2021 Fall AMC 10A Problems/Problem 13


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 $\frac{20}{64} = \boxed{\textbf{(D) } \frac{5}{16}}$.


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)

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~Education, the Study of Everything

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See Also

2021 Fall AMC 10A (ProblemsAnswer KeyResources)
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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