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Difference between revisions of "2021 Fall AMC 10A Problems/Problem 13"

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(The two solutions are very similar with 6C3. So, I deleted one solution and retained credit to both authors.)
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<math>\textbf{(A) } \frac{1}{64}\qquad\textbf{(B) } \frac{1}{6}\qquad\textbf{(C) } \frac{1}{4}\qquad\textbf{(D) } \frac{5}{16}\qquad\textbf{(E) }\frac{1}{2}</math>
 
<math>\textbf{(A) } \frac{1}{64}\qquad\textbf{(B) } \frac{1}{6}\qquad\textbf{(C) } \frac{1}{4}\qquad\textbf{(D) } \frac{5}{16}\qquad\textbf{(E) }\frac{1}{2}</math>
  
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==Solution 1==
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==Solution==
 
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>.
  
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-Aidensharp
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~Aidensharp ~countmath1
  
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==Solution 2==
 
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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 <math>BBBWWW</math>, which has a <math>\frac{1}{2^6}</math> probability of occuring. However, there are <math>\frac{6!}{3!\cdot{3!}}</math> ways to arrange the three black and three white balls, meaning that the answer is, <math>\frac{1}{64}\cdot{\frac{6!}{3!\cdot{3!}}} = \boxed{\textbf{(D)} \frac{5}{16}}</math>
 
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~~countmath1~~
 
 
==See Also==
 
==See Also==
 
{{AMC10 box|year=2021 Fall|ab=A|num-b=12|num-a=14}}
 
{{AMC10 box|year=2021 Fall|ab=A|num-b=12|num-a=14}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 19:24, 24 November 2021

Problem

Each of $6$ 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 $5$ balls?

$\textbf{(A) } \frac{1}{64}\qquad\textbf{(B) } \frac{1}{6}\qquad\textbf{(C) } \frac{1}{4}\qquad\textbf{(D) } \frac{5}{16}\qquad\textbf{(E) }\frac{1}{2}$

Solution

Note that for this restriction to be true, there must be $3$ balls of each color. There are a total of $2^6 = 64$ ways to color the balls, and there are ${6 \choose 3} = 20$ ways for three balls chosen to be painted white. Thus, the answer is $\frac{20}{64} = \boxed{\textbf{(D) } \frac{5}{16}}$.

~Aidensharp ~countmath1

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

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