Difference between revisions of "2013 AMC 8 Problems/Problem 8"

Revision as of 23:58, 27 November 2013

A fair coin is tossed 3 times. What is the probability of at least two consecutive heads?

$\textbf{(A)}\ \frac18 \qquad \textbf{(B)}\ \frac14 \qquad \textbf{(C)}\ \frac38 \qquad \textbf{(D)}\ \frac12 \qquad \textbf{(E)}\ \frac34$

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

The ways to get two consecutive heads are HHT and THH.

The way to get three consecutive heads is HHH.

Therefore, the probability of flipping at least two consecutive heads is $\boxed{\textbf{(C)}\ \frac38}$ .

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

@@ Line 11: / Line 11: @@
 The way to get three consecutive heads is HHH.
-Therefore <math>\boxed{\textbf{(C)}\ \frac38}</math> of the possible flip sequences have at least two consecutive heads, and that is the probability.
+Therefore, the probability of flipping at least two consecutive heads is <math>\boxed{\textbf{(C)}\ \frac38}</math>.
 ==See Also==
 {{AMC8 box|year=2013|num-b=7|num-a=9}}
 {{MAA Notice}}