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

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

Aicharya · Accepted Answer

Probability of obtaining total sum 7 &#160; = Probability of getting opposite faces. Probability of getting opposite faces = 2 1 6 - x 1 6 + x + 1 6 &#215; 1 6 + 1 6 &#215; 1 6 = 2 1 6 - x 1 6 + x + 1 6 &#215; 1 6 + 1 6 &#215; 1 6 = 13 96 x = 1 8

When a certain biased die is rolled, a particular face occurs with probability 1 6 - x and its opposite face…

Solution