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Difference between revisions of "2013 AMC 8 Problems/Problem 8"

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A fair coin is tossed 3 times. What is the probability of at least two consecutive heads?
 
A fair coin is tossed 3 times. What is the probability of at least two consecutive heads?
  
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<math>\textbf{(A)}\ \frac18 \qquad \textbf{(B)}\ \frac14 \qquad \textbf{(C)}\ \frac38 \qquad \textbf{(D)}\ \frac12 \qquad \textbf{(E)}\ \frac34</math>
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<math>\textbf{(A)}\frac{1}{8} \qquad \textbf{(B)}\frac{1}{4} \qquad \textbf{(C)}\frac{3}{8} \qquad \textbf{(D)}\frac{1}{2} \qquad \textbf{(E)}\frac{3}{4}</math>
  
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==Video Solution by OmegaLearn Using Casework==
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==Solution 1==
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https://youtu.be/6xNkyDgIhEE?t=44
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There are <math>2^3 = 8</math> ways to flip the coins, in order.
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There are two ways to get exactly two consecutive heads: HHT and THH.
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There is only one way to get three consecutive heads: HHH.
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Therefore, the probability of flipping at least two consecutive heads is <math>\boxed{\textbf{(C)}\frac{3}{8}}</math>.
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==Solution 2==
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Let's use [[complementary counting]]. To start with, the unfavorable outcomes (in this case, not getting 2 consecutive heads) are: TTT, HTH, and THT. The probability of these three outcomes is <math>\frac{1}{8}</math>, <math>\frac{1}{4}</math>, and <math>\frac{1}{4}</math>, respectively. So the rest is exactly the probability of flipping at least two consecutive heads: <math>1-\frac{1}{8}-\frac{1}{4}-\frac{1}{4}=\frac{3}{8}</math>. It is the answer <math>\boxed{\textbf{(C)}\frac{3}{8}}</math>.
  
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~ pi_is_3.14
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~LarryFlora
  
 
==Video Solution==
 
==Video Solution==
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https://youtu.be/2lynqd2bRZY ~savannahsolver
 
https://youtu.be/2lynqd2bRZY ~savannahsolver
 +
https://youtu.be/6xNkyDgIhEE?t=44
  
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==Solution 1==
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~ pi_is_3.14
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First, there are <math>2^3 = 8</math> ways to flip the coins, in order.
 
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The ways to get no one head are HHT and THH.
 
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The way to get three consecutive heads is HHH.
 
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Therefore, the probability of flipping at least two consecutive heads is <math>\boxed{\textbf{(C)}\ \frac38}</math>.
 
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==Solution 2==
 
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Let's figure it out using complementary counting.
 
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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 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
 
  
 
==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}}

Latest revision as of 00:39, 2 February 2023

Problem

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

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

Solution 1

There are $2^3 = 8$ ways to flip the coins, in order. There are two ways to get exactly two consecutive heads: HHT and THH. There is only one way to get three consecutive heads: HHH. Therefore, the probability of flipping at least two consecutive heads is $\boxed{\textbf{(C)}\frac{3}{8}}$.

Solution 2

Let's use complementary counting. To start with, the unfavorable outcomes (in this case, not getting 2 consecutive heads) are: TTT, HTH, and THT. The probability of these three outcomes is $\frac{1}{8}$, $\frac{1}{4}$, and $\frac{1}{4}$, respectively. So the rest is exactly the probability of flipping at least two consecutive heads: $1-\frac{1}{8}-\frac{1}{4}-\frac{1}{4}=\frac{3}{8}$. It is the answer $\boxed{\textbf{(C)}\frac{3}{8}}$.

~LarryFlora

Video Solution

https://youtu.be/2lynqd2bRZY ~savannahsolver https://youtu.be/6xNkyDgIhEE?t=44

~ pi_is_3.14

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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