Mathematics - Probability Question with Solution | TestHub

Let ${\mathrm{E}}_1,{\mathrm{E}}_2,{\mathrm{E}}_3{\mathrm{E}}_4,{\mathrm{E}}_5$ and ${\mathrm{E}}_6$ be the events of occurrence of $1,2,3,4,5$ and 6 on the dice respectively, and let E be the event

$$\begin{array}{rlr}\therefore & P\left(E_1\right)=\frac{1-k}{6};P\left(E_2\right)=\frac{1+2k}{6};P\left(E_3\right)=\frac{1-k}{6}& \end{array}$$

$$P\left(E_4\right)=\frac{1+k}{6};P\left(E_5\right)=\frac{1-2k}{6};P\left(E_6\right)=\frac{1+k}{6}and\frac{1}{9}\le P(E)\le \frac{2}{9}$$

Then, $\mathrm{E}\equiv \{(3,6),(6,3),(4,5),(5,4)\}$

Hence, $\mathrm{P}(\mathrm{E})=\mathrm{P}\left({\mathrm{E}}_3{\mathrm{E}}_6\right)+\mathrm{P}\left({\mathrm{E}}_6{\mathrm{E}}_3\right)+\mathrm{P}\left({\mathrm{E}}_4{\mathrm{E}}_5\right)+\mathrm{P}\left({\mathrm{E}}_5{\mathrm{E}}_4\right)$ $=\mathrm{P}\left({\mathrm{E}}_3\right)\mathrm{P}\left({\mathrm{E}}_6\right)+\mathrm{P}\left({\mathrm{E}}_6\right)\mathrm{P}\left({\mathrm{E}}_3\right)+\mathrm{P}\left({\mathrm{E}}_4\right)\mathrm{P}\left({\mathrm{E}}_5\right)+\mathrm{P}\left({\mathrm{E}}_5\right)\mathrm{P}\left({\mathrm{E}}_4\right)$ $=2\mathrm{P}\left({\mathrm{E}}_3\right)\mathrm{P}\left({\mathrm{E}}_6\right)+2\mathrm{P}\left({\mathrm{E}}_4\right)\mathrm{P}\left({\mathrm{E}}_5\right)$

{Since ${\mathrm{E}}_1,{\mathrm{E}}_2,{\mathrm{E}}_3,{\mathrm{E}}_4,{\mathrm{E}}_5$ and ${\mathrm{E}}_6$ are independent} $=2\left(\frac{1-\mathrm{k}}{6}\right)\left(\frac{1+\mathrm{k}}{6}\right)+2\left(\frac{1+\mathrm{k}}{6}\right)\left(\frac{1-2\mathrm{k}}{6}\right)$

$=\frac{1}{18}\left[2-\mathrm{k}-3{\mathrm{k}}^2\right]$

Since, $\frac{1}{9}\le \mathrm{P}(\mathrm{E})\le \frac{2}{9}$

$\therefore -\frac{1}{3}\le \mathrm{k}\le 0$

$\therefore$ Set of integral value of $\mathrm{k}=\{0\}$

$\therefore$ Number of integral solution of k is 1