2013 AMC 8 Problems/Problem 8
Problem
A fair coin is tossed 3 times. What is the probability of at least two consecutive heads?
Solution 1
There are ways to flip the coins, in order. There are two ways to get exactly two consecutive heads: HHT and THH. There is only one way to get three consecutive heads: HHH. Therefore, the probability of flipping at least two consecutive heads is .
Solution 2
Let's use complementary counting. To start with, the unfavorable outcomes (in this case, not getting 2 consecutive heads) are: TTT, HTH, and THT. The probability of these three outcomes is , , and , respectively. So the rest is exactly the probability of flipping at least two consecutive heads: . It is the answer .
~LarryFlora
Solution 3
We can list out all the ways to flip a coin three times: HHH,HHT,HTH,THH,HTT,THT,TTH,TTT. Out of them, only HHH,HHT,THH, have at least two consecutive heads. Since there are three ways to flip at least two consecutive heads, and eight total choices, the answer is .
~andliu766
Video Solution
https://youtu.be/2lynqd2bRZY ~savannahsolver https://youtu.be/6xNkyDgIhEE?t=44
~ pi_is_3.14
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.