Difference between revisions of "2013 AMC 8 Problems/Problem 8"
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==Problem== | ==Problem== | ||
| + | A fair coin is tossed 3 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> | ||
==Solution== | ==Solution== | ||
| + | First, there are <math>2^3 = 8</math> 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 <math>\textbf{(C)}\ \frac38</math> of the possible flip sequences have at least two consecutive heads, and that is the probability. | ||
==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 11:47, 27 November 2013
Problem
A fair coin is tossed 3 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 
of the possible flip sequences have at least two consecutive heads, and that is the probability.
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. 
