Mathematics - Probability Question with Solution | TestHub

Sample space A dice is thrown thrice, $n(s)=6\times 6\times 6$. Favorable events ${\omega }^{r_1}+{\omega }^{r_2}+{\omega }^{r_3}=0$ ie, $\left(r_1,r_2,r_3\right)$ are ordered 3-triples which can take values, $\begin{array}{llll}(1,2,3),& (1,5,3),& (4,2,3),& (4,5,3)\\ (1,2,6),& (1,5,6),& (4,2,6),& (4,5,6)\end{array}\}$ ie, 8 ordered pairs and each can be arranged in 3 ! ways $=6$ $$\begin{array}{ll}\therefore & n(E)=8\times 6\\ \Rightarrow & P(E)=\frac{8\times 6}{6\times 6\times 6}=\frac{2}{9}\end{array}$$