2013 AMC 8 Problems/Problem 8: Difference between revisions
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==Solution 2== | ==Solution 2== | ||
Let's figure it out | Let's figure it out using complementary counting. | ||
First, there are <math>2^3 = 8</math> ways to flip the coins, in order. | First, there are <math>2^3 = 8</math> ways to flip the coins, in order. | ||
Revision as of 19:20, 20 December 2021
Problem
A fair coin is tossed 3 times. What is the probability of at least two consecutive heads?
Video Solution
https://youtu.be/6xNkyDgIhEE?t=44
Solution 1
First, there are 
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 
.
Solution 2
Let's figure it out using complementary counting.
First, there are 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 these three ways of flipping is 
, 
,and , respectively. So the rest is exactly the probability of flipping at least two consecutive heads: 
. It is the answer . ----LarryFlora
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 | ||
These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. Error creating thumbnail: File missing
