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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 :
Option: 1 \frac{1}{16}
Option: 2 \frac{1}{12}
Option: 3 \frac{1}{8}
Option: 4 \frac{1}{9}

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best_answer

The following combinations are possible (1,6),(2,5),(3,4),(4,3),(5,2),(6,7)
Let The particular biased pair of faces with probability \frac{1}{6}-x\, and\: \frac{1}{6}+x
\frac{13}{96}= \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}+\frac{1}{6}\times \frac{1}{6}+\frac{1}{6}\times \frac{1}{6}+\left ( \frac{1}{6}+x \right )\left ( \frac{1}{6} -x\right )
\Rightarrow \frac{4}{36}+\frac{2}{36}-2x^{2}= \frac{13}{96}\Rightarrow \frac{1}{6}-2x^{2}= \frac{13}{96}\Rightarrow 2x^{2}= \frac{1}{6}-\frac{13}{96}= \frac{3}{96}
\Rightarrow x^{2}= \frac{1}{64}\Rightarrow x= \frac{1}{8}

Posted by

Kuldeep Maurya

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