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>( 1, 2, 3, 4, 5)</math> and multiplied together. What is the probability that the product is even?
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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>
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==Solution==
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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>.
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==See Also==
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{{AMC10 box|year=2016|ab=B|num-b=11|num-a=13}}
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{{MAA Notice}}

Revision as of 11:40, 21 February 2016

Problem

Two different numbers are selected at random from $\{1, 2, 3, 4, 5\}$ and multiplied together. What is the probability that the product is even?

$\textbf{(A)}\ 0.2\qquad\textbf{(B)}\ 0.4\qquad\textbf{(C)}\ 0.5\qquad\textbf{(D)}\ 0.7\qquad\textbf{(E)}\ 0.8$

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 $\frac{\tbinom32}{\tbinom52}=\frac3{10}$, so the answer is $1-0.3$ which is $\textbf{(D)}\ 0.7$.

See Also

2016 AMC 10B (ProblemsAnswer KeyResources)
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

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