From a bag containing 12 identical blue balls, y identical yellow balls, and no other balls, one ball will be...

1. Translate the problem requirements: We need the probability of drawing a blue ball to be less than \(\frac{2}{5}\). With 12 blue balls and y yellow balls, this means \(\frac{12}{12+y} < \frac{2}{5}\), and we want the smallest possible value of y.
2. Set up the probability inequality: Express the constraint that blue ball probability must be less than \(\frac{2}{5}\) as a mathematical inequality.
3. Solve the inequality algebraically: Cross-multiply and solve to find the minimum constraint on y.
4. Determine the minimum integer value: Since y must be a whole number of balls, find the smallest integer that satisfies our constraint.

Execution of Strategic Approach

1. Translate the problem requirements

Let's start by understanding what we're looking for in everyday terms. We have a bag with 12 blue balls and some unknown number of yellow balls (let's call this y). When we reach in and grab one ball randomly, we want the chance of getting a blue ball to be less than \(\frac{2}{5}\).

Think of it this way: if there were very few yellow balls, blue balls would make up most of the bag, so we'd be very likely to draw blue. But as we add more yellow balls, the blue balls become a smaller fraction of the total, making it less likely to draw blue.

We need to find the smallest number of yellow balls that makes the probability of drawing blue fall below \(\frac{2}{5}\).

Process Skill: TRANSLATE - Converting the probability constraint into a mathematical relationship

Mathematically, this means: \(\frac{12}{12+y} < \frac{2}{5}\), and we want the minimum value of y.

2. Set up the probability inequality

Let's think about this step by step. The probability of drawing a blue ball equals the number of blue balls divided by the total number of balls.

• Number of blue balls = 12
• Total number of balls = 12 + y
• Probability of blue = \(\frac{12}{12+y}\)

We need this probability to be less than \(\frac{2}{5}\):

\(\frac{12}{12+y} < \frac{2}{5}\)

3. Solve the inequality algebraically

Now we'll solve this inequality to find what y must be.

Starting with: \(\frac{12}{12+y} < \frac{2}{5}\)

To solve this, we'll cross-multiply. Since both denominators are positive, the inequality direction stays the same:

\(12 \times 5 < 2 \times (12+y)\)

\(60 < 2(12+y)\)

\(60 < 24 + 2y\)

Subtracting 24 from both sides:

\(60 - 24 < 2y\)

\(36 < 2y\)

Dividing both sides by 2:

\(18 < y\)

This means y must be greater than 18.

4. Determine the minimum integer value

Since we need \(y > 18\), and y must be a whole number (we can't have a fraction of a ball), the smallest integer value that satisfies this condition is y = 19.

Let's verify this works:

• With y = 18: Probability = \(\frac{12}{12+18} = \frac{12}{30} = \frac{2}{5}\) (this equals \(\frac{2}{5}\), but we need less than \(\frac{2}{5}\))
• With y = 19: Probability = \(\frac{12}{12+19} = \frac{12}{31} \approx 0.387\), which is less than \(\frac{2}{5} = 0.400\) ✓

Process Skill: APPLY CONSTRAINTS - Ensuring we find the minimum integer that satisfies the strict inequality

Final Answer

The least number of yellow balls that must be in the bag is 19.

This matches answer choice C.

Common Faltering Points

Errors while devising the approach

1. Misinterpreting the inequality direction

Students often confuse "less than \(\frac{2}{5}\)" with "at least \(\frac{2}{5}\)" or "equal to \(\frac{2}{5}\)". This leads them to set up the inequality as \(\frac{12}{12+y} \geq \frac{2}{5}\) instead of the correct \(\frac{12}{12+y} < \frac{2}{5}\), which would give them the wrong constraint and therefore the wrong answer.

2. Confusing which probability to calculate

Some students might think they need to find the probability of drawing a yellow ball instead of a blue ball. This would lead them to set up the equation as \(\frac{y}{12+y} < \frac{2}{5}\), which would give them a completely different answer since they're solving for the wrong event.

Errors while executing the approach

1. Cross-multiplication error with inequality direction

When cross-multiplying \(\frac{12}{12+y} < \frac{2}{5}\), students sometimes flip the inequality sign incorrectly, thinking that any manipulation with fractions requires flipping the sign. Since both denominators are positive, the inequality direction should remain the same, but students may write \(12 \times 5 > 2 \times (12+y)\) instead of \(12 \times 5 < 2 \times (12+y)\).

2. Arithmetic errors in algebraic manipulation

Students may make calculation mistakes when expanding \(2(12+y) = 24 + 2y\), or when solving \(36 < 2y\) may incorrectly get \(y < 18\) instead of \(y > 18\) due to rushing through the division step.

Errors while selecting the answer

1. Selecting y = 18 instead of y = 19

After correctly solving to get \(y > 18\), students might select 18 as their answer, forgetting that they need the smallest integer that is greater than 18, not greater than or equal to 18. They fail to recognize that y = 18 gives exactly \(\frac{2}{5}\) probability, which doesn't satisfy the "less than \(\frac{2}{5}\)" requirement.

2. Not verifying the boundary condition

Students might arrive at y = 19 but fail to check that y = 18 doesn't work, leading them to second-guess their answer and potentially select choice B (18) instead of the correct choice C (19), especially if they're running short on time.