Find the probability that no boy sits between two girls. 
Example. Five boys and three girls are seated at random in a row. Find the probability that no boy sits between two girls.
Solution.: $\quad n(s) = 8!$
  $n$(E) = The number of arrangement of $5$ boys and $3$ girls when $3$ girls are consecutive = $6!\times 3!$
  $$\therefore\quad \text{Required probability } = \frac{6!\times 3!}{8!} = \frac{3}{28}.$$

Source.
I think the given answer is incorrect as their might be a situation like GGBBBBBG where no boy is between two girls and it is not counted in favourable ways. According to me, required ways can be counted using total - non favourable (N.F.) ways. I calculated N. F. using the logic that at least one GBG group should not be counted. 
N.F.= $5C1 * 3C2 * 2 * 6!$ 
Total = $8!$
My ans = $13/28$
Please tell if I'm correct. 
 A: If you arrange as GGBBBBBG, there will be $5$ boys sitting between $2$ girls, but the question demands "no boys between 2 girls"
You are not puzzled by the sum, it's the language of the question which is driving you to wrong way.  Don't be upset. I also get puzzled. Recently, in an exam I didn't attend an easy probability sum for its puzzling words and did another sum which was a page long, lol.
You can do it as mentioned below:
First make all 5 boys seated, which can be done in $5!$ ways. Now you can see 6 gaps created.......(1)
$$\underline{} B \underline{} B \underline{} B \underline{} B \underline{} B \underline{}$$
Now you take group of girls together. Consider GGG.  You can make this group seat in any one of the $6$ gaps, so there are $6$ ways and you can arrange the $3$ girls among themselves in $3!$ ways , so here total no. of ways $6 \cdot 3!$
........................(2)
From (1) and (2) you can say no. of ways $= 5! \cdot 6 \cdot 3!$
If there's no restriction, then total no. of ways $=8!$.
Divide to get the probability
$$\frac{5! \cdot 6 \cdot 3!}{8!} = \frac{6! \cdot 3!}{8!} = \frac{6}{8 \cdot 7} = \frac{3}{28}$$
!Answer!
