# 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/2lynqd2bRZY ~savannahsolver

## Solution 1

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

The ways to get no one head are HHT 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 2

Let's figure it out using 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 rest 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