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

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

~ pi_is_3.14

See Also

2013 AMC 8 (ProblemsAnswer KeyResources)
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

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