A fair die is tossed repeatedly until a six is obtained. Let X denote the number of tosses required and let…

$P\left(\text{getting} 6\right)=\frac{1}{6}, P\left(\text{not getting} 6\right)=\frac{5}{6}$

$\Rightarrow \mathrm{a}=\mathrm{P}(\mathrm{X}=3)$

$\Rightarrow \mathrm{a}=\frac{5}{6}\times \frac{5}{6}\times \frac{1}{6}$

$\Rightarrow \mathrm{a}=\frac{25}{216} ...\left(i\right)$

$\Rightarrow \mathrm{b}=\mathrm{P}(\mathrm{X}\ge 3)$

$\Rightarrow \mathrm{b}=\frac{5}{6}\times \frac{5}{6}\times \frac{1}{6}+{\left(\frac{5}{6}\right)}^3·\frac{1}{6}+{\left(\frac{5}{6}\right)}^4·\frac{1}{6}+...$

We know that, ${\mathrm{S}}_{\infty }=\mathrm{a}+\mathrm{ar}+{\mathrm{ar}}^2+...\infty =\frac{\mathrm{a}}{1-\mathrm{r}}$

$\Rightarrow \mathrm{b}=\frac{\frac{25}{216}}{1-\frac{5}{6}}$

$\Rightarrow \mathrm{b}=\frac{25}{216}\times \frac{6}{1}$

$\Rightarrow \mathrm{b}=\frac{25}{36} ...\left(ii\right)$

Now, $\mathrm{P}(\mathrm{X}\ge 6)={\left(\frac{5}{6}\right)}^5·\frac{1}{6}+{\left(\frac{5}{6}\right)}^6·\frac{1}{6}+...$

$\Rightarrow \mathrm{P}(\mathrm{X}\ge 6)=\frac{{\left(\frac{5}{6}\right)}^5·\frac{1}{6}}{1-\frac{5}{6}}$

$\Rightarrow \mathrm{P}(\mathrm{X}\ge 6)={\left(\frac{5}{6}\right)}^5$

$\Rightarrow \mathrm{c}=\mathrm{P}\left(\frac{\mathrm{X}\ge 6}{\mathrm{X}>3}\right)=\frac{\mathrm{P}\left(\left(\mathrm{X}\ge 6\right)\cap \left(\mathrm{X}>3\right)\right)}{\mathrm{P}\left(\mathrm{X}>3\right)}=\frac{\mathrm{P}\left(\mathrm{X}\ge 6\right)}{\mathrm{P}\left(\mathrm{X}>3\right)}=\frac{{\left(\frac{5}{6}\right)}^5}{{\left(\frac{5}{6}\right)}^3}$

$\Rightarrow \mathrm{c}=\frac{25}{36} ...\left(iii\right)$

Using $\left(i\right), \left(ii\right) \text{and} \left(iii\right)$,

$\Rightarrow \frac{\mathrm{b}+\mathrm{c}}{\mathrm{a}}=\frac{{\displaystyle \frac{25}{36}+\frac{25}{36}}}{{\displaystyle \frac{25}{216}}}$

$\Rightarrow \frac{\mathrm{b}+\mathrm{c}}{\mathrm{a}}=\frac{{\displaystyle \frac{50}{36}}}{{\displaystyle \frac{25}{216}}}$

$\Rightarrow \frac{\mathrm{b}+\mathrm{c}}{\mathrm{a}}=12$