Difference between revisions of "2016 AMC 8 Problems/Problem 13"
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− | 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> | + | 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}=\textbf{(D)}</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 12:40, 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,
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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