2013 AMC 8 Problems/Problem 8

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}$.

2013 AMC 8 (Problems • Answer Key • Resources)
Preceded by Problem 7	Followed by Problem 9
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All AJHSME/AMC 8 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

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2013 AMC 8 Problems/Problem 8

Contents

Problem

Video Solution

Solution 2

Solution 1

See Also