Difference between revisions of "2018 AMC 8 Problems/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 15:06, 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. Unknown error_msg) 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?
