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 by OmegaLearn Using Casework

https://youtu.be/6xNkyDgIhEE?t=44

~ pi_is_3.14

Video Solution

https://youtu.be/2lynqd2bRZY ~savannahsolver

Solution 1

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

Solution 3

We know the total number of outcomes is $2^3=8$ . Then we can just list out all the possibilities that have two consecutive heads: $HHH$ , $HHT$ , and $THH$ . We can see there are three desired outcomes, which makes the probability $\boxed{\textbf{(C)}\ \frac38}$ .

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

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

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