Question

When a certain biased die is rolled, a particular face occurs with probability $\frac{1}{6}-x$ and its opposite face occurs with probability $\frac{1}{6}+x$. All other faces occur with probability $\frac{1}{6}$. Note that opposite faces sum to $7$ in any die. If $0<x<\frac{1}{6}$, and the probability of obtaining total sum $=7$, when such a die is rolled twice, is $\frac{13}{96}$, then the value of $x$ is

• A. $\frac{1}{12}$
• B. $\frac{1}{8}$
• C. $\frac{1}{16}$
• D. $\frac{1}{9}$

Answer 1

The correct option is B $\frac{1}{8}$

Probability of obtaining total sum $7=$ probability of getting opposite faces.
So, Probability of obtaining total sum $7$
$=2\{(\frac{1}{6}-x)(\frac{1}{6}+x)+\frac{1}{6}\cdot \frac{1}{6}+\frac{1}{6}\cdot \frac{1}{6}\}$
$\therefore 2(\frac{1}{36}-x^2+\frac{1}{36}+\frac{1}{36})=\frac{13}{96}$
$\Rightarrow \frac{1}{6}-2x^2=\frac{13}{96}$
$\Rightarrow 2x^2=\frac{1}{6}-\frac{13}{96}=\frac{16-13}{96}$
$\Rightarrow x^2=\frac{1}{64}$
$\therefore x=\frac{1}{8}$