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 "1980 AHSME Problems/Problem 20"

(Created page with "== Problem == A box contains <math>2</math> pennies, <math>4</math> nickels, and <math>6</math> dimes. Six coins are drawn without replacement, with each coin having an equal...")
 
(→‎Solution 1)
 
(7 intermediate revisions by 4 users not shown)
Line 11: Line 11:
 
    
 
    
  
−
== Solution ==
+
== Solution 1==
−
<math>\fbox{}</math>
+
We want the number of Successful Outcomes over the number of Total Outcomes. We want to calculate the total outcomes first. Since we have <math>12</math> coins and we need to choose <math>6</math>, we have <math>\binom{12}{6}</math> = <math>924</math> Total outcomes. For our successful outcomes, we can have <math>(1) 1</math> penny and <math>5</math> dimes, <math>2</math> nickels and <math>4</math> dimes, <math>1</math> nickel and <math>5</math> dimes, or <math>6</math> dimes.
 +
 
 +
For the case of <math>1</math> penny and <math>5</math> dimes, there are <math>\binom{6}{5}</math> ways to choose the dimes and <math>2</math> ways to choose the pennies. That is <math>6 \cdot 2 = 12</math> successful outcomes. For the case of <math>2</math> nickels and <math>4</math> dimes, we have <math>\binom{6}{4}</math> ways to choose the dimes and <math>\binom{4}{2}</math> ways to choose the nickels. We have <math>15 \cdot 6</math> = <math>90</math> successful outcomes. For the case of <math>1</math> nickel and <math>5</math> dimes, we have <math>\binom{4}{1} \cdot \binom{6}{5} = 24</math>. Lastly, we have <math>6</math> dimes and <math>0</math> nickels, and <math>0</math> pennies, so we only have one case. Therefore, we have <math>\dfrac {12 + 90 + 24 + 1}{924} = \dfrac{127}{924}</math> = <math>\boxed{C}</math>
  
 
== See also ==
 
== See also ==
 
{{AHSME box|year=1980|num-b=19|num-a=21}}   
 
{{AHSME box|year=1980|num-b=19|num-a=21}}   
  
−
[[Category: Introductory Algebra Problems]]
+
[[Category: Introductory Combinatorics Problems]]
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 13:25, 22 August 2022

Problem

A box contains $2$ pennies, $4$ nickels, and $6$ dimes. Six coins are drawn without replacement, with each coin having an equal probability of being chosen. What is the probability that the value of coins drawn is at least $50$ cents?

$\text{(A)} \ \frac{37}{924} \qquad \text{(B)} \ \frac{91}{924} \qquad \text{(C)} \ \frac{127}{924} \qquad \text{(D)}\ \frac{132}{924}\qquad \text{(E)}\ \text{none of these}$


Solution 1

We want the number of Successful Outcomes over the number of Total Outcomes. We want to calculate the total outcomes first. Since we have $12$ coins and we need to choose $6$, we have $\binom{12}{6}$ = $924$ Total outcomes. For our successful outcomes, we can have $(1) 1$ penny and $5$ dimes, $2$ nickels and $4$ dimes, $1$ nickel and $5$ dimes, or $6$ dimes.

For the case of $1$ penny and $5$ dimes, there are $\binom{6}{5}$ ways to choose the dimes and $2$ ways to choose the pennies. That is $6 \cdot 2 = 12$ successful outcomes. For the case of $2$ nickels and $4$ dimes, we have $\binom{6}{4}$ ways to choose the dimes and $\binom{4}{2}$ ways to choose the nickels. We have $15 \cdot 6$ = $90$ successful outcomes. For the case of $1$ nickel and $5$ dimes, we have $\binom{4}{1} \cdot \binom{6}{5} = 24$. Lastly, we have $6$ dimes and $0$ nickels, and $0$ pennies, so we only have one case. Therefore, we have $\dfrac {12 + 90 + 24 + 1}{924} = \dfrac{127}{924}$ = $\boxed{C}$

See also

1980 AHSME (Problems • Answer Key • Resources)
Preceded by
Problem 19 		Followed by
Problem 21
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 • 26 • 27 • 28 • 29 • 30
All AHSME 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=1980_AHSME_Problems/Problem_20&oldid=177342"