A biased die is marked 2,4,8,16,32,32; probability of face n is 1n.

Question

A biased die is marked $2,4,8,16,32,32$; probability of face $n$ is $\dfrac1n$. If thrown thrice, the probability the sum is 48 is:

Accepted Answer

Only ways to get sum 48: $(32,8,8)$ (in $3$ arrangements) and $(16,16,16)$.

$$P = 3\cdot\frac{1}{32}\cdot\frac18\cdot\frac18+\frac{1}{16}\cdot\frac{1}{16}\cdot\frac{1}{16} = \frac{3}{2^{10}}+\frac{1}{2^{12}}=\frac{13}{2^{12}}$$

Solutions · 1