$\begin {array} {1 1} (A)\;\large\frac{1}{18} & \quad (B)\;\large\frac{1}{9} \\ (C)\;\large\frac{2}{9} & \quad (D)\;\large\frac{1}{36} \end {array}$

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A dice is thrown thrice $ n(s) = 6 \times 6 \times \times 6$

Favourable events

$ W^{r_1}+ W^{r_2}+ W^{r_3}=0$

$ ( r_1, r_2, r_3)$ are ordered 3 triples which can take values.

$(1,2,3)\: (1,5,3)\: (4,2,3)\: (4,5,3)$

$ (1,2,6)\: (1,5,6)\: (4,2,6)\: (4,5,6)$

i.e, 8 ordered pairs and each can be arranged in 3! ways.

$ 3! = 6$

$ n(E)=8 \times 6$

$P(E)=\large\frac{8 \times 6}{6 \times 6\times 6}$

$ = \large\frac{2}{9}$

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