2013 AMC 8 Problems/Problem 8: Difference between revisions

Revision as of 08:03, 3 August 2021

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/6xNkyDgIhEE?t=44

Solution 2

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

The ways to get no one head are HTH 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 1

Let's figure it out by 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 the three ways of flipping is $\frac18$ , $\frac14$ ,and $\frac14$ respectively. So the left 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

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

These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. Error creating thumbnail: Unable to save thumbnail to destination

@@ Line 20: / Line 20: @@
 First, there are <math>2^3 = 8</math> ways to flip the coins, in order.
-Secondly, what we don't want are the ways not to get two consecutive heads: TTT, HTH, and THT. Therefore, the probability of flipping is <math> \frac18</math>, <math> \frac14 </math>,and <math> \frac14 </math> respectively. So the probability of flipping at least two consecutive heads is <math>1-\frac18-\frac14-\frac14 = \frac38</math>. The answer is <math>\boxed{\textbf{(C)}\ \frac38}</math>.
+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 the three ways of flipping is <math> \frac18</math>, <math> \frac14 </math>,and <math> \frac14 </math> respectively. So the left is exactly the probability of flipping at least two consecutive heads: <math>1-\frac18-\frac14-\frac14 = \frac38</math>. It is the answer <math>\boxed{\textbf{(C)}\ \frac38}</math>.  ----LarryFlora
 ==See Also==
 {{AMC8 box|year=2013|num-b=7|num-a=9}}
 {{MAA Notice}}

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