Difference between revisions of "2002 AMC 8 Problems/Problem 21"
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==Solution== | ==Solution== | ||
| − | Case 1: There are two heads, two tails. | + | Case 1: There are two heads, two tails. There are <math>\binom{4}{2} = 6</math> to choose which two tosses are heads, and the other two must be tails. |
Case 2: There are three heads, one tail. There are <math>\binom{4}{1} = 4</math> ways to choose which of the four tosses is a tail. | Case 2: There are three heads, one tail. There are <math>\binom{4}{1} = 4</math> ways to choose which of the four tosses is a tail. | ||
Revision as of 12:20, 5 July 2021
Problem
Harold tosses a coin four times. The probability that he gets at least as many heads as tails is
Solution
Case 1: There are two heads, two tails. There are 
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, 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
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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
