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2002 AMC 8 Problems/Problem 21: Difference between revisions

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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>


==Solution==
==Solution 1==
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.
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.


Case 3: There are four heads, no tails. This can only happen <math>1</math> way.
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>.


==Solution 2 (fastest)==
==Video Solution==
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>.
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>.
~chrisdiamond10


==Solution 3==
https://www.youtube.com/watch?v=4vLTPszBLeg  ~David
You can use complimentary counting to solve this problem. Because we are trying to figure out the probability that Harold gets at least as many heads as tails, when using complimentary counting, we want to find the probability that there are MORE tails than heads. That are the cases when there are <math>3</math> tails and <math>1</math> head, and all tails. For the first case (<math>3</math> tails and <math>1</math> head), there are <math>4</math> cases that it works.


-<math>HTTT</math>
==Video Solution by WhyMath==
 
https://youtu.be/q-G5nLcc4kE
-<math>THTT</math>
 
-<math>TTHT</math>
 
-<math>TTTH</math>
For the second case (all tails), there is <math>1</math> case that works. Which is, of course <math>TTTT</math>.
 
So in total, there is <math>4+1=5</math> cases that there are more tails then heads out of <math>2^4=16</math> possibilities. Since we are doing complimentary counting, we need to subtract <math>5/16</math> from one. Hence the answer is <math>\boxed{\text{(E)}\ \frac{11}{16}}</math>.


==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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