Difference between revisions of "2002 AMC 8 Problems/Problem 21"
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==Problem== | ==Problem== | ||
| − | + | Harold tosses a coin four times. The probability that he gets at least as many heads as tails is | |
| − | Harold tosses | ||
<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 | + | 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 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 | + | Case 3: There are four heads and no tails. This can only happen <math>1</math> way. |
There are a total of <math>2^4=16</math> possible configurations, giving a probability of <math>\frac{6+4+1}{16} = \boxed{\text{(E)}\ \frac{11}{16}}</math>. | There are a total of <math>2^4=16</math> possible configurations, giving a probability of <math>\frac{6+4+1}{16} = \boxed{\text{(E)}\ \frac{11}{16}}</math>. | ||
| + | |||
| + | ==Video Solution== | ||
| + | |||
| + | https://www.youtube.com/watch?v=4vLTPszBLeg ~David | ||
| + | |||
| + | ==Video Solution by WhyMath== | ||
| + | https://youtu.be/q-G5nLcc4kE | ||
==See Also== | ==See Also== | ||
{{AMC8 box|year=2002|num-b=20|num-a=22}} | {{AMC8 box|year=2002|num-b=20|num-a=22}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Latest revision as of 13:33, 29 October 2024
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 
.
Video Solution
https://www.youtube.com/watch?v=4vLTPszBLeg ~David
Video Solution by WhyMath
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. 
