When a certain biased die is rolled,

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

  1. $\frac{1}{16}$

  2. $\frac{1}{8}$

  3. $\frac{1}{9}$

  4. $\frac{1}{12}$


Correct Option: , 2

Solution:

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)

$\mathrm{X}=\frac{1}{8}$

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