A certain junior class has 1,000 students and a certain senior class has 800 students. Among these students, there are...

1. Translate the problem requirements: We need to find the probability that when we randomly select 1 student from the junior class (1,000 students) and 1 student from the senior class (800 students), these 2 students turn out to be siblings. We know there are exactly 60 sibling pairs where each pair has 1 junior and 1 senior.
2. Count total possible selection outcomes: Determine how many different ways we can select 1 junior and 1 senior student, regardless of whether they're siblings or not.
3. Count favorable outcomes (sibling pairs): Identify exactly how many ways we can select 1 junior and 1 senior who are actually siblings.
4. Apply basic probability formula: Calculate the probability as the ratio of favorable outcomes to total possible outcomes.

Execution of Strategic Approach

1. Translate the problem requirements

Let's break down what we're being asked to find. We have two classes of students - 1,000 juniors and 800 seniors. Among all these students, there are exactly 60 pairs of siblings, where each sibling pair consists of one junior and one senior.

The question is asking: if we randomly pick one student from the junior class and one student from the senior class, what's the chance that these two students happen to be siblings?

Think of it like this: imagine you're blindfolded and randomly point to one name on the junior class roster and one name on the senior class roster. What are the odds that you've accidentally picked two students who are brother and sister?

Process Skill: TRANSLATE - Converting the word problem into a clear probability question

2. Count total possible selection outcomes

First, let's figure out how many different ways we can pick one junior and one senior student.

For the junior student: we have 1,000 choices
For the senior student: we have 800 choices

Since we're picking one from each class independently, we multiply these numbers together. This is like saying "for each of the 1,000 juniors I could pick, there are 800 different seniors I could pair them with."

Total possible ways = \(1,000 \times 800 = 800,000\)

So there are 800,000 different combinations of picking one junior and one senior.

3. Count favorable outcomes (sibling pairs)

Now let's count how many of these 800,000 combinations would actually result in picking siblings.

We're told there are exactly 60 sibling pairs. Each pair has one junior and one senior. So there are exactly 60 ways to pick one junior and one senior such that they are siblings.

Think about it this way: if I pick any one of the 60 juniors who has a sibling in the senior class, there's exactly one senior (their sibling) that would make them a sibling pair. Similarly, if I pick any one of the 60 seniors who has a sibling in the junior class, there's exactly one junior that would complete the sibling pair.

Favorable outcomes = 60

4. Apply basic probability formula

Probability is simply the number of favorable outcomes divided by the total number of possible outcomes.

Probability = Favorable outcomes ÷ Total possible outcomes
Probability = \(60 \div 800,000\)

Let's simplify this fraction:
\(\frac{60}{800,000} = \frac{6}{80,000} = \frac{3}{40,000}\)

Looking at our answer choices, we can see that \(\frac{3}{40,000}\) matches choice A exactly.

Final Answer

The probability that the 2 students selected will be a sibling pair is \(\frac{3}{40,000}\).

This corresponds to answer choice A.

To verify: out of 800,000 possible ways to select one junior and one senior, only 60 of those ways result in selecting siblings, giving us the probability of \(\frac{60}{800,000} = \frac{3}{40,000}\).

Common Faltering Points

Errors while devising the approach

Faltering Point 1: Misunderstanding what constitutes a favorable outcome
Students might think there are 120 favorable outcomes instead of 60, reasoning that since there are 60 sibling pairs, they could pick the junior first (60 ways) OR the senior first (60 ways). This double-counts the same sibling pairs and shows confusion about what "selecting one from each class" means.

Faltering Point 2: Confusing conditional vs. unconditional probability
Students might try to calculate \(P(\text{senior is sibling | junior selected}) \times P(\text{junior is sibling})\), thinking they need to find the probability that a randomly selected junior has a sibling AND then that sibling gets selected. This overcomplicates the straightforward counting approach needed.

Errors while executing the approach

Faltering Point 1: Incorrect calculation of total possible outcomes
Students might calculate \(1,000 + 800 = 1,800\) instead of \(1,000 \times 800 = 800,000\), confusing the addition principle with the multiplication principle. This happens when they think about selecting "one student from 1,800 total" rather than "one from each class independently."

Faltering Point 2: Arithmetic errors in fraction simplification
When simplifying \(\frac{60}{800,000}\), students might make errors like incorrectly canceling zeros or making division mistakes, potentially arriving at fractions like \(\frac{1}{13,333}\) or \(\frac{6}{80,000}\) without further simplification to \(\frac{3}{40,000}\).

Errors while selecting the answer

Faltering Point 1: Stopping simplification too early
Students might correctly calculate \(\frac{60}{800,000}\) and simplify it to \(\frac{6}{80,000}\) but fail to notice this can be further reduced to \(\frac{3}{40,000}\). Since \(\frac{6}{80,000}\) doesn't match any answer choice exactly, they might incorrectly select the "closest" option rather than completing the simplification.