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

m (Latex)
m (Solution)
Line 11: Line 11:
 
The way to get three consecutive heads is HHH.
 
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==
 
==See Also==
 
{{AMC8 box|year=2013|num-b=7|num-a=9}}
 
{{AMC8 box|year=2013|num-b=7|num-a=9}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 22:58, 27 November 2013

Problem

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

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

Solution

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 $\boxed{\textbf{(C)}\ \frac38}$.

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

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