frac{50}{100} = \frac{1}{2}\)

4. Enumerate favorable scenarios

From step 2, we know the sum is odd when we have:

Case 1: All three balls are odd (OOO)

• This can happen in exactly 1 way: Odd, Odd, Odd

Case 2: Exactly one ball is odd (OEE, EOE, EEO)

• This can happen in exactly 3 ways:
  • First ball odd, second even, third even (OEE)
  • First ball even, second odd, third even (EOE)
  • First ball even, second even, third odd (EEO)

Process Skill: CONSIDER ALL CASES - Systematically listing all scenarios that give us an odd sum

5. Apply probability multiplication and addition

Now we calculate the probability of each favorable scenario:

Case 1: All three odd (OOO)
Probability = \(\frac{1}{2} \times \frac{1}{2} \times \frac{1}{2} = \frac{1}{8}\)

Case 2: Exactly one odd

• OEE: \(\frac{1}{2} \times \frac{1}{2} \times \frac{1}{2} = \frac{1}{8}\)
• EOE: \(\frac{1}{2} \times \frac{1}{2} \times \frac{1}{2} = \frac{1}{8}\)
• EEO: \(\frac{1}{2} \times \frac{1}{2} \times \frac{1}{2} = \frac{1}{8}\)

Total for Case 2 = \(\frac{1}{8} + \frac{1}{8} + \frac{1}{8} = \frac{3}{8}\)

Total probability of odd sum:
Case 1 + Case 2 = \(\frac{1}{8} + \frac{3}{8} = \frac{4}{8} = \frac{1}{2}\)

6. Final Answer

The probability that the sum of three numbers drawn with replacement is odd is \(\frac{1}{2}\).

This matches answer choice C: \(\frac{1}{2}\).

This result makes intuitive sense - there's exactly a 50-50 chance that any sum of three randomly chosen numbers will be odd, which demonstrates the beautiful symmetry in probability theory.

Common Faltering Points

Errors while devising the approach

1. Misunderstanding when sums are odd

Many students incorrectly assume that if you pick mostly odd numbers, the sum will be odd. They might think "2 odd + 1 even = odd sum" without working through the actual addition rules. The key insight that escapes them is that Odd + Odd = Even, so having two odd numbers actually creates an even partial sum.

2. Forgetting the "with replacement" condition

Students may overlook that balls are drawn "with replacement," meaning each draw is independent with the same probabilities. Some might mistakenly think the probability changes after each draw (like it would without replacement), leading them to create more complex probability calculations than needed.

3. Attempting to enumerate specific number combinations

Instead of recognizing this as an odd/even pattern problem, students might try to list specific combinations of numbers that sum to odd values (like 1+2+4=7, 3+5+9=17, etc.). This approach is unnecessarily complex and time-consuming for 100 balls, when the elegant solution only requires tracking odd/even patterns.

Errors while executing the approach

1. Miscounting favorable scenarios for "exactly one odd"

When calculating Case 2 (exactly one odd number), students often miss one of the three arrangements: OEE, EOE, or EEO. They might only count 2 arrangements instead of all 3, leading to a probability of \(\frac{2}{8}\) instead of \(\frac{3}{8}\) for this case, and a final answer of \(\frac{3}{8}\) instead of \(\frac{1}{2}\).

2. Incorrect probability calculations for individual draws

Students might miscount how many odd vs even numbers exist from 1 to 100. They could incorrectly assume there are 49 odd and 51 even numbers (thinking 50 is odd), or make other counting errors that change the basic probability from \(\frac{1}{2}\) to something else like \(\frac{49}{100}\).

3. Mixing up which cases produce odd sums

Even after correctly identifying the odd/even addition rules, students might incorrectly categorize which scenarios yield odd sums. For example, they might think "two odds and one even" gives an odd sum, when actually it gives an even sum, leading them to calculate probabilities for the wrong cases.

Errors while selecting the answer

No likely faltering points - once students correctly calculate \(\frac{4}{8}\), the simplification to \(\frac{1}{2}\) and matching to answer choice C is straightforward.