2016 AMC 10A Problems/Problem 12
Contents
Problem
Three distinct integers are selected at random between and , inclusive. Which of the following is a correct statement about the probability that the product of the three integers is odd?
Solution 1
For the product to be odd, all three factors have to be odd. The probability of this is .
, but and are slightly less than . Thus, the whole product is slightly less than , so .
Solution 2
For the product to be odd, all three factors have to be odd. There are a total of ways to choose 3 numbers at random, and there are to choose 3 odd numbers. Therefore, the probability of choosing 3 odd numbers is . Simplifying this, we obtain , which is slightly less than , so our answer is .
Solution 3
The probability that the product is odd, allowing duplication of the integers, is just . Since forbidding duplication reduces the probability of all three integers being odd, we see and our answer is .
Video Solution (CREATIVE THINKING)
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Video Solution
https://youtu.be/dHY8gjoYFXU?t=300
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See Also
2016 AMC 10A (Problems • Answer Key • Resources) | ||
Preceded by Problem 11 |
Followed by Problem 13 | |
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All AMC 10 Problems and Solutions |
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