3 dice are thrown simultaneously ,find the probability that -
(1) all of them show the same face.
(2) all shows distinct faces.
(3) two of them show the same face.

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Please find below the solution to the asked query:

$$\begin{gathered} \left(1\right) \\ \mathrm{As} \mathrm{there} \mathrm{are} 6 \mathrm{faces} \mathrm{on} \mathrm{dice}, \mathrm{hence} \\ \mathrm{Probabilty} \mathrm{of} \mathrm{getting} \mathrm{a} \mathrm{face} \mathrm{on} \mathrm{dice}=\frac{1}{6} \\ \mathrm{So} \mathrm{we} \mathrm{want} \mathrm{to} \mathrm{have} \mathrm{same} \mathrm{face} \left(\mathrm{Any} \mathrm{one} \mathrm{of} \mathrm{the} 6 \mathrm{faces}\right) \mathrm{on} \mathrm{three} \mathrm{dice}. \\ \mathrm{Required} \mathrm{Probabilty}{=}^6{\mathrm{C}}_1 \left(\frac{1}{6}\right)\times \left(\frac{1}{6}\right)\times \left(\frac{1}{6}\right) \\ =\frac{6}{6\times 6\times 6}=\frac{1}{36} \left(\mathrm{Answer}\right) \\ \left(2\right) \\ \mathrm{As} \mathrm{now} \mathrm{we} \mathrm{want} \mathrm{three} \mathrm{distinct} \mathrm{numbers}, \mathrm{first} \mathrm{dice} \mathrm{has} 6 \mathrm{choices}, \mathrm{second} \mathrm{dice} \\ \mathrm{has} 5 \mathrm{choices} \mathrm{while} \mathrm{third} \mathrm{dice} \mathrm{has} 4 \mathrm{choices}. \\ \mathrm{Required} \mathrm{Probabilty}=\frac{6\times 5\times 4}{6\times 6\times 6}=\frac{\mathbf{5}}{\mathbf{9}}\mathbf{ }\left(\mathbf{Answer}\right) \\ \left(3\right) \mathrm{First} \mathrm{we} \mathrm{chooces} \mathrm{two} \mathrm{dice} \mathrm{from} \mathrm{three} \mathrm{dice}. \mathrm{Then} \mathrm{we} \mathrm{we} \mathrm{will} \mathrm{choose} 1 \mathrm{number} \\ \mathrm{from} 6 \mathrm{numbers} \mathrm{which} \mathrm{we} \mathrm{can} \mathrm{to} \mathrm{see} \mathrm{on} \mathrm{two} \mathrm{dice}, \mathrm{then} \mathrm{we} \mathrm{will} \mathrm{choose}, 1 \mathrm{number} \\ \mathrm{from} \mathrm{remaining} 5 \mathrm{numbers} \mathrm{for} \mathrm{third} \mathrm{dice}. \\ \Rightarrow \mathrm{Required} \mathrm{Probabilty}=\frac{{ }^3{\mathrm{C}}_2{\times }^6{\mathrm{C}}_1{\times }^5{\mathrm{C}}_1}{6\times 6\times 6} \\ =\frac{{ }^3{\mathrm{C}}_1{\times }^6{\mathrm{C}}_1{\times }^5{\mathrm{C}}_1}{6\times 6\times 6} \\ =\frac{ 3\times 6\times 5}{6\times 6\times 6} \\ \mathbf{=}\frac{\mathbf{5}}{\mathbf{12}}\mathbf{ }\left(\mathbf{Answer}\right) \end{gathered}$$

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