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Difference between revisions of "2004 AMC 10A Problems/Problem 10"
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==Solution==
 
==Solution==
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There are 4 ways that the same number of heads will be obtained; 0, 1, 2, or 3 heads.
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There are <math>4</math> ways that the same number of heads will be obtained; <math>0</math>, <math>1</math>, <math>2</math>, or <math>3</math> heads.
  
−
The probability of both getting 0 heads is <math>\left(\frac12\right)^3{3\choose0}\left(\frac12\right)^4{4\choose0}=\frac1{128}</math>.
+
The probability of both getting <math>0</math> heads is <math>\left(\frac12\right)^3{3\choose0}\left(\frac12\right)^4{4\choose0}=\frac1{128}</math>
  
−
The probability of both getting 1 head is
+
The probability of both getting <math>1</math> head is
 
<math>\left(\frac12\right)^3{3\choose1}\left(\frac12\right)^4{4\choose1}=\frac{12}{128}</math>
 
<math>\left(\frac12\right)^3{3\choose1}\left(\frac12\right)^4{4\choose1}=\frac{12}{128}</math>
  
−
The probability of both getting 2 heads is
+
The probability of both getting <math>2</math> heads is
 
<math>\left(\frac12\right)^3{3\choose2}\left(\frac12\right)^4{4\choose2}=\frac{18}{128}</math>
 
<math>\left(\frac12\right)^3{3\choose2}\left(\frac12\right)^4{4\choose2}=\frac{18}{128}</math>
  
−
The probability of both getting 3 heads is
+
The probability of both getting <math>3</math> heads is
 
<math>\left(\frac12\right)^3{3\choose3}\left(\frac12\right)^4{4\choose3}=\frac{4}{128}</math>
 
<math>\left(\frac12\right)^3{3\choose3}\left(\frac12\right)^4{4\choose3}=\frac{4}{128}</math>
  
 
Therefore, the probabiliy of flipping the same number of heads is:
 
Therefore, the probabiliy of flipping the same number of heads is:
−
<math>\frac{1+12+18+4}{128}=\frac{35}{128}\Rightarrow\mathrm{(D)}</math>
+
<math>\frac{1+12+18+4}{128}=\frac{35}{128}\Rightarrow\boxed{\mathrm{(D)}\ \frac{35}{128}}</math>
  
 
== See also ==
 
== See also ==

Revision as of 23:33, 20 July 2014

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?

$\mathrm{(A) \ } \frac{29}{128} \qquad \mathrm{(B) \ } \frac{23}{128} \qquad \mathrm{(C) \ } \frac14 \qquad \mathrm{(D) \ } \frac{35}{128} \qquad \mathrm{(E) \ } \frac12$

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: $\frac{1+12+18+4}{128}=\frac{35}{128}\Rightarrow\boxed{\mathrm{(D)}\ \frac{35}{128}}$

See also

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

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