Difference between revisions of "2004 AMC 10A Problems/Problem 10"
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==Solution== | ==Solution== | ||
| − | There are 4 ways that the same number of heads will be obtained; 0, 1, 2, or 3 heads. | + | 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 
is flipped three times and coin 
is flipped four times. What is the probability that the number of heads obtained from flipping the two fair coins is the same?
Solution
There are 
ways that the same number of heads will be obtained; 
, 
, 
, or 
heads.
The probability of both getting heads is 
The probability of both getting head is
The probability of both getting heads is
The probability of both getting heads is
Therefore, the probabiliy of flipping the same number of heads is:
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 | ||
These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. 
