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? | ||
| − | <math> \mathrm{(A) \ } \frac{ | + | <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== | ||
| − | 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> |
| + | |||
| + | == 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 == | == See also == | ||
| Line 25: | Line 35: | ||
[[Category:Introductory Combinatorics Problems]] | [[Category:Introductory Combinatorics Problems]] | ||
| + | {{MAA Notice}} | ||
Latest revision as of 20:49, 15 April 2024
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:
Video Solution by OmegaLearn
https://youtu.be/IRyWOZQMTV8?t=1518
~ pi_is_3.14
Video Solution
Education, the Study of Everything
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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
