Difference between revisions of "2013 AMC 8 Problems/Problem 8"
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A fair coin is tossed 3 times. What is the probability of at least two consecutive heads? | A fair coin is tossed 3 times. What is the probability of at least two consecutive heads? | ||
| − | <math>\textbf{(A)}\ | + | <math>\textbf{(A)}\frac{1}{8} \qquad \textbf{(B)}\frac{1}{4} \qquad \textbf{(C)}\frac{3}{8} \qquad \textbf{(D)}\frac{1}{2} \qquad \textbf{(E)}\frac{3}{4}</math> |
| − | ==Solution== | + | ==Solution 1== |
| − | + | There are <math>2^3 = 8</math> 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 <math>\boxed{\textbf{(C)}\frac{3}{8}}</math>. | ||
| − | The | + | ==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 <math>\frac{1}{8}</math>, <math>\frac{1}{4}</math>, and <math>\frac{1}{4}</math>, respectively. So the rest is exactly the probability of flipping at least two consecutive heads: <math>1-\frac{1}{8}-\frac{1}{4}-\frac{1}{4}=\frac{3}{8}</math>. It is the answer <math>\boxed{\textbf{(C)}\frac{3}{8}}</math>. | ||
| − | + | ~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 <math>\boxed{\textbf{(C)}\frac{3}{8}}</math>. | ||
| + | |||
| + | ~andliu766 | ||
| + | |||
| + | ==Video Solution== | ||
| + | https://youtu.be/2lynqd2bRZY ~savannahsolver | ||
| + | https://youtu.be/6xNkyDgIhEE?t=44 | ||
| + | |||
| + | ~ pi_is_3.14 | ||
==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}} | ||
Latest revision as of 12:59, 5 June 2024
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. 
