Difference between revisions of "2016 AMC 8 Problems/Problem 13"
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==Solution== | ==Solution== | ||
The product can only be <math>0</math> if one of the numbers is 0. Once we chose <math>0</math>, there are <math>5</math> ways we can chose the second number, or <math>6-1</math>. There are <math>\dbinom{6}{2}</math> ways we can chose <math>2</math> numbers randomly, and that is <math>15</math>. So, <math>\frac{5}{15}=\frac{1}{3}</math> so the answer is <math>\boxed{\textbf{(D)} \, \frac{1}{3}}</math> | The product can only be <math>0</math> if one of the numbers is 0. Once we chose <math>0</math>, there are <math>5</math> ways we can chose the second number, or <math>6-1</math>. There are <math>\dbinom{6}{2}</math> ways we can chose <math>2</math> numbers randomly, and that is <math>15</math>. So, <math>\frac{5}{15}=\frac{1}{3}</math> so the answer is <math>\boxed{\textbf{(D)} \, \frac{1}{3}}</math> | ||
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+ | ==Solution 2== | ||
+ | There are a total of <math>36</math> possibilities. We want <math>0</math> so one of the multiples is <math>0</math>. There are <math>6</math> possibilities where <math>0</math> is chosen for the first number and there are <math>6</math> ways for <math>0</math> to be chosen as the second number. We seek <math>\frac {6+6}{36}=\frac {1}{3}</math> | ||
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{{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 16:19, 23 November 2016
Two different numbers are randomly selected from the set and multiplied together. What is the probability that the product is ?
Solution
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. We want so one of the multiples 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
2016 AMC 8 (Problems • Answer Key • Resources) | ||
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Followed by Problem 14 | |
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Solution 2
There are a total of possibilities. We want so one of the multiples 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