From a group of 8 volunteers, including Andrew and Karen, 4 people are to be selected at random to organize...

1. Translate the problem requirements: We need to find the probability that when selecting 4 people from 8 volunteers, Andrew is selected AND Karen is not selected simultaneously.
2. Identify the constraint structure: Since Andrew must be in and Karen must be out, we're really selecting 3 more people from the remaining 6 volunteers to join Andrew.
3. Count favorable outcomes: Calculate how many ways we can select 3 people from the 6 remaining volunteers (excluding Andrew and Karen).
4. Count total possible outcomes: Calculate how many ways we can select 4 people from all 8 volunteers without any restrictions.
5. Calculate probability ratio: Divide favorable outcomes by total outcomes to get the final probability.

Execution of Strategic Approach

1. Translate the problem requirements

Let's break down what we need to find in simple terms. We have 8 volunteers total, and we're selecting 4 of them for a charity event. We want the probability that Andrew gets selected AND Karen does NOT get selected in the same group of 4.

Think of it this way: we need Andrew to be "in" and Karen to be "out" when the final 4 people are chosen. This is a specific condition we're looking for among all the possible ways to select 4 people.

Process Skill: TRANSLATE - Converting the probability language into clear selection criteria

2. Identify the constraint structure

Here's the key insight that makes this problem much simpler: if Andrew MUST be selected and Karen MUST NOT be selected, then we've already made decisions about 2 of our 8 volunteers.

This means we really need to select 3 more people from the remaining 6 volunteers (the 8 original volunteers minus Andrew and Karen). Andrew is automatically in our group of 4, and Karen is automatically excluded.

So our selection process becomes: Andrew + 3 people chosen from the other 6 volunteers.

Process Skill: SIMPLIFY - Recognizing that constraints reduce the problem to a simpler selection

3. Count favorable outcomes

Now let's count how many ways we can select those 3 additional people from the 6 remaining volunteers.

We can think about this step by step:

• For the first additional spot: 6 choices
• For the second additional spot: 5 choices
• For the third additional spot: 4 choices

But since the order doesn't matter (we just want a group of 3), we need to divide by the number of ways to arrange 3 people, which is \(3! = 6\).

Favorable outcomes = \((6 \times 5 \times 4) \div 6 = 120 \div 6 = 20\)

Using combination notation: \(\mathrm{C(6,3)} = 20\)

4. Count total possible outcomes

Now let's count the total number of ways to select 4 people from all 8 volunteers, with no restrictions.

Using the same logic:

• For the first spot: 8 choices
• For the second spot: 7 choices
• For the third spot: 6 choices
• For the fourth spot: 5 choices

Since order doesn't matter, we divide by \(4! = 24\).

Total outcomes = \((8 \times 7 \times 6 \times 5) \div 24 = 1,680 \div 24 = 70\)

Using combination notation: \(\mathrm{C(8,4)} = 70\)

5. Calculate probability ratio

Probability = Favorable outcomes ÷ Total outcomes

Probability = \(20 \div 70 = \frac{2}{7}\)

Let's verify this makes sense: we're looking for a specific condition (Andrew in, Karen out) among many possibilities, so getting \(\frac{2}{7}\) (less than \(\frac{1}{2}\)) seems reasonable.

Final Answer

The probability that Andrew will be selected and Karen will not be selected is \(\frac{2}{7}\).

This matches answer choice D: \(\frac{2}{7}\).

Common Faltering Points

Errors while devising the approach

1. Misinterpreting the condition as "either Andrew OR Karen": Students often misread the problem and think they need to find the probability that either Andrew is selected OR Karen is not selected, rather than both conditions happening simultaneously (Andrew IN AND Karen OUT).

2. Not recognizing the constraint simplification: Many students attempt to use complex conditional probability formulas instead of recognizing that when Andrew must be selected and Karen must not be selected, the problem reduces to simply choosing 3 people from the remaining 6 volunteers.

3. Confusing the selection structure: Students may incorrectly think they need to calculate \(\mathrm{P(Andrew\,selected)} \times \mathrm{P(Karen\,not\,selected)}\) as independent events, failing to recognize this is a single selection scenario with specific constraints.

Errors while executing the approach

1. Combination calculation errors: Students frequently make arithmetic mistakes when calculating \(\mathrm{C(6,3)}\) or \(\mathrm{C(8,4)}\), especially when computing factorials manually (like confusing \(\frac{6!}{3! \times 3!}\) or \(\frac{8!}{4! \times 4!}\)).

2. Double-counting or order confusion: Some students forget to divide by the factorial when counting combinations, treating the problem as permutations instead, leading to incorrect counts like \(6 \times 5 \times 4 = 120\) for favorable outcomes instead of 20.

3. Fraction simplification mistakes: When arriving at \(\frac{20}{70}\), students may incorrectly simplify the fraction, perhaps getting \(\frac{1}{3}\) or other incorrect ratios instead of properly reducing to \(\frac{2}{7}\).

Errors while selecting the answer

1. Selecting the complement probability: Students might calculate correctly but then select \(\frac{5}{7}\) (the probability that the condition does NOT occur) instead of \(\frac{2}{7}\), confusing themselves about what they actually calculated.

2. Choosing a nearby incorrect fraction: Students may recognize their answer should be close to \(\frac{2}{7} \approx 0.286\) but mistakenly select answer choice D (\(\frac{27}{70}\)) which also simplifies to a similar-looking fraction, not realizing that \(\frac{27}{70}\) doesn't reduce to \(\frac{2}{7}\).