Difference between revisions of "2017 AMC 8 Problems/Problem 10"
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There are <math>\binom{5}{3}</math> possible groups of cards that can be selected. If <math>4</math> is the largest card selected, then the other two cards must be either <math>1</math>, <math>2</math>, or <math>3</math>, for a total <math>\binom{3}{2}</math> groups of cards. Then the probability is just <math>{\frac{{\binom{3}{2}}}{{\binom{5}{3}}}} = \boxed{{\textbf{(C) }} {\frac{3}{10}}}</math> | There are <math>\binom{5}{3}</math> possible groups of cards that can be selected. If <math>4</math> is the largest card selected, then the other two cards must be either <math>1</math>, <math>2</math>, or <math>3</math>, for a total <math>\binom{3}{2}</math> groups of cards. Then the probability is just <math>{\frac{{\binom{3}{2}}}{{\binom{5}{3}}}} = \boxed{{\textbf{(C) }} {\frac{3}{10}}}</math> | ||
| − | ==See Also== | + | ==See Also:== |
{{AMC8 box|year=2017|num-b=9|num-a=11}} | {{AMC8 box|year=2017|num-b=9|num-a=11}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Revision as of 17:41, 14 October 2019
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?
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. 
