A special dice is so constructed that the probabilities of throwing 1,2,3,4,5 and 6 are $ (1-mathrm{k}) 6,(1+2 mathrm{k}) / 6,(1-mathrm{k}) / 6,(1+mathrm{k}) / 6,(1-2 mathrm{k}) / 6 text { and } quad(1+mathrm{k}) $ respectively. If two such dice are thrown and the probability of getting a sum equal to 9 lies between $1 / 9$ and $2 / 9$. Then. the number of integral solutions of k is

A special dice is so constructed that the probabilities of throwing 1,2,3,4,5 and 6 are <img alt="\mathrm{(1-k) 6,(1+2 k) / 6, (1-k) / 6,(1+k) / 6, (1-2 k) / 6 \: and \quad(1+k) / 6}" src="https://

A special dice is so constructed that the probabilities of throwing 1,2,3,4,5 and 6 are $\mathrm{(1-k) 6,(1+2 k) / 6, (1-k) / 6,(1+k) / 6, (1-2 k) / 6 \: and \quad(1+k) / 6}$ respectively. If two such dice are thrown and the probability of getting a sum equal to 9 lies between $\mathrm{1 / 9\: and\: 2 / 9}$ . Then. the number of integral solutions of $\mathrm{k}$ is
Option: 1
$2$

Option: 2
$3$

Option: 3
$1$

Option: 4
$0$

Answers (1)

Let $\mathrm{E_1, E_2, E_3, E_4, E_5 \: and\: E_6}$ be the events of occurrence of 1,2,3,4,5 and 6 on the dice respectively, and let $\mathrm{E}$ be the event of getting a sum of numbers equal to 9 .

$\mathrm{\therefore P\left(E_1\right)=\frac{1-k}{6} ; P\left(E_2\right)=\frac{1+2 k}{6} ; P\left(E_3\right)=\frac{1-k}{6} ; }$

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

and $\mathrm{ \frac{1}{9} \leq P(E) \leq \frac{2}{9}}$

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

Hence, $\mathrm{P(E)=P\left(E_3 E_6\right)+P\left(E_6 E_3\right)+P\left(E_4 E_5\right)+P\left(E_5 E_4\right)}$

$\mathrm{=P\left(E_3\right) P\left(E_6\right)+P\left(E_6\right) P\left(E_3\right)+P\left(E_4\right) P\left(E_5\right)+P\left(E_5\right) P\left(E_4\right)}$