Six cards numbered from 1 to 6 are placed in an empty bowl. First one card is drawn and then...

1. Translate the problem requirements: We need to find \(\mathrm{P(at\ least\ one\ card\ is\ 5\ |\ sum\ equals\ 8)}\). This is a conditional probability where we're only considering cases where the sum is 8, and among those, counting how many include at least one 5.
2. Identify all outcomes that satisfy the condition: List all possible ways two cards (with replacement) can sum to 8, since these form our restricted sample space.
3. Count favorable outcomes within the restricted space: Among the outcomes that sum to 8, count how many include at least one card numbered 5.
4. Calculate the conditional probability: Divide the number of favorable outcomes by the total number of outcomes that satisfy the given condition (sum = 8).

Execution of Strategic Approach

1. Translate the problem requirements

Let's break down what we're actually looking for. We have a special situation here - we're not looking at all possible card draws. Instead, we already know that the sum of the two cards equals 8. Given this information, we want to find the probability that at least one of those cards is numbered 5.

This is what mathematicians call a conditional probability. Think of it like this: imagine we've already filtered out all the cases where the sum doesn't equal 8. Now, among only those remaining cases where the sum does equal 8, what fraction includes at least one card numbered 5?

Process Skill: TRANSLATE - Converting the conditional probability language into a concrete counting problem

2. Identify all outcomes that satisfy the condition

Since we're only interested in cases where the sum equals 8, let's list all the ways two cards (drawn with replacement) can add up to 8.

Remember, we have cards numbered 1, 2, 3, 4, 5, and 6. Let's systematically find all pairs (first card, second card) that sum to 8:

• (2, 6): \(\mathrm{2 + 6 = 8}\)
• (3, 5): \(\mathrm{3 + 5 = 8}\)
• (4, 4): \(\mathrm{4 + 4 = 8}\)
• (5, 3): \(\mathrm{5 + 3 = 8}\)
• (6, 2): \(\mathrm{6 + 2 = 8}\)

Notice that (2, 6) and (6, 2) are different outcomes since order matters - the first number represents the first draw, and the second number represents the second draw.

So we have exactly 5 ways to get a sum of 8. This becomes our restricted sample space.

Process Skill: CONSIDER ALL CASES - Systematically listing all possible ways to achieve the sum of 8

3. Count favorable outcomes within the restricted space

Now, among these 5 outcomes that sum to 8, let's count how many include at least one card numbered 5:

• (2, 6): No 5 here
• (3, 5): Yes! This includes a 5 (as the second card)
• (4, 4): No 5 here
• (5, 3): Yes! This includes a 5 (as the first card)
• (6, 2): No 5 here

So out of our 5 possible outcomes that sum to 8, exactly 2 of them include at least one card numbered 5.

4. Calculate the conditional probability

Now we can find our answer using simple counting:

Probability = (Number of favorable outcomes) ÷ (Total outcomes in our restricted space)
Probability = \(\mathrm{2 ÷ 5 = 2/5}\)

This means that when we know the sum equals 8, there's a \(\mathrm{2/5}\) chance that at least one of the cards is numbered 5.

Final Answer

The probability that one of the two cards drawn is numbered 5, given that their sum is 8, equals \(\mathrm{2/5}\).

This matches answer choice D.

Common Faltering Points

Errors while devising the approach

Many students fail to recognize this as a conditional probability problem. They might try to calculate P(at least one card is 5) for all possible draws, then separately calculate P(sum equals 8), leading them down a complex path involving the formula \(\mathrm{P(A|B) = P(A∩B)/P(B)}\). The key insight missed is that we can solve this more directly by working only within the restricted sample space where the sum equals 8.

Students might incorrectly assume this is sampling without replacement, thinking that if the first card drawn is a 5, then there's no 5 available for the second draw. This would lead them to incorrectly eliminate certain combinations like (5,3) in their analysis, significantly affecting their final answer.

Some students interpret this phrase as "exactly one card is numbered 5" rather than "at least one card is numbered 5." This misinterpretation would cause them to exclude cases where both cards could be 5 (though in this specific problem, having both cards as 5 would give a sum of 10, not 8, so this error wouldn't affect the final answer, but it represents faulty reasoning).

Errors while executing the approach

When listing all ways to get a sum of 8, students often forget that order matters in sequential draws. They might list only (2,6), (3,5), and (4,4), missing that (6,2) and (5,3) are distinct outcomes. This would give them only 3 total outcomes instead of 5, leading to an incorrect final probability.

Students might make simple addition errors when determining which pairs sum to 8. For example, they might incorrectly include (1,6) thinking \(\mathrm{1+6=8}\), or exclude (3,5) by miscalculating \(\mathrm{3+5}\). Such errors would change both the total number of favorable outcomes and the total sample space.

Errors while selecting the answer

Some students might correctly identify that 3 out of 5 outcomes do NOT include a 5, then mistakenly select \(\mathrm{3/5}\) as their final answer instead of recognizing that they need \(\mathrm{1 - 3/5 = 2/5}\). This type of error occurs when students lose track of what they're actually calculating toward the end of their solution process.