A group of students comprising $3$ girls and $5$ boys went for a picnic. During a game they were arranged in a circle then the probability that each boy has one girl on at least one side is ______.
Step-by-Step Solution
Key Concept: Arrange boys first in a circle, then place girls in gaps between them to ensure each boy has at least one girl adjacent.
Total arrangements of 8 people in a circle is $(8-1)! = 7! = 5040$. For each boy to have at least one girl on his side, we need girls to be distributed such that no two boys are adjacent. With 3 girls and 5 boys, we arrange 5 boys in a circle: $(5-1)! = 4! = 24$ ways. Then 3 girls must be placed in the 5 gaps between consecutive boys such that each selected gap has exactly one girl. We choose 3 gaps from 5: $inom{5}{3} = 10$ ways, and arrange girls in them: $3! = 6$ ways. Favorable outcomes = $24 imes 10 imes 6 = 1440$. Probability = $ rac{1440}{5040} = rac{2}{14} = rac{1}{7} \approx 0.142857$.
Correct Answer: 0.142857