Difference between revisions of "2016 AMC 10B Problems/Problem 12"
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==Solution 1== | ==Solution 1== | ||
| − | The product will be even if at least one selected number is even, and odd if none are. Using complementary counting, the chance that both numbers are odd is <math>\frac{\tbinom32}{\tbinom52}=\frac3{10}</math>, so the answer is <math>1-0.3</math> which is <math>\textbf{(D)} | + | The product will be even if at least one selected number is even, and odd if none are. Using complementary counting, the chance that both numbers are odd is <math>\frac{\tbinom32}{\tbinom52}=\frac3{10}</math>, so the answer is <math>1-0.3</math> which is <math>\boxed{\textbf{(D) }0.7}</math> |
==Solution 2== | ==Solution 2== | ||
Revision as of 21:47, 15 April 2022
Problem
Two different numbers are selected at random from 
and multiplied together. What is the probability that the product is even?
Solution 1
The product will be even if at least one selected number is even, and odd if none are. Using complementary counting, the chance that both numbers are odd is 
, so the answer is 
which is 
Solution 2
There are 
cases to get an even number. Case 1: Even 
Even and Case 2: Odd Even. Thus, to get an Even Even, you get 
. And to get Odd Even, you get 
. 
which is 
and the answer is .
Video Solution
https://youtu.be/tUpKpGmOwDQ - savannahsolver
https://youtu.be/IRyWOZQMTV8?t=933 - pi_is_3.14
See Also
| 2016 AMC 10B (Problems • Answer Key • Resources) | ||
| Preceded by Problem 11 |
Followed by Problem 13 | |
| 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 AMC 10 Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
