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

Revision as of 21:52, 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

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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

@@ Line 9: / Line 9: @@
 ===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>
-===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: <math>0</math> is the first number chosen
-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.
-Case 2: <math>0</math> is the second number chosen
-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.
-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.
-~[https://artofproblemsolving.com/wiki/index.php/User:Cxsmi cxsmi]
 ==Video Solution (CREATIVE THINKING!!!)==

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