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Difference between revisions of "2016 AMC 8 Problems/Problem 13"

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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>?
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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>?
  
 
<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>
  
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==Solution==
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== Solutions==
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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>\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>
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===Solution 2 (Complementary Counting)===
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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>
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==Video Solution (CREATIVE THINKING!!!)==
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https://youtu.be/cRsvq0BH4MI
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~Education, the Study of Everything
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==Video Solution by OmegaLearn ==
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https://youtu.be/6xNkyDgIhEE?t=357
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~ pi_is_3.14
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==Video Solution==
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https://youtu.be/jDeS4A6N-nE
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 +
~savannahsolver
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==See Also==
 
{{AMC8 box|year=2016|num-b=12|num-a=14}}
 
{{AMC8 box|year=2016|num-b=12|num-a=14}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 22:52, 17 May 2024

Problem

Two different numbers are randomly selected from the set $\{ - 2, -1, 0, 3, 4, 5\}$ and multiplied together. What is the probability that the product is $0$?

$\textbf{(A) }\dfrac{1}{6}\qquad\textbf{(B) }\dfrac{1}{5}\qquad\textbf{(C) }\dfrac{1}{4}\qquad\textbf{(D) }\dfrac{1}{3}\qquad \textbf{(E) }\dfrac{1}{2}$

Solutions

Solution 2 (Complementary Counting)

Because the only way the product of the two numbers is $0$ is if one of the numbers we choose is $0,$ we calculate the probability of NOT choosing a $0.$ We get $\frac{5}{6} \cdot \frac{4}{5} = \frac{2}{3}.$ Therefore our answer is $1 - \frac{2}{3} = \boxed{\textbf{(D)} \ \frac{1}{3}}.$

Video Solution (CREATIVE THINKING!!!)

https://youtu.be/cRsvq0BH4MI

~Education, the Study of Everything


Video Solution by OmegaLearn

https://youtu.be/6xNkyDgIhEE?t=357

~ pi_is_3.14

Video Solution

https://youtu.be/jDeS4A6N-nE

~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

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