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Difference between revisions of "2004 AMC 10A Problems/Problem 10"

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Coin <math>A</math> is flipped three times and coin <math>B</math> is flipped four times.  What is the probability that the number of heads obtained from flipping the two fair coins is the same?
 
Coin <math>A</math> is flipped three times and coin <math>B</math> is flipped four times.  What is the probability that the number of heads obtained from flipping the two fair coins is the same?
  
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<math> \mathrm{(A) \ } \frac{29}{128} \qquad \mathrm{(B) \ } \frac{23}{128} \qquad \mathrm{(C) \ } \frac14 \qquad \mathrm{(D) \ } \frac{35}{128} \qquad \mathrm{(E) \ } \frac12  </math>
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<math> \mathrm{(A) \ } \frac{19}{128} \qquad \mathrm{(B) \ } \frac{23}{128} \qquad \mathrm{(C) \ } \frac14 \qquad \mathrm{(D) \ } \frac{35}{128} \qquad \mathrm{(E) \ } \frac12  </math>
  
 
==Solution==
 
==Solution==
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<math>\frac{1+12+18+4}{128}=\frac{35}{128}\Rightarrow\boxed{\mathrm{(D)}\ \frac{35}{128}}</math>
 
<math>\frac{1+12+18+4}{128}=\frac{35}{128}\Rightarrow\boxed{\mathrm{(D)}\ \frac{35}{128}}</math>
  
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== Video Solution ==
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== Video Solution by OmegaLearn ==
 
https://youtu.be/IRyWOZQMTV8?t=1518
 
https://youtu.be/IRyWOZQMTV8?t=1518
  

Latest revision as of 20:49, 15 April 2024

Problem

Coin $A$ is flipped three times and coin $B$ is flipped four times. What is the probability that the number of heads obtained from flipping the two fair coins is the same?

$\mathrm{(A) \ } \frac{19}{128} \qquad \mathrm{(B) \ } \frac{23}{128} \qquad \mathrm{(C) \ } \frac14 \qquad \mathrm{(D) \ } \frac{35}{128} \qquad \mathrm{(E) \ } \frac12$

Solution

There are $4$ ways that the same number of heads will be obtained; $0$, $1$, $2$, or $3$ heads.

The probability of both getting $0$ heads is $\left(\frac12\right)^3{3\choose0}\left(\frac12\right)^4{4\choose0}=\frac1{128}$

The probability of both getting $1$ head is $\left(\frac12\right)^3{3\choose1}\left(\frac12\right)^4{4\choose1}=\frac{12}{128}$

The probability of both getting $2$ heads is $\left(\frac12\right)^3{3\choose2}\left(\frac12\right)^4{4\choose2}=\frac{18}{128}$

The probability of both getting $3$ heads is $\left(\frac12\right)^3{3\choose3}\left(\frac12\right)^4{4\choose3}=\frac{4}{128}$

Therefore, the probabiliy of flipping the same number of heads is: $\frac{1+12+18+4}{128}=\frac{35}{128}\Rightarrow\boxed{\mathrm{(D)}\ \frac{35}{128}}$

Video Solution by OmegaLearn

https://youtu.be/IRyWOZQMTV8?t=1518

~ pi_is_3.14

Video Solution

https://youtu.be/bq34-DyUpEU

Education, the Study of Everything

See also

2004 AMC 10A (Problems • Answer Key • Resources)
Preceded by
Problem 9 		Followed by
Problem 11
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All AMC 10 Problems and Solutions

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