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(B) Let the pair of opposite faces be $(a, b)$ where $a+b=7$. The probability of getting $a$ is $P(a) = \frac{1}{6}-x$ and $P(b) = \frac{1}{6}+x$. For

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$\frac{1}{8}$

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(B) Let the pair of opposite faces be $(a, b)$ where $a+b=7$. The probability of getting $a$ is $P(a) = \frac{1}{6}-x$ and $P(b) = \frac{1}{6}+x$. For the other two pairs of opposite faces,the probability of each face is $\frac{1}{6}$.The sum of two rolls is $7$ if the outcomes are $(1,6), (6,1), (2,5), (5,2), (3,4), (4,3)$.Probability of sum $7 = 2[P(1)P(6) + P(2)P(5) + P(3)P(4)]$.Assuming $(1,6)$ are the faces with probabilities $\frac{1}{6}-x$ and $\frac{1}{6}+x$,then $P(2)=P(5)=\frac{1}{6}$ and $P(3)=P(4)=\frac{1}{6}$.Sum probability $= 2[(\frac{1}{6}-x)(\frac{1}{6}+x) + (\frac{1}{6})(\frac{1}{6}) + (\frac{1}{6})(\frac{1}{6})] = \frac{13}{96}$.$2[(\frac{1}{36}-x^2) + \frac{1}{36} + \frac{1}{36}] = \frac{13}{96}$.$2[\frac{3}{36}-x^2] = \frac{13}{96} \Rightarrow \frac{1}{6}-2x^2 = \frac{13}{96}$.$2x^2 = \frac{1}{6}-\frac{13}{96} = \frac{16-13}{96} = \frac{3}{96} = \frac{1}{32}$.$x^2 = \frac{1}{64} \Rightarrow x = \frac{1}{8}$.

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