Difference between revisions of "2002 AMC 8 Problems/Problem 21"
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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>. | ||
| − | ==Solution 2== | + | ==Solution 2 (fastest)== |
We want the probability of at least two heads out of <math>4</math>. We can do this a faster way by noticing that the probabilities are symmetric around two heads. | We want the probability of at least two heads out of <math>4</math>. We can do this a faster way by noticing that the probabilities are symmetric around two heads. | ||
Define <math>P(n)</math> as the probability of getting <math>n</math> heads on <math>4</math> rolls. Now our desired probability is <math>\frac{1-P(2)}{2} +P(2)</math> | Define <math>P(n)</math> as the probability of getting <math>n</math> heads on <math>4</math> rolls. Now our desired probability is <math>\frac{1-P(2)}{2} +P(2)</math> | ||
We can easily calculate <math>P(2)</math> similarly as above solution, so plugging this in gives us <math>\boxed{\text{(E)}\ \frac{11}{16}}</math>. | We can easily calculate <math>P(2)</math> similarly as above solution, so plugging this in gives us <math>\boxed{\text{(E)}\ \frac{11}{16}}</math>. | ||
~chrisdiamond10 | ~chrisdiamond10 | ||
| + | |||
==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}} | ||
Revision as of 13:08, 29 December 2020
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 
.
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 
similarly as above solution, so 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. 
