# 2013 AMC 8 Problems/Problem 8

## Problem

A fair coin is tossed 3 times. What is the probability of at least two consecutive heads? $\textbf{(A)}\frac{1}{8} \qquad \textbf{(B)}\frac{1}{4} \qquad \textbf{(C)}\frac{3}{8} \qquad \textbf{(D)}\frac{1}{2} \qquad \textbf{(E)}\frac{3}{4}$

## 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

~ pi_is_3.14

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