Difference between revisions of "2021 Fall AMC 10A Problems/Problem 13"
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==Solution 1== | ==Solution 1== | ||
| − | Note that for this restriction to be true, there must be <math>3</math> balls of each color. There are a total of <math>2^6 = 64</math> ways to color the balls, and there are <math>{6 \choose 3} = 20</math> ways for three balls chosen to be painted white. Thus, the answer is <math>\frac{20}{64} = \boxed{\textbf{(D) } \frac{5}{16}}</math>. | + | Note that for this restriction to be true, there must be <math>3</math> balls of each color. There are a total of <math>2^6 = 64</math> ways to color the balls, and there are <math>{6 \choose 3} = 20</math> ways for three balls chosen to be painted white. Thus, the answer is <math>\frac{20}{64} = \boxed{\textbf{(D) } \frac{5}{16}}</math>. |
-Aidensharp | -Aidensharp | ||
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~Steven Chen (www.professorchenedu.com) | ~Steven Chen (www.professorchenedu.com) | ||
| − | ==Video Solution | + | ==Video Solution == |
https://youtu.be/95aRRJ3jMqo | https://youtu.be/95aRRJ3jMqo | ||
~Education, the Study of Everything | ~Education, the Study of Everything | ||
| − | |||
==Video Solution == | ==Video Solution == | ||
Latest revision as of 15:38, 10 April 2024
Contents
Problem
Each of 
balls is randomly and independently painted either black or white with equal probability. What is the probability that every ball is different in color from more than half of the other 
balls?
Solution 1
Note that for this restriction to be true, there must be 
balls of each color. There are a total of 
ways to color the balls, and there are 
ways for three balls chosen to be painted white. Thus, the answer is 
.
-Aidensharp
Solution 2
For this restriction to be upheld, there must be three black and three white balls. One such way for this to occur is the arrangement 
, which has a 
probability of occuring. However, there are 
ways to arrange the three black and three white balls, meaning that the answer is, 
.
~Benedict T (countmath1)
Solution 3
To get every ball different in color from more than half of the other 5 balls, we must have 3 black balls and 3 white balls.
Following from the binomial theorem, this happens with probability
![\[ \binom{6}{3} \left( \frac{1}{2} \right)^3 \left( 1 - \frac{1}{2} \right)^{6-3} = \frac{5}{16} . \]](http://latex.artofproblemsolving.com/d/d/1/dd15f1f8a9850aa66568b488c77a4f76f92cbf57.png)
Therefore, the answer is 
.
~Steven Chen (www.professorchenedu.com)
Video Solution
~Education, the Study of Everything
Video Solution
https://youtu.be/zq3UPu4nwsE?t=707
~IceMatrix
Video Solution by WhyMath
~savannahsolver
See Also
| 2021 Fall AMC 10A (Problems • Answer Key • Resources) | ||
| Preceded by Problem 12 |
Followed by Problem 14 | |
| 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. 
