Difference between revisions of "2016 AMC 10B Problems/Problem 12"
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==Solution 2== | ==Solution 2== | ||
| − | There are <math>2</math> cases to get an even number. Case 1: | + | There are <math>2</math> cases to get an even number. Case 1: <math>\text{Even} \times \text{Even}</math> and Case 2: <math>\text{Odd} \times \text{Even}</math>. Thus, to get an <math>\text{Even} \times \text{Even}</math>, you get <math>\frac {\binom {2}{2}}{\binom {5}{2}}= \frac {1}{10}</math>. And to get <math>\text{Odd} \times \text{Even}</math>, you get <math>\frac {\binom {3}{1}}{\binom {5}{2}}= \frac {6}{10}</math>. <math>\frac {1}{10}+\frac {6}{10}=\frac {7}{10}</math> which is <math>0.7</math> and the answer is <math>\boxed{\textbf{(D) }0.7}</math>. |
==Video Solution== | ==Video Solution== | ||
Revision as of 09:00, 19 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: 
and Case 2: 
. Thus, to get an , you get 
. And to get , 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. 
