Difference between revisions of "2016 AMC 10B Problems/Problem 12"
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==Problem== | ==Problem== | ||
| − | Two different numbers are selected at random from <math> | + | Two different numbers are selected at random from <math>\{1, 2, 3, 4, 5\}</math> and multiplied together. What is the probability that the product is even? |
<math>\textbf{(A)}\ 0.2\qquad\textbf{(B)}\ 0.4\qquad\textbf{(C)}\ 0.5\qquad\textbf{(D)}\ 0.7\qquad\textbf{(E)}\ 0.8</math> | <math>\textbf{(A)}\ 0.2\qquad\textbf{(B)}\ 0.4\qquad\textbf{(C)}\ 0.5\qquad\textbf{(D)}\ 0.7\qquad\textbf{(E)}\ 0.8</math> | ||
| + | |||
| + | ==Solution== | ||
| + | It will be even if at least one selected number is even, and odd if none are. Using complementary counting, the chance that both 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>. | ||
| + | |||
| + | ==See Also== | ||
| + | {{AMC10 box|year=2016|ab=B|num-b=11|num-a=13}} | ||
| + | {{MAA Notice}} | ||
Revision as of 11:40, 21 February 2016
Problem
Two different numbers are selected at random from 
and multiplied together. What is the probability that the product is even?
Solution
It will be even if at least one selected number is even, and odd if none are. Using complementary counting, the chance that both are odd is 
, so the answer is 
which is 
.
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. 
