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