Difference between revisions of "2016 AMC 8 Problems/Problem 13"
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==Solution 2== | ==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>. | 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== | ||
+ | We can use complementary counting counting to solve this problem | ||
{{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 19:13, 4 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
We can use complementary counting counting to solve this problem
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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