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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 1==
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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}</math> so the answer is  <math>\boxed{\textbf{(D)} \, \frac{1}{3}}</math>.
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===Solution 1===
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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>.
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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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===Solution 3 (Casework)===
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There are two different cases in which the product is zero; either the first number we select is zero, or the second one is. We consider these cases separately.
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Case 1: <math>0</math> is the first number chosen
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There is a <math>\frac{1}{6}</math> chance of selecting zero as the first number. At this point, the product will be zero no matter the choice of the second number, so there is a <math>\frac{1}{6}</math> of getting the desired product in this case.
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Case 2: <math>0</math> is the second number chosen
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There is a <math>\frac{5}{6}</math> chance of choosing a number that is NOT zero as the first number. From there, there is a <math>\frac{1}{5}</math> chance of picking zero from the remaining 5 numbers. Thus, there is a <math>\frac{5}{6} \cdot \frac{1}{5} = \frac{1}{6}</math> chance of getting a product of 0 in this case.
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Adding the probabilities from the two distinct cases up, we find that there is a <math>\frac{1}{6} + \frac{1}{6} = \boxed{\textbf{(D)} \ \frac{1}{3}}</math> chance of getting a product of zero.
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~[https://artofproblemsolving.com/wiki/index.php/User:Cxsmi cxsmi]
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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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 +
 
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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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==Solution 2==
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==Video Solution==
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There are a total of <math>30</math> possibilities, because the numbers are different. 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>.
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https://youtu.be/jDeS4A6N-nE
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==Solution 3(Complementary Counting)==
 
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Because the only way the product is 0 is if a number we chose is 0 we calculate the probability of NOT choosing a 0. We get 5/6*4/5=2/3. 1-2/3=1/3 <math>\boxed{\textbf{(D)} \, \frac{1}{3}}</math>
 
  
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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 18:33, 20 January 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 1

The product can only be $0$ if one of the numbers is $0$. Once we chose $0$, there are $5$ ways we can chose the second number, or $6-1$. There are $\dbinom{6}{2}$ ways we can chose $2$ numbers randomly, and that is $15$. So, $\frac{5}{15}=\frac{1}{3}$ so the answer is $\boxed{\textbf{(D)} \, \frac{1}{3}}$.

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 $1 - \frac{2}{3} = \boxed{\textbf{(D)} \ \frac{1}{3}}.$

Solution 3 (Casework)

There are two different cases in which the product is zero; either the first number we select is zero, or the second one is. We consider these cases separately.

Case 1: $0$ is the first number chosen

There is a $\frac{1}{6}$ chance of selecting zero as the first number. At this point, the product will be zero no matter the choice of the second number, so there is a $\frac{1}{6}$ of getting the desired product in this case.


Case 2: $0$ is the second number chosen

There is a $\frac{5}{6}$ chance of choosing a number that is NOT zero as the first number. From there, there is a $\frac{1}{5}$ chance of picking zero from the remaining 5 numbers. Thus, there is a $\frac{5}{6} \cdot \frac{1}{5} = \frac{1}{6}$ chance of getting a product of 0 in this case.

Adding the probabilities from the two distinct cases up, we find that there is a $\frac{1}{6} + \frac{1}{6} = \boxed{\textbf{(D)} \ \frac{1}{3}}$ chance of getting a product of zero.

~cxsmi

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
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