Difference between revisions of "2002 AMC 8 Problems/Problem 21"
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===Solution=== | ===Solution=== | ||
| − | Case 1: There are two heads, two tails. The number of ways to choose which two tosses are heads is <math> | + | Case 1: There are two heads, two tails. The number of ways to choose which two tosses are heads is <math>\binom{4}{2} = 6</math>, and the other two must be tails. |
| − | Case 2: There are three heads, one tail. There are <math> | + | 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 3: There are four heads, no tails. This can only happen <math>1</math> way. | Case 3: There are four heads, no tails. This can only happen <math>1</math> way. | ||
Revision as of 22:56, 11 November 2018
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. The number of ways to choose which two tosses are heads is 
, 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 
.
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. 
