Difference between revisions of "2016 AMC 8 Problems/Problem 13"
m (→Solution) |
m (→Solution 3(Complementary Counting)) |
||
(12 intermediate revisions by 5 users not shown) | |||
Line 3: | Line 3: | ||
<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== | + | ==Solution 1== |
− | 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> | + | 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 2== | ||
+ | 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>. | ||
+ | ==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> | ||
+ | |||
+ | |||
{{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 23:09, 31 March 2020
Two different numbers are randomly selected from the set and multiplied together. What is the probability that the product is ?
Solution 1
The product can only be if one of the numbers is 0. 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
There are a total of possibilities, because the numbers are different. We want to be the product so one of the numbers is . There are possibilities where is chosen for the first number and there are ways for to be chosen as the second number. We seek .
Solution 3 (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
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.