If x and y are each randomly chosen from the integers 1, 2, 3, 4, and 5, what is the...

1. Translate the problem requirements: We need to find the probability that when two integers are randomly chosen from \(\{1, 2, 3, 4, 5\}\), their product exceeds 10. This means we need favorable outcomes divided by total possible outcomes.
2. Determine the sample space: Count all possible ways to choose x and y, recognizing that each variable can independently take any of the 5 values.
3. Identify favorable outcomes systematically: List all pairs (x,y) where the product \(\mathrm{xy} > 10\) by examining each possible value methodically.
4. Calculate the probability ratio: Apply the basic probability formula using our counts of favorable outcomes versus total outcomes.

Execution of Strategic Approach

1. Translate the problem requirements

Let's break down what this problem is really asking. We have two variables, x and y, and each can be any of the numbers 1, 2, 3, 4, or 5. We're picking these numbers randomly, meaning each number has an equal chance of being selected.

The key question is: when we multiply our two chosen numbers together, what's the chance that this product will be greater than 10?

In simple terms, we need to find:

• How many ways can we get a product greater than 10?
• How many total ways can we pick the two numbers?
• Then divide the first by the second to get our probability.

Process Skill: TRANSLATE - Converting the probability language into concrete counting tasks

2. Determine the sample space

Since x can be any of 5 values (1, 2, 3, 4, 5) and y can independently be any of 5 values (1, 2, 3, 4, 5), we can think of this like filling in a grid.

For every choice of x, there are 5 possible choices for y. Since x itself has 5 possibilities, our total number of outcomes is:
\(5 \times 5 = 25\) total possible pairs (x, y)

To visualize this, imagine all possible pairs: (1,1), (1,2), (1,3), (1,4), (1,5), (2,1), (2,2), and so on, all the way to (5,5). That gives us exactly 25 different combinations.

3. Identify favorable outcomes systematically

Now we need to find which pairs give us a product greater than 10. Let's go through this methodically by checking each possible value of x:

When x = 1:

• \(1 \times 1 = 1\) (not > 10)
• \(1 \times 2 = 2\) (not > 10)
• \(1 \times 3 = 3\) (not > 10)
• \(1 \times 4 = 4\) (not > 10)
• \(1 \times 5 = 5\) (not > 10)

No favorable outcomes when x = 1.

When x = 2:

• \(2 \times 1 = 2\) (not > 10)
• \(2 \times 2 = 4\) (not > 10)
• \(2 \times 3 = 6\) (not > 10)
• \(2 \times 4 = 8\) (not > 10)
• \(2 \times 5 = 10\) (not > 10, since we need greater than 10)

No favorable outcomes when x = 2.

When x = 3:

• \(3 \times 1 = 3\) (not > 10)
• \(3 \times 2 = 6\) (not > 10)
• \(3 \times 3 = 9\) (not > 10)
• \(3 \times 4 = 12\) ✓ (this works!)
• \(3 \times 5 = 15\) ✓ (this works!)

2 favorable outcomes when x = 3.

When x = 4:

• \(4 \times 1 = 4\) (not > 10)
• \(4 \times 2 = 8\) (not > 10)
• \(4 \times 3 = 12\) ✓ (this works!)
• \(4 \times 4 = 16\) ✓ (this works!)
• \(4 \times 5 = 20\) ✓ (this works!)

3 favorable outcomes when x = 4.

When x = 5:

• \(5 \times 1 = 5\) (not > 10)
• \(5 \times 2 = 10\) (not > 10)
• \(5 \times 3 = 15\) ✓ (this works!)
• \(5 \times 4 = 20\) ✓ (this works!)
• \(5 \times 5 = 25\) ✓ (this works!)

3 favorable outcomes when x = 5.

Total favorable outcomes: \(0 + 0 + 2 + 3 + 3 = 8\)

Process Skill: CONSIDER ALL CASES - Systematically checking every possibility to avoid missing any favorable outcomes

4. Calculate the probability ratio

Now we can apply the basic probability formula:

Probability = Number of favorable outcomes ÷ Total number of possible outcomes

Probability = \(8 ÷ 25 = \frac{8}{25}\)

Let's verify this makes sense: we found 8 pairs that give us a product greater than 10 out of 25 total possible pairs. This gives us \(\frac{8}{25}\), which matches answer choice (D).

4. Final Answer

The probability that the product xy will be greater than 10 is \(\frac{8}{25}\).

This corresponds to answer choice (D) \(\frac{8}{25}\).

Common Faltering Points

Errors while devising the approach

1. Misunderstanding the independence of variables

Students may incorrectly assume that x and y must be different values, thinking "randomly chosen" means "chosen without replacement." This leads them to believe there are only \(5×4 = 20\) possible outcomes instead of \(5×5 = 25\), since they exclude cases like (1,1), (2,2), etc.

2. Confusing "greater than 10" with "greater than or equal to 10"

Students might misread the constraint and include cases where \(\mathrm{xy} = 10\) (like \(2×5\) or \(5×2\)) as favorable outcomes, leading to an incorrect count of favorable cases.

Errors while executing the approach

1. Systematic counting errors when checking all cases

When going through each value of x systematically, students may accidentally skip certain combinations or double-count others. For example, they might forget to check (3,4) or accidentally count (4,3) twice when checking both x=3 and x=4 cases.

2. Arithmetic mistakes in multiplication

Students may make basic calculation errors when computing products like \(3×4\), \(4×5\), or \(5×4\), leading to incorrect identification of which pairs satisfy \(\mathrm{xy} > 10\).

3. Incorrect addition of favorable outcomes

After correctly identifying favorable cases in each group \((0 + 0 + 2 + 3 + 3)\), students may make errors when adding these numbers together, potentially getting 7 or 9 instead of 8.

Errors while selecting the answer

1. Choosing the complement probability

Students may correctly calculate that 8 out of 25 cases have \(\mathrm{xy} > 10\), but then mistakenly select the probability that \(\mathrm{xy} ≤ 10\), which would be \(\frac{17}{25}\). Since this exact fraction doesn't appear in the choices, they might incorrectly convert or approximate it.