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"

(→‎Solution 2 (regular probability))
(→‎Solution 2 (regular probability))
Line 13: Line 13:
  
 
==Solution 2 (regular probability)==
 
==Solution 2 (regular probability)==
−
P (no 5)= <math>\frac{4}{5} * \frac{3]{4} * \frac{2}{3} = \frac{2}{5} this is the fraction of total cases with no fives.
+
P (no 5)= <math>\frac{4}{5} * \frac{3]{4} * \frac{2}{3} = \frac{2}{5}</math> this is the fraction of total cases with no fives.
−
p (no 4 and no 5)= </math>\frac{3}{5} * \frac{2}{4} * \frac{1}{3}= \frac{6}{60} = \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.
+
p (no 4 and no 5)= <math>\frac{3}{5} * \frac{2}{4} * \frac{1}{3}= \frac{6}{60} = \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}$ (C)
+
<math>\frac{2}{5} - \frac{1}{10} = \frac{3}{10}</math> (C)
  
 
Video here:
 
Video here:

Revision as of 22:23, 7 November 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?

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

Video Solution

https://youtu.be/OOdK-nOzaII?t=1237

Solution

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)= $\frac{4}{5} * \frac{3]{4} * \frac{2}{3} = \frac{2}{5}$ (Error compiling LaTeX. Unknown error_msg) 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 fraction of no fours and no fives from no fives.

$\frac{2}{5} - \frac{1}{10} = \frac{3}{10}$ (C)

Video here: https://youtu.be/M9kj4ztWbwo

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=136996"