Problem 32 · AMC 8 Stretch
Core
Counting & Probability
Number Theory
or-process-addlogical-reasoning
The numbers 7, 8, 11, 12, and 15 are written on 5 slips of paper and mixed in a hat. Two slips are picked (without replacement). Find the probability that the sum of the two numbers is odd.
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Answer: 3/5
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Hint 1 of 4
A sum is odd exactly when one number is odd and the other is even. Sort the five numbers into odds and evens.
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Hint 2 of 4
Count how many are odd and how many are even. An odd sum needs exactly one of each.
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Hint 3 of 4
Favorable pairs = (number of odds) x (number of evens). Total pairs = number of ways to pick 2 of 5.
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Approach: Count favorable pairs over total pairs
- A sum is odd only when one number is odd and the other is even. Among {7, 8, 11, 12, 15}, the odds are 7, 11, 15 (three) and the evens are 8, 12 (two).
- Favorable pairs (one odd, one even): \(3 \times 2 = 6\). Total ways to pick 2 of 5 slips: 10.
- So \(P(\text{odd sum}) = \frac{6}{10} = \frac{3}{5}\).
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