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>? | + | == Problem == |
+ | |||
+ | 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> | ||
− | ==Solution 1== | + | == Solutions== |
− | 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>. | + | |
+ | ===Solution 1=== | ||
+ | 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>. | ||
− | ==Solution 2 | + | ===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> | 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. | ||
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+ | 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!!!)== | ||
+ | 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== | ||
{{AMC8 box|year=2016|num-b=12|num-a=14}} | {{AMC8 box|year=2016|num-b=12|num-a=14}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 18:33, 20 January 2024
Contents
Problem
Two different numbers are randomly selected from the set and multiplied together. What is the probability that the product is ?
Solutions
Solution 1
The product can only be if one of the numbers is . Once we chose , there are ways we can chose the second number, or . There are ways we can chose numbers randomly, and that is . So, so the answer is .
Solution 2 (Complementary Counting)
Because the only way the product of the two numbers is is if one of the numbers we choose is we calculate the probability of NOT choosing a We get Therefore our answer is
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: is the first number chosen
There is a 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 of getting the desired product in this case.
Case 2: is the second number chosen
There is a chance of choosing a number that is NOT zero as the first number. From there, there is a chance of picking zero from the remaining 5 numbers. Thus, there is a 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 chance of getting a product of zero.
Video Solution (CREATIVE THINKING!!!)
~Education, the Study of Everything
Video Solution by OmegaLearn
https://youtu.be/6xNkyDgIhEE?t=357
~ pi_is_3.14
Video Solution
~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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