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

Solution 3

We can list out all the ways to flip a coin three times: HHH,HHT,HTH,THH,HTT,THT,TTH,TTT. Out of them, only HHH,HHT,THH, have at least two consecutive heads. Since there are three ways to flip at least two consecutive heads, and eight total choices, the answer is $\boxed{\textbf{(C)}\frac{3}{8}}$ .

~andliu766

Video Solution

https://youtu.be/2lynqd2bRZY ~savannahsolver https://youtu.be/6xNkyDgIhEE?t=44

~ pi_is_3.14

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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