Difference between revisions of "2016 AMC 8 Problems/Problem 13"
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| + | == Problem == | ||
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Two different numbers are randomly selected from the set <math>{ - 2, -1, 0, 3, 4, 5}</math> and multiplied together. What is the probability that the product is <math>0</math>? | Two different numbers are randomly selected from the set <math>{ - 2, -1, 0, 3, 4, 5}</math> and multiplied together. What is the probability that the product is <math>0</math>? | ||
<math>\textbf{(A) }\dfrac{1}{6}\qquad\textbf{(B) }\dfrac{1}{5}\qquad\textbf{(C) }\dfrac{1}{4}\qquad\textbf{(D) }\dfrac{1}{3}\qquad \textbf{(E) }\dfrac{1}{2}</math> | <math>\textbf{(A) }\dfrac{1}{6}\qquad\textbf{(B) }\dfrac{1}{5}\qquad\textbf{(C) }\dfrac{1}{4}\qquad\textbf{(D) }\dfrac{1}{3}\qquad \textbf{(E) }\dfrac{1}{2}</math> | ||
| − | ==Solution 1== | + | == Solutions== |
| + | |||
| + | ===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 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>. | ||
| − | ==Solution 2== | + | ===Solution 2=== |
There are a total of <math>30</math> possibilities, because the two numbers that being multiplied are being picked at the same time, so there are <math>5</math> possibilities that zero is being chosen because another number is already being chosen. 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 two numbers that being multiplied are being picked at the same time, so there are <math>5</math> possibilities that zero is being chosen because another number is already being chosen. 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 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> | 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> | ||
Revision as of 02:42, 16 January 2021
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 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 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 
| 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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