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If \(\mathrm{x}\) and \(\mathrm{y}\) are each randomly chosen from the integers 1, 2, 3, 4, and 5, what is the probability that the product \(\mathrm{xy}\) will be greater than \(10\)?
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:
Process Skill: TRANSLATE - Converting the probability language into concrete counting tasks
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.
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:
No favorable outcomes when x = 1.
When x = 2:
No favorable outcomes when x = 2.
When x = 3:
2 favorable outcomes when x = 3.
When x = 4:
3 favorable outcomes when x = 4.
When x = 5:
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
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).
The probability that the product xy will be greater than 10 is \(\frac{8}{25}\).
This corresponds to answer choice (D) \(\frac{8}{25}\).
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.
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.
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.
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 \(\mathrm{xy} > 10\).
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.
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.