Difference between revisions of "2018 AMC 8 Problems/Problem 11"

(Problem 11)
(Problem 11)
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==Problem 11==
 
==Problem 11==
 
Abby, Bridget, and four of their classmates will be seated in two rows of three for a group picture, as shown.
 
Abby, Bridget, and four of their classmates will be seated in two rows of three for a group picture, as shown.
\begin{eqnarray*}
+
<math>\begin{eqnarray*}
 
\text{X}&\quad\text{X}\quad&\text{X} \\
 
\text{X}&\quad\text{X}\quad&\text{X} \\
 
\text{X}&\quad\text{X}\quad&\text{X}  
 
\text{X}&\quad\text{X}\quad&\text{X}  
\end{eqnarray*}
+
\end{eqnarray*}</math>
 
If the seating positions are assigned randomly, what is the probability that Abby and Bridget are adjacent to each other in the same row or the same column?
 
If the seating positions are assigned randomly, what is the probability that Abby and Bridget are adjacent to each other in the same row or the same column?
  
 
<math>\textbf{(A) } \frac{1}{3} \qquad \textbf{(B) } \frac{2}{5} \qquad \textbf{(C) } \frac{7}{15} \qquad \textbf{(D) } \frac{1}{2} \qquad \textbf{(E) } \frac{2}{3}</math>
 
<math>\textbf{(A) } \frac{1}{3} \qquad \textbf{(B) } \frac{2}{5} \qquad \textbf{(C) } \frac{7}{15} \qquad \textbf{(D) } \frac{1}{2} \qquad \textbf{(E) } \frac{2}{3}</math>
{{AMC8 box|year=2018|num-b=10|num-a=12}}
+
<math>{{AMC8 box|year=2018|num-b=10|num-a=12}}</math>

Revision as of 14:06, 21 November 2018

Problem 11

Abby, Bridget, and four of their classmates will be seated in two rows of three for a group picture, as shown. $\begin{eqnarray*} \text{X}&\quad\text{X}\quad&\text{X} \\ \text{X}&\quad\text{X}\quad&\text{X} \end{eqnarray*}$ (Error compiling LaTeX. ! Missing \endgroup inserted.) If the seating positions are assigned randomly, what is the probability that Abby and Bridget are adjacent to each other in the same row or the same column?

$\textbf{(A) } \frac{1}{3} \qquad \textbf{(B) } \frac{2}{5} \qquad \textbf{(C) } \frac{7}{15} \qquad \textbf{(D) } \frac{1}{2} \qquad \textbf{(E) } \frac{2}{3}$ ${{AMC8 box|year=2018|num-b=10|num-a=12}}$

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