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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.  
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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 the three ways of flipping is <math> \frac18</math>, <math> \frac14 </math>,and <math> \frac14 </math> , respectively. So the left 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 left 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
  
 
==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 08:04, 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 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 left 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

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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