2016 AMC 8 Problems/Problem 13
Two different numbers are randomly selected from the set and multiplied together. What is the probability that the product is ?
Solution 1
The product can only be if one of the numbers is 0. 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
There are a total of possibilities, because the numbers are different. We want to be the product so one of the numbers is . There are possibilities where is chosen for the first number and there are ways for to be chosen as the second number. We seek .
Solution 3(Complementary Counting)
Because the only way the product is 0 is if a number we chose is 0 we calculate the probability of NOT choosing a 0. We get 5/6*4/5=2/3. 1-2/3=1/3
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.