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

Revision as of 22:40, 11 December 2020

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

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

2016 AMC 10B (Problems • Answer Key • Resources)
Preceded by Problem 11	Followed by Problem 13
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All AMC 10 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

@@ Line 7: / Line 7: @@
 ==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>.
+==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==

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