Difference between revisions of "2002 AMC 8 Problems/Problem 21"
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==Problem== | ==Problem== | ||
| − | Harold tosses | + | Harold tosses me four times. The probability that he gets at least as many heads as tails is |
<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> | ||
Revision as of 20:24, 8 November 2016
Problem
Harold tosses me 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. 
