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

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==Solution==
 
==Solution==
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Since <math>AB = AC</math>, then <math>\triangle ABC</math> is isosceles, so <math>BD = CD</math>. Therefore, the coordinates of <math>C</math> are <math>(-1 - 3, 3 + 6) = \boxed{\textbf{(C) } (-4,9)}</math>.
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==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 13:50, 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

Since $AB = AC$, then $\triangle ABC$ is isosceles, so $BD = CD$. Therefore, the coordinates of $C$ are $(-1 - 3, 3 + 6) = \boxed{\textbf{(C) } (-4,9)}$.

See Also

2017 AMC 10B (ProblemsAnswer KeyResources)
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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