Question

Eight people are sitting around a circular table, each holding a fair coin. All eight people flip their coins and those who flip heads stand while those who flip tails remain seated. What is the probability that no two adjacent people will stand?

$\dfrac{47}{256}$

$\dfrac{3}{16}$

$\dfrac{49}{256}$

$\dfrac{25}{128}$

$\dfrac{51}{256}$

Solution:

There are $2^8 = 256$ equally likely outcomes. Count the arrangements of standers (heads) with no two adjacent around the circle of $8$ seats, grouped by how many people stand.

The number of ways to choose $k$ non-adjacent seats from a circle of $n$ is $\dfrac{n}{n-k}\dbinom{n-k}{k}.$ For $n = 8$ this gives $1,\; 8,\; 20,\; 16,\; 2$ for $k = 0,1,2,3,4,$ and more than $4$ standers is impossible without an adjacency.

The total is $1 + 8 + 20 + 16 + 2 = 47,$ so the probability is $\dfrac{47}{256}.$

Thus, the correct answer is A.

2015 AMC 12A Problem 17

Solution:

Problem 17 in Other Years