Difference between revisions of "2016 AMC 10B Problems/Problem 12"
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<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== | + | ==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)}\ 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>. | ||
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==Video Solution== | ==Video Solution== | ||
| − | https://youtu.be/tUpKpGmOwDQ | + | https://youtu.be/tUpKpGmOwDQ - savannahsolver |
| − | + | https://youtu.be/IRyWOZQMTV8?t=933 - pi_is_3.14 | |
==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 02:32, 16 February 2021
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 2 cases to get an even number. Case 1: even*even and Case 2: odd*even. Thus, to get an even*even, you get (2C2)/(5C2)= 1/10. And to get odd*even, you get [(3C1)*(2C1)]/(5C2) = 6/10. 1/10 + 6/10 yields 0.7 solution D
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. 
