2002 AMC 8 Problems/Problem 21: Difference between revisions
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<math> \text{(A)}\ \frac{5}{16}\qquad\text{(B)}\ \frac{3}{8}\qquad\text{(C)}\ \frac{1}{2}\qquad\text{(D)}\ \frac{5}{8}\qquad\text{(E)}\ \frac{11}{16} </math> | <math> \text{(A)}\ \frac{5}{16}\qquad\text{(B)}\ \frac{3}{8}\qquad\text{(C)}\ \frac{1}{2}\qquad\text{(D)}\ \frac{5}{8}\qquad\text{(E)}\ \frac{11}{16} </math> | ||
==Solution== | ==Solution 1== | ||
Case 1: There are two heads and two tails. There are <math>\binom{4}{2} = 6</math> ways to choose which two tosses are heads, and the other two must be tails. | Case 1: There are two heads and two tails. There are <math>\binom{4}{2} = 6</math> ways to choose which two tosses are heads, and the other two must be tails. | ||
Revision as of 13:49, 21 December 2023
Problem
Harold tosses a coin four times. The probability that he gets at least as many heads as tails is
Solution 1
Case 1: There are two heads and two tails. There are 
ways to choose which two tosses are heads, and the other two must be tails.
Case 2: There are three heads, one tail. There are 
ways to choose which of the four tosses is a tail.
Case 3: There are four heads and no tails. This can only happen 
way.
There are a total of 
possible configurations, giving a probability of 
.
Solution 2 (fastest)
We want the probability of at least two heads out of 
. We can do this a faster way by noticing that the probabilities are symmetric around two heads.
Define 
as the probability of getting 
heads on rolls. Now our desired probability is 
.
We can easily calculate 
because there are ways to get 
heads and tails, and there are total ways to flip these coins, giving 
, and plugging this in gives us 
.
~chrisdiamond10
Solution 3
You can use complimentary counting to solve this problem. Because we are trying to figure out the probability that Harold gets at least as many heads as tails, when using complimentary counting, we want to find the probability that there are MORE tails than heads. That are the cases when there are 
tails and head, and all tails. For the first case ( tails and head), there are cases that it works.
-
-
-
-
For the second case (all tails), there is case that works. Which is, of course 
.
So in total, there is 
cases that there are more tails then heads out of possibilities. Since we are doing complimentary counting, we need to subtract 
from one. Hence the answer is .
Video Solution
https://www.youtube.com/watch?v=4vLTPszBLeg ~David
See Also
| 2002 AMC 8 (Problems • Answer Key • Resources) | ||
| Preceded by Problem 20 |
Followed by Problem 22 | |
| 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: Unable to save thumbnail to destination
