Question

A ordinary cube has four plane faces, one face marked 2 and another face marked 3, find the probability of getting a total of 7 in 5 throws.

Answer 1

$$\begin{gathered} \text{A cube has total 6 faces.} \\ \text{Total possible outcomes in 5 throws}=6\times 6\times 6\times 6\times 6={\left(6\right)}^5 \\ \text{The o}\mathrm{nly}\text{way of getting 7 is by getting two 2s and one 3.} \\ \text{Total possible ways}=\frac{{P^5}_3}{2!} \\ =\frac{5\times 4\times 3\times 2\times 1}{2\times 1\times 2\times 1} \\ =30 \\ \mathrm{Now}, \\ P\left(\text{getting 7 in 5 throws}\right)=\frac{30}{6^5}=\frac{5}{6^4} \end{gathered}$$