A biased die: one face has probability 16-x, its opposite face 16+x, others 16.

Question

A biased die: one face has probability $\dfrac16-x$, its opposite face $\dfrac16+x$, others $\dfrac16$. For $0<x<\dfrac16$, probability of sum 7 on two throws is $\dfrac{13}{96}$. Find $x$.

Accepted Answer

$$P(\text{sum}=7) = 2\left[\left(\frac16-x\right)\left(\frac16+x\right)+\left(\frac16\right)^2+\left(\frac16\right)^2\right] = 2\left[\frac{1}{36}-x^2+\frac{2}{36}\right]=\frac{13}{96}$$

$$\frac16-2x^2 = \frac{13}{96} \implies x^2=\frac{1}{64}\implies x=\frac18$$

Solutions · 1