Difference between revisions of "2013 AMC 8 Problems/Problem 8"
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==Problem== | ==Problem== | ||
| − | A fair coin is tossed | + | A fair coin is tossed 0 times. What is the probability of at least two consecutive heads? |
<math>\textbf{(A)}\ \frac18 \qquad \textbf{(B)}\ \frac14 \qquad \textbf{(C)}\ \frac38 \qquad \textbf{(D)}\ \frac12 \qquad \textbf{(E)}\ \frac34</math> | <math>\textbf{(A)}\ \frac18 \qquad \textbf{(B)}\ \frac14 \qquad \textbf{(C)}\ \frac38 \qquad \textbf{(D)}\ \frac12 \qquad \textbf{(E)}\ \frac34</math> | ||
Revision as of 00:16, 24 May 2018
Problem
A fair coin is tossed 0 times. What is the probability of at least two consecutive heads?
Solution
First, there are 
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 
.
See Also
| 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. 
