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

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==Problem==
 
==Problem==
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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>
  
 
==Solution==
 
==Solution==
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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 two consecutive heads are HHT and THH.
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The way to get three consecutive heads is HHH.
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Therefore <math>\textbf{(C)}\ \frac38</math> of the possible flip sequences have at least two consecutive heads, and that is the probability.
  
 
==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 12:47, 27 November 2013

Problem

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

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

Solution

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

The ways to get two consecutive heads are HHT and THH.

The way to get three consecutive heads is HHH.

Therefore $\textbf{(C)}\ \frac38$ of the possible flip sequences have at least two consecutive heads, and that is the probability.

See Also

2013 AMC 8 (ProblemsAnswer KeyResources)
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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