2016 AMC 8 Problems/Problem 13
Contents
Problem
Two different numbers are randomly selected from the set and multiplied together. What is the probability that the product is ?
Solutions
Solution 1
The product can only be if one of the numbers is . Once we chose , there are ways we can chose the second number, or . There are ways we can chose numbers randomly, and that is . So, so the answer is .
Solution 2 (Complementary Counting)
Because the only way the product of the two numbers is is if one of the numbers we choose is we calculate the probability of NOT choosing a We get Therefore our answer is
Solution 3 (Casework)
There are two different cases in which the product is zero; either the first number we select is zero, or the second one is. We consider these cases separately.
Case 1: is the first number chosen
There is a chance of selecting zero as the first number. At this point, the product will be zero no matter the choice of the second number, so there is a of getting the desired product in this case.
Case 2: is the second number chosen
There is a chance of choosing a number that is NOT zero as the first number. From there, there is a chance of picking zero from the remaining 5 numbers. Thus, there is a chance of getting a product of 0 in this case.
Adding the probabilities from the two distinct cases up, we find that there is a chance of getting a product of zero.
Video Solution (CREATIVE THINKING!!!)
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Video Solution by OmegaLearn
https://youtu.be/6xNkyDgIhEE?t=357
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Video Solution
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See Also
2016 AMC 8 (Problems • Answer Key • Resources) | ||
Preceded by Problem 12 |
Followed by Problem 14 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | ||
All AJHSME/AMC 8 Problems and Solutions |
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