Difference between revisions of "2017 AMC 8 Problems/Problem 10"
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<math>\textbf{(A) }\frac{1}{10}\qquad\textbf{(B) }\frac{1}{5}\qquad\textbf{(C) }\frac{3}{10}\qquad\textbf{(D) }\frac{2}{5}\qquad\textbf{(E) }\frac{1}{2}</math> | <math>\textbf{(A) }\frac{1}{10}\qquad\textbf{(B) }\frac{1}{5}\qquad\textbf{(C) }\frac{3}{10}\qquad\textbf{(D) }\frac{2}{5}\qquad\textbf{(E) }\frac{1}{2}</math> | ||
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| + | ==Video Solution== | ||
| + | https://youtu.be/OOdK-nOzaII?t=1237 | ||
==Solution== | ==Solution== | ||
Revision as of 21:03, 12 August 2020
Problem 10
A box contains five cards, numbered 1, 2, 3, 4, and 5. Three cards are selected randomly without replacement from the box. What is the probability that 4 is the largest value selected?
Video Solution
https://youtu.be/OOdK-nOzaII?t=1237
Solution
There are 
possible groups of cards that can be selected. If 
is the largest card selected, then the other two cards must be either 
, 
, or 
, for a total 
groups of cards. Then the probability is just 
See Also:
| 2017 AMC 8 (Problems • Answer Key • Resources) | ||
| Preceded by Problem 9 |
Followed by Problem 11 | |
| 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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
