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Difference between revisions of "2016 AMC 8 Problems/Problem 13"
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===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>
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==Video Solution (CREATIVE THINKING!!!)==
 +
https://youtu.be/cRsvq0BH4MI
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 +
~Education, the Study of Everything
 +
  
 
==Video Solution by OmegaLearn ==
 
==Video Solution by OmegaLearn ==

Revision as of 13:07, 4 April 2023

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

There are a total of $30$ possibilities, because the two numbers that being multiplied are being picked at the same time, so there are $5$ possibilities that zero is being chosen because another number is already being chosen. We want $0$ to be the product so one of the numbers is $0$. There are $5$ possibilities where $0$ is chosen for the first number and there are $5$ ways for $0$ to be chosen as the second number. We seek $\boxed{\textbf{(D)} \, \frac{1}{3}}$.

Solution 3 (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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