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?

$\dfrac{47}{256}$

$\dfrac{3}{16}$

$\dfrac{49}{256}$

$\dfrac{25}{128}$

$\dfrac{51}{256}$

Answer: A

Concepts:arrangements with restrictions basic probability casework

Difficulty rating: 1970

Solution:

Count the possible sets of people who stand. For $0$ and $1$ people standing, there are $1$ and $8$ possibilities.

For $2$ people standing, choose any pair and subtract the $8$ adjacent pairs: $\binom82-8=20$ .

For $3$ people standing, first choose one standing person. Among the remaining five non-neighbor seats, $10$ pairs are possible, but $4$ of those pairs are adjacent, leaving $6$ . This counts each final set three times, so there are $8\cdot6/3=16$ possibilities.

For $4$ people standing, the only possibilities are the two alternating sets. Thus the number of favorable coin-flip outcomes is $1+8+20+16+2=47$ . Since all $2^8=256$ outcomes are equally likely, the probability is $\frac{47}{256}$ .

Thus, A is the correct answer.

2015 AMC 10A Problem 22

Answer: A

Difficulty rating: 1970

Solution:

Problem 22 in Other Years