There are 10 solid-colored balls in a box, including 1 green ball and 1 yellow ball. If 3 of the...

1. Translate the problem requirements: We need to find the probability of selecting exactly 3 balls where the green ball is included but the yellow ball is not included
2. Determine total possible outcomes: Calculate how many ways we can choose 3 balls from 10 total balls
3. Identify favorable outcomes structure: Break down what makes an outcome favorable - must include the green ball, must exclude the yellow ball, and need 2 more balls from the remaining 8
4. Calculate favorable outcomes: Count the ways to select the remaining 2 balls from the 8 balls that are neither green nor yellow
5. Apply probability formula: 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 the problem is asking in plain English. We have a box with 10 balls total. Among these, there's exactly 1 green ball and 1 yellow ball. We're going to pick 3 balls randomly without putting any back.

The key requirement is that our 3 chosen balls must include the green ball but NOT include the yellow ball. Think of it like this: we want the green ball to definitely be in our selection, but we want to avoid picking the yellow ball entirely.

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

2. Determine total possible outcomes

Now let's figure out how many different ways we can choose 3 balls from the 10 total balls, regardless of color.

Imagine we're picking balls one by one. For the first ball, we have 10 choices. For the second ball, we have 9 remaining choices. For the third ball, we have 8 remaining choices. However, since the order doesn't matter (picking balls 1, 2, 3 is the same as picking 3, 1, 2), we need to account for this.

The total number of ways = \((10 \times 9 \times 8) \div (3 \times 2 \times 1) = 720 \div 6 = 120\)

Technically, this is \(\mathrm{C(10,3)} = 120\) total possible outcomes.

3. Identify favorable outcomes structure

For an outcome to be favorable, it must satisfy these conditions:

• The green ball MUST be included (this is guaranteed in our favorable cases)
• The yellow ball MUST NOT be included (so we can't pick from this ball)
• We need to pick 2 more balls to complete our set of 3

Since we've used up the green ball (included) and excluded the yellow ball, we have \(10 - 1 - 1 = 8\) remaining balls to choose from for our final 2 selections.

Process Skill: APPLY CONSTRAINTS - Systematically accounting for the inclusion and exclusion requirements

4. Calculate favorable outcomes

Now we need to count how many ways we can pick the remaining 2 balls from the 8 balls that are neither green nor yellow.

Using the same logic as before: we can pick the first of these 2 balls in 8 ways, and the second in 7 ways. Since order doesn't matter, we divide by 2.

Number of ways to pick 2 balls from 8 = \((8 \times 7) \div (2 \times 1) = 56 \div 2 = 28\)

Technically, this is \(\mathrm{C(8,2)} = 28\) favorable outcomes.

5. Apply probability formula

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

Probability = Favorable outcomes ÷ Total outcomes = \(28 \div 120\)

To simplify this fraction:
\(28 \div 120 = 7 \div 30\) (dividing both numerator and denominator by 4)

Therefore, the probability is \(\frac{7}{30}\).

4. Final Answer

The probability that 3 randomly chosen balls will include the green ball but not the yellow ball is \(\frac{7}{30}\).

This matches answer choice B: \(\frac{7}{30}\).

Common Faltering Points

Errors while devising the approach

1. Misinterpreting the constraint "include green but not yellow"

Students often misread this as "include green AND yellow" or "include green OR yellow but not both." The key requirement is that green MUST be selected while yellow MUST be avoided. This constraint fundamentally changes how we count favorable outcomes.

2. Attempting to use conditional probability incorrectly

Some students try to calculate \(\mathrm{P(no\ yellow\ |\ green\ selected)}\) without properly setting up the conditional probability framework. This leads to incorrect probability calculations and overcomplicated approaches when a direct counting method is more straightforward.

3. Confusing this with a "with replacement" scenario

Students may forget that this is sampling "without replacement" and try to treat each ball selection as independent events with fixed probabilities, rather than recognizing this as a combinations problem where we're selecting 3 balls simultaneously.

Errors while executing the approach

1. Incorrect calculation of remaining balls available

After accounting for the green ball (must include) and yellow ball (must exclude), students often miscalculate that there are 9 remaining balls instead of 8. They forget to subtract both the green AND yellow balls from the total, leading to \(\mathrm{C(9,2)} = 36\) instead of \(\mathrm{C(8,2)} = 28\).

2. Arithmetic errors in combination calculations

When calculating \(\mathrm{C(8,2)}\), students may compute \((8 \times 7) = 56\) correctly but then forget to divide by \(2!\), giving 56 instead of 28. Or they may make basic multiplication errors when computing \(\mathrm{C(10,3)} = 120\).

3. Fraction simplification mistakes

Students may correctly get \(\frac{28}{120}\) but then incorrectly simplify it. Common errors include simplifying to \(\frac{7}{40}\) (dividing by 4 incorrectly) or \(\frac{14}{60}\) (dividing by 2 instead of 4), rather than the correct \(\frac{7}{30}\).

Errors while selecting the answer

1. Selecting the decimal equivalent instead of the fraction

Students may calculate \(\frac{7}{30} \approx 0.233\) and then try to match this to decimal approximations of the answer choices rather than recognizing that \(\frac{7}{30}\) directly matches option B.

2. Choosing the complement probability

After calculating the correct probability of \(\frac{7}{30}\), some students mistakenly select the complement \((1 - \frac{7}{30} = \frac{23}{30})\), thinking the question asked for the probability that the condition is NOT satisfied.