The probability that no two adjacent people will stand is 47/256. This is a problem that represents a combination.
A combination is a grouping of items where order does not matter. The formula of combination is
[tex]_{n}C_{r} = \frac{n!} {r! (n-r)!}[/tex]
Where
Given the case eight people sit around a circular table, flip coins, and those who flip heads will stand while those who flip tails will remain seated.
What is the probability that no two adjacent people will stand?
First, we count the total standing arrangements. They will be 2⁸ = 256 arrangements. n(S) = 256.
The number of arrangements according to the number of people standing.
Case 1: 0 people standing = 1 arrangement.
Case 2: 1 people standing = [tex]_{8}C_{1}[/tex] = [tex]\frac{8!} {1! 7!}[/tex] = 8 arrangements.
Case3: 2 people standing = [tex]_{8}C_{2}[/tex] = [tex]\frac{8!} {2! 6!}[/tex] = [tex]\frac{7 \times 8} {2}[/tex] = 28 arrangements, but no two people are next to each other. There are 8 arrangements that two people in this case are standing next to each other. So, it will be 28 - 8 = 20 arrangements.
Case 4: 3 people standing = [tex]_{8}C_{3}[/tex] = [tex]\frac{8!} {3! 5!}[/tex] = [tex]\frac{6 \times 7 \times 8} {6}[/tex] = 56 arrangements. In this case, three people standing are next to each other = [tex]_{8}C_{1}[/tex] = 8 arrangements. Two people standing are next to each other and the third person is not = [tex]_{8}C_{1} \times _{4}C_{1}[/tex] = 8 × 4 = 32. So, it will be 56 - 8 - 32 = 16 arrangements.
Case 5: 4 people standing but no two adjacent people = 2 arrangements.
The number of arrangements that no two adjacent people will stand.
n(A) = 1 + 8 + 20 + 16 + 2 = 47
The probability that no two adjacent people will stand is
P(A) = n(A)/n(S)
P(A) = 47/256
Hence, the probability that no two adjacent people will stand is 47/256.
Learn more about finding probability with combination here:
brainly.com/question/3901018
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