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

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<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\mathrm{(D)}</math>
  
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==See Also==
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== See also ==
 
{{AMC10 box|year=2004|ab=A|num-b=9|num-a=11}}
 
{{AMC10 box|year=2004|ab=A|num-b=9|num-a=11}}
  
 
[[Category:Introductory Combinatorics Problems]]
 
[[Category:Introductory Combinatorics Problems]]

Revision as of 01:39, 11 September 2007

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

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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