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 "2013 AMC 8 Problems/Problem 8"

m
(→‎Solution 2)
Line 27: Line 27:
 
First, there are <math>2^3 = 8</math> ways to flip the coins, in order.  
 
First, there are <math>2^3 = 8</math> ways to flip the coins, in order.  
 
Secondly, what we don't want are the ways without getting two consecutive heads: TTT, HTH, and THT. Then we can find out the probability of these three ways of flipping is <math> \frac18</math>, <math> \frac14 </math>,and <math> \frac14 </math> , respectively. So the rest is exactly the probability of flipping at least two consecutive heads: <math>1-\frac18-\frac14-\frac14 = \frac38</math>. It is the answer <math>\boxed{\textbf{(C)}\ \frac38}</math>.  ----LarryFlora
 
Secondly, what we don't want are the ways without getting two consecutive heads: TTT, HTH, and THT. Then we can find out the probability of these three ways of flipping is <math> \frac18</math>, <math> \frac14 </math>,and <math> \frac14 </math> , respectively. So the rest is exactly the probability of flipping at least two consecutive heads: <math>1-\frac18-\frac14-\frac14 = \frac38</math>. It is the answer <math>\boxed{\textbf{(C)}\ \frac38}</math>.  ----LarryFlora
 +
 +
==Solution 3==
 +
We know the total number of outcomes is <math>2^3=8</math>. Then we can just list out all the possibilities that have two consecutive heads: <math>HHH</math>, <math>HHT</math>, and <math>THH</math>. We can see there are three desired outcomes, which makes the probability <math>\boxed{\textbf{(C)}\ \frac38}</math>.
  
 
==See Also==
 
==See Also==
 
{{AMC8 box|year=2013|num-b=7|num-a=9}}
 
{{AMC8 box|year=2013|num-b=7|num-a=9}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 16:36, 12 January 2023

Problem

A fair coin is tossed 3 times. What is the probability of at least two consecutive heads?

$\textbf{(A)}\ \frac18 \qquad \textbf{(B)}\ \frac14 \qquad \textbf{(C)}\ \frac38 \qquad \textbf{(D)}\ \frac12 \qquad \textbf{(E)}\ \frac34$

Video Solution by OmegaLearn Using Casework

https://youtu.be/6xNkyDgIhEE?t=44

~ pi_is_3.14

Video Solution

https://youtu.be/2lynqd2bRZY ~savannahsolver

Solution 1

First, there are $2^3 = 8$ ways to flip the coins, in order.

The ways to get no one head are HHT and THH.

The way to get three consecutive heads is HHH.

Therefore, the probability of flipping at least two consecutive heads is $\boxed{\textbf{(C)}\ \frac38}$.

Solution 2

Let's figure it out using complementary counting.

First, there are $2^3 = 8$ ways to flip the coins, in order. Secondly, what we don't want are the ways without getting two consecutive heads: TTT, HTH, and THT. Then we can find out the probability of these three ways of flipping is $\frac18$, $\frac14$,and $\frac14$ , respectively. So the rest is exactly the probability of flipping at least two consecutive heads: $1-\frac18-\frac14-\frac14 = \frac38$. It is the answer $\boxed{\textbf{(C)}\ \frac38}$. ----LarryFlora

Solution 3

We know the total number of outcomes is $2^3=8$. Then we can just list out all the possibilities that have two consecutive heads: $HHH$, $HHT$, and $THH$. We can see there are three desired outcomes, which makes the probability $\boxed{\textbf{(C)}\ \frac38}$.

See Also

2013 AMC 8 (Problems • Answer Key • Resources)
Preceded by
Problem 7 		Followed by
Problem 9
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=2013_AMC_8_Problems/Problem_8&oldid=186441"