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Let's break down what this problem is asking in plain English:
We have 8 people total, and two of them have names: George and Nina. We need to pick 3 people for a project team. But there's a catch - we want George on the team, but we definitely don't want Nina.
Think of it like this: if George must be on the team, we've already filled 1 spot out of 3. Since Nina can't be on the team, she's completely out of consideration. So we're really just picking 2 more people from the remaining 6 people (the 8 original people minus George and Nina).
Process Skill: TRANSLATE - Converting the probability language into a simpler selection problem
Now let's think about our constraints more clearly:
• George is automatically selected (1 person chosen)
• Nina is automatically excluded
• We have 6 remaining people to consider
• We need to choose 2 more people from these 6
This transforms our problem from "select 3 people from 8 with conditions" to "select 2 people from 6 with no conditions." Much simpler!
Process Skill: APPLY CONSTRAINTS - Recognizing how constraints simplify the problem structure
Favorable outcomes are the ways we can successfully form our team (George + 2 others from the remaining 6 people).
How many ways can we choose 2 people from 6 people? Let's think about this step by step:
• For the first additional person, we have 6 choices
• For the second additional person, we have 5 choices
• But wait - the order doesn't matter! Choosing person A then person B is the same as choosing person B then person A
So we calculate: \(\mathrm{(6 \times 5) \div 2 = 30 \div 2 = 15}\)
This gives us 15 favorable outcomes.
Technical notation: \(\mathrm{C(6,2) = 6!/(2! \times 4!) = 15}\)
Total possible outcomes means all the ways we can choose any 3 people from the original 8 people, with no restrictions.
Using the same logic:
• First person: 8 choices
• Second person: 7 choices
• Third person: 6 choices
• Since order doesn't matter, we divide by the number of ways to arrange 3 people: \(\mathrm{3 \times 2 \times 1 = 6}\)
So we get: \(\mathrm{(8 \times 7 \times 6) \div 6 = 336 \div 6 = 56}\)
This gives us 56 total possible outcomes.
Technical notation: \(\mathrm{C(8,3) = 8!/(3! \times 5!) = 56}\)
Probability is simply the ratio of favorable outcomes to total possible outcomes.
We found:
• Favorable outcomes = 15
• Total possible outcomes = 56
Therefore: Probability = \(\mathrm{15/56}\)
Looking at our answer choices, this matches choice C exactly.
The probability that the 3 people selected will include George but not Nina is \(\mathrm{15/56}\).
This corresponds to answer choice C.
Quick verification: Our answer makes intuitive sense. Since George must be included and Nina must be excluded, we're essentially choosing 2 from 6 remaining people, which should be more restrictive than random selection, leading to a probability less than 1/2. Indeed, \(\mathrm{15/56 \approx 0.27}\), which seems reasonable.
Faltering Point 1: Misinterpreting the constraint "include George but not Nina"
Students often struggle with compound constraints and may interpret this as two separate conditions rather than one unified requirement. They might try to calculate P(George selected) and P(Nina not selected) separately and then multiply them, which is incorrect because these events are not independent in this context. The correct approach recognizes this as a single constraint that simultaneously requires George's inclusion and Nina's exclusion.
Faltering Point 2: Not recognizing the constraint simplification
Many students miss the key insight that when George must be selected and Nina must be excluded, the problem transforms from "choose 3 from 8 with conditions" to "choose 2 from 6 without conditions." Instead, they attempt complex conditional probability calculations or use inclusion-exclusion principles unnecessarily, making the problem much more complicated than needed.
Faltering Point 3: Confusing favorable outcomes with conditional probability setup
Students may incorrectly set up the problem using conditional probability notation like \(\mathrm{P(George\ selected\ |\ Nina\ not\ selected)}\) or attempt to use formulas for dependent events, when this is actually a straightforward counting problem that can be solved using basic combinations once the constraints are properly understood.
Faltering Point 1: Combination formula calculation errors
Students frequently make arithmetic mistakes when calculating combinations, especially with \(\mathrm{C(8,3) = 56}\). Common errors include forgetting to divide by the factorial (calculating \(\mathrm{8\times7\times6 = 336}\) instead of \(\mathrm{336\div6 = 56}\)) or making basic multiplication errors. Some students also confuse combinations with permutations and don't divide by the appropriate factorial.
Faltering Point 2: Incorrect counting of remaining people
After establishing that George is selected and Nina is excluded, students sometimes incorrectly count the remaining people as 7 instead of 6, forgetting that both George AND Nina must be removed from consideration. This leads them to calculate \(\mathrm{C(7,2) = 21}\) instead of the correct \(\mathrm{C(6,2) = 15}\) for favorable outcomes.
Faltering Point 1: fraction reversal or misplacement
Students may correctly calculate 15 favorable outcomes and 56 total outcomes but then write the probability as \(\mathrm{56/15}\) instead of \(\mathrm{15/56}\), confusing which number should be the numerator and which should be the denominator in the probability fraction.