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Difference between revisions of "2017 AMC 10B Problems/Problem 9"

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There are two ways that the contestant can win.
 
There are two ways that the contestant can win.
  
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Case 1: They guess all three right. This can only happen <math>\frac1/{3} * \frac1/{3} * \frac1/{3} = \frac1/{27}</math> of the time.
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Case 1: They guess all three right. This can only happen <math>1/3 * 1/3 * 1/3 = 1/27</math> of the time.
  
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Case 2: They guess only two right. We pick one of the questions to get wrong, <math>3</math>, and this can happen <math>\frac1/{3} * \frac1/{3} * \frac2/{3}</math> of the time. Thus, <math>\frac2/{27} * 3</math> = <math>\frac6/{27}</math>.
+
Case 2: They guess only two right. We pick one of the questions to get wrong, <math>3</math>, and this can happen <math>1/3 * 1/3 * 2/3</math> of the time. Thus, <math>2/27 * 3</math> = <math>6/27</math>.
  
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So, in total the two cases combined equals <math>\frac1/{27} + \frac6/{27}</math> = <math>\boxed{\textbf{(D)}\ \frac7/{27]}</math>.
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So, in total the two cases combined equals <math>1/27 + 6/27</math> = <math>\boxed{\textbf{(D)}\ 7/27}</math>.
  
 
==See Also==
 
==See Also==
 
{{AMC10 box|year=2017|ab=B|num-b=8|num-a=10}}
 
{{AMC10 box|year=2017|ab=B|num-b=8|num-a=10}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 13:40, 16 February 2017

Problem

A radio program has a quiz consisting of $3$ multiple-choice questions, each with $3$ choices. A contestant wins if he or she gets $2$ or more of the questions right. The contestant answers randomly to each question. What is the probability of winning?

$\textbf{(A)}\ \frac{1}{27}\qquad\textbf{(B)}\ \frac{1}{9}\qquad\textbf{(C)}\ \frac{2}{9}\qquad\textbf{(D)}\ \frac{7}{27}\qquad\textbf{(E)}\ \frac{1}{2}$

Solution

There are two ways that the contestant can win.

Case 1: They guess all three right. This can only happen $1/3 * 1/3 * 1/3 = 1/27$ of the time.

Case 2: They guess only two right. We pick one of the questions to get wrong, $3$, and this can happen $1/3 * 1/3 * 2/3$ of the time. Thus, $2/27 * 3$ = $6/27$.

So, in total the two cases combined equals $1/27 + 6/27$ = $\boxed{\textbf{(D)}\ 7/27}$.

See Also

2017 AMC 10B (Problems • Answer Key • Resources)
Preceded by
Problem 8 		Followed by
Problem 10
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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