Difference between revisions of "2016 AMC 8 Problems/Problem 13"
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===Solution 1=== | ===Solution 1=== | ||
| − | 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 <math>0</math>. 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>. |
===Solution 2=== | ===Solution 2=== | ||
Revision as of 22:22, 16 May 2022
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
There are a total of 
possibilities, because the two numbers that being multiplied are being picked at the same time, so there are possibilities that zero is being chosen because another number is already being chosen. 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 
Video Solution
~savannahsolver
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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
