Dear Student,

$$\begin{gathered} Probability of Selecting 2 Red balls and 2 Green balls from total 10 balls = \frac{{}^{6}C_2\times {}^{4}C_2}{{}^{10}C_4} \\ {}^{6}C_2 = \frac{6!}{2! \left(6-2\right)!} = \frac{6!}{2! 4!} = 15 \\ {}^{4}C_2 = \frac{4!}{2! \left(4-2\right)!} = \frac{4!}{2! 2!} = 6 \\ {}^{10}C_4 = \frac{10!}{4! \left(10-4\right)!} = \frac{10!}{4! 6!} = \frac{10\times 9\times 8\times 7}{4\times 3\times 2}= 210 \\ \frac{{}^{6}C_2\times {}^{4}C_2}{{}^{10}C_4} = \frac{15\times 6}{210} =\frac{3}{7} \\ \mathit{P}\mathit{r}\mathit{o}\mathit{b}\mathit{a}\mathit{b}\mathit{i}\mathit{l}\mathit{i}\mathit{t}\mathit{y}\mathbf{ }\left(\mathbf{2}\mathbf{ }\mathbf{Red}\mathbf{ }\mathbf{balls}\mathbf{ }\mathbf{and}\mathbf{ }\mathbf{2}\mathbf{ }\mathbf{Green}\mathbf{ }\mathbf{balls}\mathbf{ }\mathbf{are}\mathbf{ }\mathbf{selected}\right)\mathbf{ }\mathbf{=}\mathbf{ }\frac{\mathbf{3}}{\mathbf{7}} \end{gathered}$$

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