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 "2000 AMC 8 Problems/Problem 21"

m (→‎Solution)
(→‎Solution 2)
 
Line 17: Line 17:
  
 
Thus, since all possible coin flips of 3 coins are equally likely, the probability is <math>\boxed{(B) \frac38}</math>.
 
Thus, since all possible coin flips of 3 coins are equally likely, the probability is <math>\boxed{(B) \frac38}</math>.
−
 
−
==Solution 2==
 
−
 
−
Let <math>K(n)</math> be the probability that Keiko gets <math>n</math> heads, and let <math>E(n)</math> be the probability that Ephriam gets <math>n</math> heads.
 
−
 
−
<math>K(0) = \frac{1}{2}</math>
 
−
 
−
<math>K(1) = \frac{1}{2}</math>
 
−
 
−
<math>K(2) = 0</math>  (Keiko only has one penny!)
 
−
 
−
<math>E(0) = \frac{1}{2}\cdot\frac{1}{2} = \frac{1}{4}</math>
 
−
 
−
<math>E(1) = \frac{1}{2}\cdot\frac{1}{2} + \frac{1}{2}\cdot\frac{1}{2} = 2\cdot\frac{1}{4} = \frac{1}{2}</math>  (because Ephraim can get HT or TH)
 
−
 
−
<math>E(2) = \frac{1}{2}\cdot\frac{1}{2} = \frac{1}{4}</math>
 
−
 
−
The probability that Keiko gets <math>0</math> heads and Ephriam gets <math>0</math> heads is <math>K(0)\cdot E(0)</math>.  Similarly for <math>1</math> head and <math>2</math> heads.  Thus, we have:
 
−
 
−
<math>P = K(0)\cdot E(0) + K(1)\cdot E(1) + K(2)\cdot E(2)</math>
 
−
 
−
<math>P = \frac{1}{2}\cdot\frac{1}{4} + \frac{1}{2}\cdot\frac{1}{2} + 0</math>
 
−
 
−
<math>P = \frac{3}{8}</math>
 
−
 
−
Thus the answer is <math>\boxed{B}</math>.
 
  
 
== Video Solution ==
 
== Video Solution ==

Latest revision as of 10:02, 12 June 2024

Problem

Keiko tosses one penny and Ephraim tosses two pennies. The probability that Ephraim gets the same number of heads that Keiko gets is

$\text{(A)}\ \frac{1}{4}\qquad\text{(B)}\ \frac{3}{8}\qquad\text{(C)}\ \frac{1}{2}\qquad\text{(D)}\ \frac{2}{3}\qquad\text{(E)}\ \frac{3}{4}$

Solution

Divide it into $2$ cases:

1) Keiko and Ephriam both get $0$ heads: This means that they both roll all tails, so there is only $1$ way for this to happen.

2) Keiko and Ephriam both get $1$ head: For Keiko, there is only $1$ way for this to happen because he is only flipping 1 penny, but for Ephriam, there are 2 ways since there are $2$ choices for when he can flip the head. So, in total there are $2 \cdot 1 = 2$ ways for this case.

Thus, in total there are $3$ ways that work. Since there are $2$ choices for each coin flip (Heads or Tails), there are $2^3 = 8$ total ways of flipping 3 coins.

Thus, since all possible coin flips of 3 coins are equally likely, the probability is $\boxed{(B) \frac38}$.

Video Solution

https://youtu.be/a_Tfeb_6dqE Soo, DRMS, NM

https://www.youtube.com/watch?v=mLrtRuJuYI4 ~David

See Also

2000 AMC 8 (Problems • Answer Key • Resources)
Preceded by
Problem 20 		Followed by
Problem 22
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=2000_AMC_8_Problems/Problem_21&oldid=220395"