A fair die is tossed repeatedly until a six is obtained. Let $X$ denote the number of tosses required and let $a=P(X=3), b=P(X \geqslant 3)$ and $c=P(X \geqslant 6 \mid X\gt 3)$. Then $\frac{b+c}{a}$ is equal to $\underline{\hspace{2cm}}$.