Do you think its easy ?????????

Total number of ways to choose $3$ balls: $\space N = \begin{pmatrix} 12 \\ 3 \end{pmatrix} = \dfrac{12\times 11 \times 10}{3\times 2 \times 1} = 220$

The number of ways to choose $3$ green balls: $\space N_g = \begin{pmatrix} 5 \\ 3 \end{pmatrix} = \dfrac{5\times 4 \times 3}{3\times 2 \times 1} = 10$

The number of ways to choose $3$ yellow balls: $\space N_y = \begin{pmatrix} 4 \\ 3 \end{pmatrix} = \begin{pmatrix} 4 \\ 1 \end{pmatrix} = 4$

The number of ways to choose $3$ while balls: $\space N_w = \begin{pmatrix} 3 \\ 3 \end{pmatrix} = \begin{pmatrix} 3 \\ 0 \end{pmatrix} = 1$

The number of ways of choosing $3$ balls of not the same colour: $\space N - N_g - N_y - N_w = 220 - 10 -4 -1 = 205$

The probability of choosing $3$ balls of NOT the same colour: $\space P = \dfrac {205}{220} = \boxed{0.93}$ to the nearest $2$ decimal places.

P(not getting same colour) = 1-p(same colour) P(all green)=5/12 *4/11 *3/10 =1/22 P(all yellow)=4/12 *3/11 *2/10=1/55 P(all white)=3/12 *2/11 *1/10=1/220 P(same colour)=3/44 So P(not getting same colour) = 41/44=0.93