A certain basket contains 10 apples, 7 of which are red and 3 are green. If 3 different apples are...

1. Translate the problem requirements: We need to find the probability of getting exactly 2 red apples and 1 green apple when randomly selecting 3 apples from a basket containing 7 red and 3 green apples (10 total).
2. Count total possible ways to select 3 apples: Determine how many different combinations of 3 apples can be chosen from the 10 available apples.
3. Count favorable outcomes for our specific requirement: Calculate how many ways we can select exactly 2 red apples from 7 red apples AND exactly 1 green apple from 3 green apples.
4. Apply probability formula: Divide favorable outcomes by total possible outcomes to get the final probability.

Execution of Strategic Approach

1. Translate the problem requirements

Let's break down what we're looking for in everyday language. We have a basket with 10 apples total - 7 red ones and 3 green ones. We're going to reach in and grab 3 apples at random. We want to know: what's the chance that when we look at our 3 apples, we'll have exactly 2 red apples and exactly 1 green apple?

Think of it like this: we want to end up with a specific "mix" of apples in our hand - not all red, not all green, but exactly 2 red and 1 green.

Process Skill: TRANSLATE - Converting the problem language into a clear mathematical understanding of what we need to find

2. Count total possible ways to select 3 apples

Let's think about this step by step. If we're picking 3 apples from 10 total apples, how many different combinations could we end up with?

Imagine labeling each apple with a number from 1 to 10. We could pick apples {1,2,3} or {1,2,4} or {1,3,5}, etc. What matters is which specific apples we get, not the order we pick them in.

Using our combination formula, the total number of ways to choose 3 apples from 10 is:
\(\mathrm{C(10,3)} = \frac{10!}{3! \times 7!} = \frac{10 \times 9 \times 8}{3 \times 2 \times 1} = \frac{720}{6} = 120\)

So there are 120 different possible combinations of 3 apples we could select.

3. Count favorable outcomes for our specific requirement

Now let's count how many of those 120 combinations give us exactly what we want: 2 red apples and 1 green apple.

To get this specific mix, we need to:

• Choose 2 red apples from the 7 available red apples
• AND choose 1 green apple from the 3 available green apples

For the red apples: How many ways can we pick 2 red apples from 7?
\(\mathrm{C(7,2)} = \frac{7!}{2! \times 5!} = \frac{7 \times 6}{2 \times 1} = \frac{42}{2} = 21\) ways

For the green apples: How many ways can we pick 1 green apple from 3?
\(\mathrm{C(3,1)} = \frac{3!}{1! \times 2!} = \frac{3}{1} = 3\) ways

Since we need both conditions to happen together (2 red AND 1 green), we multiply:
Favorable outcomes = \(21 \times 3 = 63\)

Process Skill: APPLY CONSTRAINTS - Recognizing that we need to satisfy multiple conditions simultaneously

4. Apply probability formula

Now we can find our probability using the basic formula:

Probability = Number of favorable outcomes / Total number of possible outcomes
Probability = \(\frac{63}{120}\)

Let's simplify this fraction by finding the greatest common divisor. Both 63 and 120 are divisible by 3:
\(63 \div 3 = 21\)
\(120 \div 3 = 40\)

So our probability is \(\frac{21}{40}\).

Final Answer

The probability of selecting exactly 2 red apples and 1 green apple is \(\frac{21}{40}\).

Looking at our answer choices, this matches option D: \(\frac{21}{40}\).

To verify this makes sense: \(\frac{21}{40} = 0.525\) or 52.5%. Since we have more red apples than green apples in our basket, it's reasonable that getting 2 red and 1 green (which favors red) would have a fairly high probability.

Common Faltering Points

Errors while devising the approach

1. Misunderstanding the selection constraint

Students often misinterpret "2 red and 1 green" as meaning "at least 2 red and at least 1 green" rather than "exactly 2 red and exactly 1 green." This leads them to calculate additional scenarios like selecting all 3 red apples, thinking it satisfies the condition since it includes "2 red."

2. Confusing combination vs. permutation

Since the problem asks about randomly selecting apples, students may think the order matters and use permutations instead of combinations. They might calculate \(\mathrm{P(10,3)}\) for total outcomes instead of \(\mathrm{C(10,3)}\), leading to incorrect denominators and final answers.

3. Attempting to use conditional probability incorrectly

Some students try to solve this as a sequential selection problem, calculating the probability of picking red, then red, then green in that specific order, without accounting for all possible arrangements of 2 red and 1 green apples.

Errors while executing the approach

1. Arithmetic errors in combination calculations

Students frequently make computational mistakes when calculating combinations, especially with \(\mathrm{C(10,3)} = \frac{10\times9\times8}{3\times2\times1}\). They might incorrectly calculate this as 90 instead of 120, or make similar errors with \(\mathrm{C(7,2)}\) and \(\mathrm{C(3,1)}\).

2. Forgetting to multiply the individual combination results

After correctly calculating \(\mathrm{C(7,2)} = 21\) and \(\mathrm{C(3,1)} = 3\), students may add these values (\(21 + 3 = 24\)) instead of multiplying them (\(21 \times 3 = 63\)) to find the total favorable outcomes, misunderstanding that both conditions must occur simultaneously.

3. Improper fraction simplification

When simplifying \(\frac{63}{120}\), students might make errors in finding the greatest common divisor or in the division process, potentially arriving at incorrect simplified fractions like \(\frac{21}{30}\) instead of \(\frac{21}{40}\).

Errors while selecting the answer

1. Selecting the unsimplified fraction equivalent

If students correctly calculate \(\frac{63}{120}\) but fail to simplify it, they might look for this fraction among the answer choices. Not finding it, they may either guess or incorrectly convert it to a decimal and try to match it with the closest-looking option rather than recognizing that \(\frac{63}{120} = \frac{21}{40}\).

2. Choosing a partial calculation result

Students who calculated intermediate steps correctly might accidentally select an answer that represents just one part of their calculation. For example, they might choose \(\frac{7}{40}\) (option A) if they somehow confused their intermediate result of \(\mathrm{C(7,2)} = 21\) with the total red apples (7) in their final fraction.