Difference between revisions of "2016 AMC 10B Problems/Problem 12"
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==Solution== | ==Solution== | ||
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)}\ 0.7</math>. | 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)}\ 0.7</math>. | ||
| + | |||
| + | ==Video Solution== | ||
| + | https://youtu.be/tUpKpGmOwDQ | ||
| + | |||
| + | ~savannahsolver | ||
==See Also== | ==See Also== | ||
{{AMC10 box|year=2016|ab=B|num-b=11|num-a=13}} | {{AMC10 box|year=2016|ab=B|num-b=11|num-a=13}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Revision as of 17:22, 7 December 2020
Contents
Problem
Two different numbers are selected at random from 
and multiplied together. What is the probability that the product is even?
Solution
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 
.
Video Solution
~savannahsolver
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. 
