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Difference between revisions of "2016 AMC 10B Problems/Problem 12"

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==Problem==
 
==Problem==
  
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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?
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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 1==
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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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==Solution 2==
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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
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==Video Solution==
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https://youtu.be/tUpKpGmOwDQ - savannahsolver
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https://youtu.be/IRyWOZQMTV8?t=933 - pi_is_3.14
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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 02:32, 16 February 2021

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

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

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