Difference between revisions of "2018 AMC 8 Problems/Problem 11"
(→Problem 11) |
(→Problem 11) |
||
| Line 9: | Line 9: | ||
<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}} | {{AMC8 box|year=2018|num-b=10|num-a=12}} | ||
| + | |||
| + | ==Solution== | ||
Revision as of 17:17, 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.
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?
| 2018 AMC 8 (Problems • Answer Key • Resources) | ||
| Preceded by Problem 10 |
Followed by Problem 12 | |
| 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 AJHSME/AMC 8 Problems and Solutions | ||
