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Difference between revisions of "2016 AMC 8 Problems/Problem 13"
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===Solution 1===
 
===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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1. Identify the total number of ways to select two different numbers from the set:
  
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===Solution 2===
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The set has 6 elements. The number of ways to choose 2 different numbers from 6 is given by the combination formula: <math>\binom{6}{2} = \frac{6 \times 5}{2 \times 1} = 15</math>.
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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>.
 
  
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===Solution 3 (Complementary Counting)===
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2. Identify the favorable outcomes:
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For the product to be zero, one of the chosen numbers must be zero. The set contains one zero (0). To have a product of zero, we need to choose 0 and any other number from the remaining five numbers <math>-2, -1, 3, 4, 5</math>.
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The number of ways to choose 0 and one other number from the remaining five is 5.
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3. Calculate the probability:
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The probability is the number of favorable outcomes divided by the total number of outcomes: <math>\text{Probability} = \frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \frac{5}{15} = \frac{1}{3}</math>.
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Thus, the probability that the product is <math>0</math> is <math>\boxed{\textbf{(D)} \ \frac{1}{3}}.</math>
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~GeometryMystery
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===Solution 2 (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
 +
 +
~Education, the Study of Everything
 +
  
 
==Video Solution by OmegaLearn ==
 
==Video Solution by OmegaLearn ==

Latest revision as of 21:14, 10 June 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

1. Identify the total number of ways to select two different numbers from the set:

The set has 6 elements. The number of ways to choose 2 different numbers from 6 is given by the combination formula: $\binom{6}{2} = \frac{6 \times 5}{2 \times 1} = 15$.

2. Identify the favorable outcomes:

For the product to be zero, one of the chosen numbers must be zero. The set contains one zero (0). To have a product of zero, we need to choose 0 and any other number from the remaining five numbers $-2, -1, 3, 4, 5$.

The number of ways to choose 0 and one other number from the remaining five is 5.

3. Calculate the probability:

The probability is the number of favorable outcomes divided by the total number of outcomes: $\text{Probability} = \frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \frac{5}{15} = \frac{1}{3}$.

Thus, the probability that the product is $0$ is $\boxed{\textbf{(D)} \ \frac{1}{3}}.$

~GeometryMystery

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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