Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "2017 AMC 8 Problems/Problem 10"

(i added my own complementary counting solution)
m (→‎Solution 1 (combinoton))
 
(33 intermediate revisions by 11 users not shown)
Line 5: Line 5:
 
<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>
  
−
==Video Solution==
+
==Solution 1 (combinatianol)==
−
https://youtu.be/OOdK-nOzaII?t=1237
 
  
−
==Solution==
+
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>.
  
−
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>.
  
−
==Solution 2 (regular probability)==
+
==Video Solution (CREATIVE THINKING!!!)==
−
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.
+
https://youtu.be/P-K9AEAuhNY
−
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 fraction of no fours and no fives from no fives. <math>\frac{2}{5} - \frac{1}{10} = \frac{3}{10}</math> (C)
+
 
 +
~Education, the Study of Everything
 +
 
 +
==Video Solutions==
 +
*https://youtu.be/OOdK-nOzaII?t=1237
 +
*https://youtu.be/M9kj4ztWbwo
 +
 
 +
https://youtu.be/FN9qkU62a9U
  
−
Video here:
+
~savannahsolver
−
https://youtu.be/M9kj4ztWbwo
 
  
−
==Solution 3 (Complementary Probability)==
+
*https://www.youtube.com/watch?v=F935tcVcXvY  ~David
−
Using complementary counting, <math>P_\textbf{4 is largest} = 1-P_\textbf{5 is largest} - P_\textbf{3 is largest} = 1- \frac{\dbinom{2}{4}}{\dbinom{5}{3}} - \frac{\dbinom{2}{2}}{\dbinom{5}{3}} = 1- \frac{6}{10} - \frac{1}{10} = \boxed{{\frac{3}{10}}{\textbf{(C)}}}</math>
 
  
 
==See Also:==
 
==See Also:==

Latest revision as of 23:22, 3 November 2024

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?

$\textbf{(A) }\frac{1}{10}\qquad\textbf{(B) }\frac{1}{5}\qquad\textbf{(C) }\frac{3}{10}\qquad\textbf{(D) }\frac{2}{5}\qquad\textbf{(E) }\frac{1}{2}$

Solution 1 (combinatianol)

There are $\binom{5}{3}$ possible groups of cards that can be selected. If $4$ is the largest card selected, then the other two cards must be either $1$, $2$, or $3$, for a total $\binom{3}{2}$ groups of cards. Then, the probability is just ${\frac{{\dbinom{3}{2}}}{{\dbinom{5}{3}}}} = \boxed{{\textbf{(C) }} {\frac{3}{10}}}$.

Solution 2 (regular probability)

P (no 5)= $\frac{4}{5}$ * $\frac{3}{4}$ * $\frac{2}{3}$ = $\frac{2}{5}$. This is the fraction of total cases with no fives. p (no 4 and no 5)= $\frac{3}{5}$ * $\frac{2}{4}$ * $\frac{1}{3}$ = $\frac{6}{60}$ = $\frac{1}{10}$. This is the intersection of no fours and no fives. Subtract the fraction of no fours and no fives from that of no fives. $\frac{2}{5} - \frac{1}{10} = \frac{3}{10} = \boxed{{\textbf{(C) }} {\frac{3}{10}}}$.

Video Solution (CREATIVE THINKING!!!)

https://youtu.be/P-K9AEAuhNY

~Education, the Study of Everything

Video Solutions

https://youtu.be/FN9qkU62a9U

~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. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2017_AMC_8_Problems/Problem_10&oldid=231517"