Of the 5 assembly lines in a certain plant, 2 do not meet a certain safety standard. If 2 of...

1. Translate the problem requirements: We have 5 assembly lines total, where 2 do NOT meet safety standards (so 3 DO meet standards). We're selecting 2 lines randomly and want the probability that BOTH selected lines meet the safety standard.
2. Count total possible selection outcomes: Determine how many different ways we can choose 2 assembly lines from the 5 available lines.
3. Count favorable outcomes: Determine how many ways we can choose 2 assembly lines from only the 3 lines that meet the safety standard.
4. Calculate probability ratio: Apply the fundamental probability principle by dividing favorable outcomes by total outcomes to get our final answer.

Execution of Strategic Approach

1. Translate the problem requirements

Let's break down what we know in plain English:

• We have 5 assembly lines in total at this plant
• 2 of these lines do NOT meet the safety standard
• This means 3 lines DO meet the safety standard (since \(5 - 2 = 3\))
• We're randomly picking 2 lines for inspection
• We want to find the probability that BOTH selected lines meet the safety standard

Think of it like having 5 boxes: 3 are "good" (meet standards) and 2 are "bad" (don't meet standards). We're picking 2 boxes randomly and want both to be "good".

Process Skill: TRANSLATE - Converting the problem language into clear mathematical understanding

2. Count total possible selection outcomes

Now let's figure out how many different ways we can choose 2 assembly lines from our 5 total lines.

Imagine the 5 lines are labeled A, B, C, D, and E. If we pick 2 at random, our possible pairs are:

• (A,B), (A,C), (A,D), (A,E)
• (B,C), (B,D), (B,E)
• (C,D), (C,E)
• (D,E)

Counting these up: \(4 + 3 + 2 + 1 = 10\) total ways to select 2 lines from 5.

This makes sense because we're choosing combinations (order doesn't matter), and we can verify this using the combination formula: \(\mathrm{C(5,2)} = \frac{5!}{2! \times 3!} = \frac{5 \times 4}{2 \times 1} = 10\).

3. Count favorable outcomes

Now we need to count how many ways we can choose 2 assembly lines from only the 3 "good" lines (the ones that meet the safety standard).

Let's say the 3 good lines are A, B, and C. The ways to choose 2 from these 3 are:

• (A,B), (A,C), (B,C)

That's exactly 3 favorable outcomes.

We can verify: \(\mathrm{C(3,2)} = \frac{3!}{2! \times 1!} = \frac{3 \times 2 \times 1}{2 \times 1 \times 1} = 3\).

4. Calculate probability ratio

Now we apply the fundamental probability principle:

Probability = (Number of favorable outcomes)/(Total number of possible outcomes)

In plain English: Out of all 10 possible ways to select 2 lines, exactly 3 of these ways give us both lines meeting the safety standard.

Probability = \(\frac{3}{10}\)

Let's double-check this makes sense: \(\frac{3}{10} = 0.3 = 30\%\), which seems reasonable since we have more "good" lines (3) than "bad" lines (2), but we need both selections to be good.

Final Answer

The probability that both randomly selected assembly lines meet the safety standard is \(\frac{3}{10}\).

Looking at our answer choices, this matches choice C. \(\frac{3}{10}\), which confirms our solution is correct.

Common Faltering Points

Errors while devising the approach

• Misinterpreting which lines meet the standard: Students may confuse the given information and think that 2 lines DO meet the safety standard instead of understanding that 2 lines do NOT meet it. This leads them to believe only 2 lines are "good" instead of 3, fundamentally changing their calculation approach.
• Choosing the wrong probability method: Students might attempt to solve this using sequential probability (picking one line, then another) instead of recognizing this as a combinations problem. This can lead to more complex calculations with conditional probabilities that aren't necessary.
• Confusing combinations with permutations: Students may think the order of selection matters (permutations) rather than recognizing that we're simply choosing 2 lines where order doesn't matter (combinations). This would lead them to use the wrong counting formulas.

Errors while executing the approach

• Incorrect combination calculations: Students may make arithmetic errors when calculating \(\mathrm{C(5,2)}\) or \(\mathrm{C(3,2)}\), such as forgetting to divide by the factorial in the denominator or making basic multiplication/division mistakes that give wrong totals for possible or favorable outcomes.
• Using wrong numbers in the combination formula: Even if students understand they need combinations, they might use \(\mathrm{C(2,2)}\) for favorable outcomes (thinking about the 2 bad lines instead of the 3 good ones) or mix up which numbers represent total lines versus good lines.

Errors while selecting the answer

• Calculating the complement probability: Students might correctly find that the probability of both lines being good is \(\frac{3}{10}\), but then mistakenly think the question asks for the probability that both lines do NOT meet the standard, leading them to select \(\frac{1}{10}\) (which is \(\frac{\mathrm{C(2,2)}}{\mathrm{C(5,2)}} = \frac{1}{10}\)) instead of \(\frac{3}{10}\).
• Fraction simplification confusion: Students may arrive at equivalent fractions like \(\frac{6}{20}\) through incorrect intermediate steps and then simplify to get \(\frac{3}{10}\), coincidentally matching the right answer despite flawed calculations, or conversely, fail to recognize that their correct decimal 0.3 equals \(\frac{3}{10}\).