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

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==Solution 2==
 
==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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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>.
 +
 
 +
~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
 +
<cmath>
 +
\[
 +
\binom{6}{3} \left( \frac{1}{2} \right)^3 \left( 1 - \frac{1}{2} \right)^{6-3}
 +
= \frac{5}{16} .
 +
\]
 +
</cmath>
 +
 
 +
Therefore, the answer is <math>\boxed{\textbf{(D) }\frac{5}{16}}</math>.
 +
 
 +
~Steven Chen (www.professorchenedu.com)
 +
==Video Solution ==
 +
https://youtu.be/95aRRJ3jMqo
 +
 
 +
~Education, the Study of Everything
 +
 
 +
==Video Solution ==
 +
https://youtu.be/zq3UPu4nwsE?t=707
 +
 
 +
~IceMatrix
 +
 
 +
==Video Solution by WhyMath==
 +
https://youtu.be/5m8zYiOagCY
 +
 
 +
~savannahsolver
  
−
~~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}}

Latest revision as of 15:38, 10 April 2024

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 1

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

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 $BBBWWW$, which has a $\frac{1}{2^6}$ probability of occuring. However, there are $\frac{6!}{3!\cdot{3!}}$ ways to arrange the three black and three white balls, meaning that the answer is, $\frac{1}{64}\cdot{\frac{6!}{3!\cdot{3!}}} = \boxed{\textbf{(D) } \frac{5}{16}}$.

~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 \[ \binom{6}{3} \left( \frac{1}{2} \right)^3 \left( 1 - \frac{1}{2} \right)^{6-3} = \frac{5}{16} . \]

Therefore, the answer is $\boxed{\textbf{(D) }\frac{5}{16}}$.

~Steven Chen (www.professorchenedu.com)

Video Solution

https://youtu.be/95aRRJ3jMqo

~Education, the Study of Everything

Video Solution

https://youtu.be/zq3UPu4nwsE?t=707

~IceMatrix

Video Solution by WhyMath

https://youtu.be/5m8zYiOagCY

~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

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