If 5 boys and 3 girls sit randomly around a circular table, the probability that there will be at least one boy sitting between any two girls is

Total no of ways to arrange 8 people around a circular table is 7! ways.

let No of boys between G1 and G2 be X , G2 and G3 be Y , G3 and G1 be Z . Then as total boys are 5 , it implies X+Y+Z = 5.
But given at least one boy between any two girls i.e $$X\ \ge1,\ \ Y\ge1\ ,\ \ Z\ge1$$: 
(X,Y, Z) = (1,2,2) or (2,1,2) or (2,2,1) ~ 3 ways.
(X,Y, Z) = (1,3,1) or (3,1,1) or (1,1,3) ~ 3 ways.

No of ways of arranging 3 girls (G1, G2, G3) on a circular table is 2! ways = 2 ways . After arranging girls , we need to now place 5 boys in between them = (6 triplets) x 5! ways = 6! ways.

Total ways of arranging 8 members such that there is at least one boy b/w any two girls = 2x6! ways .

Therefore , probability is $$\dfrac{\ 2\times\ 6!}{7!}$$ = $$\dfrac{2}{7}$$