Question

A special dice is so constructed that the probabilities of throwing $1,2,3,4,5$ and $6$ are $\frac{1-k}{6},\frac{1+2k}{6},\frac{1-k}{6},\frac{1+k}{6},\frac{1-2k}{6}$ and $\frac{1+k}{6}$ respectively. If two such dice are thrown and the probability of getting a sum equal to $9$ lies between $\frac{1}{9}$ and $\frac{2}{9}$, find the set of integral values of $k$.

• A. Set of integral values of $k=0$
• B. Set of integral values of $k=1$
• C. Set of integral values of $k=0,1$
• D. Set of integral values of $k=\varphi$

Answer 1

The correct option is A Set of integral values of $k=0$

For a sum of $9$ the two dice may show : $(3,6),(4,5),(5,4)$ or $(6,3)$

$\Rightarrow$ the probability is $\left(\frac{1-k}{6}\right)\left(\frac{1+k}{6}\right)+\left(\frac{1+k}{6}\right)\left(\frac{1-2k}{6}\right)+\left(\frac{1-2k}{6}\right)\left(\frac{1+k}{6}\right)+\left(\frac{1+k}{6}\right)\left(\frac{1-k}{6}\right)$$=\frac{1}{18}(2-k-{3k}^2)$

This probability must lie between $\frac{1}{9}$ and $\frac{2}{9}$ $\Rightarrow$ $\frac{1}{9}\le \frac{1}{18}(2-k-{3k}^2)\le \frac{2}{9}$
$\Rightarrow$$3k^2+k\le 0$ and $3k^2+k+2\ge 0$
$\Rightarrow$ $k\in [\frac{-1}{3},0]$ and $k\in (-\infty ,\infty )$
Since $k$ can only take integer values,we get $k=0$