MCQ
When a certain biased die is rolled, a particular face occurs with probability $\frac{1}{6}-\mathrm{x}$ and its opposite face occurs with probability $\frac{1}{6}+\mathrm{x}$. 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 $\mathrm{sum}=7$, when such a die is rolled twice, is $\frac{13}{96}$, then the value of $x$ is:
  • A
    $\frac{1}{16}$
  • $\frac{1}{8}$
  • C
    $\frac{1}{9}$
  • D
    $\frac{1}{12}$

Answer

Correct option: B.
$\frac{1}{8}$
b
Probability of obtaining total sum $7=$ probability of getting opposite faces.

Probability of getting opposite faces

$=2\left[\left(\frac{1}{6}-x\right)\left(\frac{1}{6}+x\right)+\frac{1}{6} \times \frac{1}{6}+\frac{1}{6} \times \frac{1}{6}\right]$

$\Rightarrow 2\left[\left(\frac{1}{6}-x\right)\left(\frac{1}{6}+x\right)+\frac{1}{6} \times \frac{1}{6}+\frac{1}{6} \times \frac{1}{6}\right]=\frac{13}{96}$

(given)

$x=\frac{1}{8}$

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