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Difference between revisions of "2002 AMC 8 Problems/Problem 21"

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<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>
  
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==Solution==
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==Solution 1==
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Case 1: There are two heads, 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.
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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>\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.
  
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Case 3: There are four heads, no tails. This can only happen <math>1</math> way.
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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>.
  
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==Solution 2 (fastest)==
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==Video Solution==
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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.
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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>.
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https://www.youtube.com/watch?v=4vLTPszBLeg  ~David
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We can easily calculate <math>P(2)</math> because there are <math>\binom{4}{2} = 6</math> ways to get <math>2</math> heads and <math>2</math> tails, and there are <math>2^4=16</math> total ways to flip these coins, giving <math>P(2)=\frac{6}{16}=\frac{3}{8}</math>, and plugging this in gives us <math>\boxed{\text{(E)}\ \frac{11}{16}}</math>.
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~chrisdiamond10
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==Video Solution by WhyMath==
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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

$\text{(A)}\ \frac{5}{16}\qquad\text{(B)}\ \frac{3}{8}\qquad\text{(C)}\ \frac{1}{2}\qquad\text{(D)}\ \frac{5}{8}\qquad\text{(E)}\ \frac{11}{16}$

Solution 1

Case 1: There are two heads and two tails. There are $\binom{4}{2} = 6$ 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 $\binom{4}{1} = 4$ ways to choose which of the four tosses is a tail.

Case 3: There are four heads and no tails. This can only happen $1$ way.

There are a total of $2^4=16$ possible configurations, giving a probability of $\frac{6+4+1}{16} = \boxed{\text{(E)}\ \frac{11}{16}}$.

Video Solution

https://www.youtube.com/watch?v=4vLTPszBLeg ~David

Video Solution by WhyMath

https://youtu.be/q-G5nLcc4kE

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

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