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