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> | ||
| − | ==Solution== | + | ==Solution 1 (combinatianol)== |
| − | ==See Also== | + | 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{{\dbinom{3}{2}}}{{\dbinom{5}{3}}}} = \boxed{{\textbf{(C) }} {\frac{3}{10}}}</math>. |
| + | |||
| + | ==Solution 2 (regular probability)== | ||
| + | P (no 5)= <math>\frac{4}{5}</math> * <math>\frac{3}{4}</math> * <math>\frac{2}{3}</math> = <math>\frac{2}{5}</math>. This is the fraction of total cases with no fives. | ||
| + | p (no 4 and no 5)= <math>\frac{3}{5}</math> * <math>\frac{2}{4}</math> * <math>\frac{1}{3}</math> = <math>\frac{6}{60}</math> = <math>\frac{1}{10}</math>. This is the intersection of no fours and no fives. Subtract the fraction of no fours and no fives from that of no fives. <math>\frac{2}{5} - \frac{1}{10} = \frac{3}{10} = \boxed{{\textbf{(C) }} {\frac{3}{10}}}</math>. | ||
| + | |||
| + | ==Video Solution (CREATIVE THINKING!!!)== | ||
| + | https://youtu.be/P-K9AEAuhNY | ||
| + | |||
| + | ~Education, the Study of Everything | ||
| + | |||
| + | ==Video Solutions== | ||
| + | *https://youtu.be/OOdK-nOzaII?t=1237 | ||
| + | *https://youtu.be/M9kj4ztWbwo | ||
| + | |||
| + | https://youtu.be/FN9qkU62a9U | ||
| + | |||
| + | ~savannahsolver | ||
| + | |||
| + | *https://www.youtube.com/watch?v=F935tcVcXvY ~David | ||
| + | |||
| + | ==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}} | ||
Latest revision as of 23:22, 3 November 2024
Contents
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 1 (combinatianol)
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 
.
Solution 2 (regular probability)
P (no 5)= 
* 
* 
= 
. This is the fraction of total cases with no fives.
p (no 4 and no 5)= 
* 
* 
= 
= 
. This is the intersection of no fours and no fives. Subtract the fraction of no fours and no fives from that of no fives. 
.
Video Solution (CREATIVE THINKING!!!)
~Education, the Study of Everything
Video Solutions
~savannahsolver
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. 
