Total number of arrangements $=15!$

Boys in these four gaps be$2x+1, 2y+1, 2z+1$ and $2t+1$, then
$2x+1+2y+1+2z+1+2t+1=10$
$\Rightarrow x+y+z+t=3$    
Where $x, y, z$, t are integers and $0\le x\le 3,$
$0\le y\le 3, 0\le z\le 3, 0\le t\le 3$
$\therefore$The number of ways of selecting positions for boys
$=$ coefficient of$x^3$ in ${\left(1+x+x^2+x^3\right)}^4$
$=$coefficient of $x^3$ in ${\left(\frac{1-x^4}{1-x}\right)}^4$
$=$ coefficient of $x^{3 }$in ${\left(1-x^4\right)}^4(1-x{)}^{-4}={}^{6}C_3=20$
$\therefore$ Number of arrangements of boys and girls with given condition $= 20 \times 20! \times 5!$

$\therefore$ Required probability $\mathrm{=}\frac{20\times 10\mathrm{!}\times 5\mathrm{!}}{15\mathrm{!}}=\frac{20}{3003}$