Difference between revisions of "2016 AMC 8 Problems/Problem 13"
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==Solution 2== | ==Solution 2== | ||
There are a total of <math>30</math> possibilities, because the numbers are different. We want <math>0</math> to be the product so one of the numbers is <math>0</math>. There are <math>5</math> possibilities where <math>0</math> is chosen for the first number and there are <math>5</math> ways for <math>0</math> to be chosen as the second number. We seek <math>\boxed{\textbf{(D)} \, \frac{1}{3}}</math>. | There are a total of <math>30</math> possibilities, because the numbers are different. We want <math>0</math> to be the product so one of the numbers is <math>0</math>. There are <math>5</math> possibilities where <math>0</math> is chosen for the first number and there are <math>5</math> ways for <math>0</math> to be chosen as the second number. We seek <math>\boxed{\textbf{(D)} \, \frac{1}{3}}</math>. | ||
− | ==Solution 3(Complementary Counting)== | + | ==Solution 3 (Complementary Counting)== |
− | Because the only way the product is 0 is if | + | Because the only way the product of the two numbers is <math>0</math> is if one of the numbers we choose is <math>0,</math> we calculate the probability of NOT choosing a <math>0.</math> We get <math>\frac{5}{6} \cdot \frac{4}{5} = \frac{2}{3}.</math> Therefore our answer is <math>1 - \frac{2}{3} = \boxed{\textbf{(D)} \ \frac{1}{3}}.</math> |
{{AMC8 box|year=2016|num-b=12|num-a=14}} | {{AMC8 box|year=2016|num-b=12|num-a=14}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 23:09, 31 March 2020
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 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
2016 AMC 8 (Problems • Answer Key • Resources) | ||
Preceded by Problem 12 |
Followed by Problem 14 | |
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All AJHSME/AMC 8 Problems and Solutions |
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