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 "2004 AMC 10A Problems/Problem 10"

m (There was a typo. The title of section "Solution" was wrong.)
(→‎Problem)
Line 1: Line 1:
 +
==Video Solution==
 +
https://youtu.be/bq34-DyUpEU
 +
 +
Education, the Study of Everything
 +
 +
 
==Problem==
 
==Problem==
 
Coin <math>A</math> is flipped three times and coin <math>B</math> is flipped four times.  What is the probability that the number of heads obtained from flipping the two fair coins is the same?
 
Coin <math>A</math> is flipped three times and coin <math>B</math> is flipped four times.  What is the probability that the number of heads obtained from flipping the two fair coins is the same?

Revision as of 17:13, 30 December 2020

Video Solution

https://youtu.be/bq34-DyUpEU

Education, the Study of Everything


Problem

Coin $A$ is flipped three times and coin $B$ is flipped four times. What is the probability that the number of heads obtained from flipping the two fair coins is the same?

$\mathrm{(A) \ } \frac{29}{128} \qquad \mathrm{(B) \ } \frac{23}{128} \qquad \mathrm{(C) \ } \frac14 \qquad \mathrm{(D) \ } \frac{35}{128} \qquad \mathrm{(E) \ } \frac12$

Solution

There are $4$ ways that the same number of heads will be obtained; $0$, $1$, $2$, or $3$ heads.

The probability of both getting $0$ heads is $\left(\frac12\right)^3{3\choose0}\left(\frac12\right)^4{4\choose0}=\frac1{128}$

The probability of both getting $1$ head is $\left(\frac12\right)^3{3\choose1}\left(\frac12\right)^4{4\choose1}=\frac{12}{128}$

The probability of both getting $2$ heads is $\left(\frac12\right)^3{3\choose2}\left(\frac12\right)^4{4\choose2}=\frac{18}{128}$

The probability of both getting $3$ heads is $\left(\frac12\right)^3{3\choose3}\left(\frac12\right)^4{4\choose3}=\frac{4}{128}$

Therefore, the probabiliy of flipping the same number of heads is: $\frac{1+12+18+4}{128}=\frac{35}{128}\Rightarrow\boxed{\mathrm{(D)}\ \frac{35}{128}}$

See also

2004 AMC 10A (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 AMC 10 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=2004_AMC_10A_Problems/Problem_10&oldid=141084"