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Difference between revisions of "2013 AMC 8 Problems/Problem 8"
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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 not to get two consecutive heads: TTT, HTH, and THT. Therefore, the probability of flipping is <math> \frac18</math>, <math> \frac14 </math>,and <math> \frac14 </math> respectively. So the probability of flipping at least two consecutive heads is <math>1-</math>\frac18<math>-</math>\frac14<math>-</math>\frac14<math>=</math>\boxed{\textbf{(C)}\ \frac38}$.
+
Secondly, what we don't want are the ways not to get two consecutive heads: TTT, HTH, and THT. Therefore, the probability of flipping is <math> \frac18</math>, <math> \frac14 </math>,and <math> \frac14 </math> respectively. So the probability of flipping at least two consecutive heads is <math>1-\frac18-\frac14-\frac14 =</math>\boxed{\textbf{(C)}\ \frac38}$.
  
 
==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 07:58, 3 August 2021

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

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

Solution 2

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

The ways to get no one head are HTH 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 1

Let's figure it out by complementary counting.

First, there are $2^3 = 8$ ways to flip the coins, in order. Secondly, what we don't want are the ways not to get two consecutive heads: TTT, HTH, and THT. Therefore, the probability of flipping is $\frac18$, $\frac14$,and $\frac14$ respectively. So the probability of flipping at least two consecutive heads is $1-\frac18-\frac14-\frac14 =$\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

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