If x=2, y= -1 and z=3 find the value of the following: 1/3x^2y+5/6y^2z^3-2y

$$\\If x=2,\;\; y= -1\;\; and\;\; z=3 \\\\
$find the value of the following: $\\\\
\frac{1}{3}x^2y+\frac{5}{6}y^2z^3-2y\\\\
=\frac{1}{3}*2^2(-1)+\frac{5}{6}(-1)^2*3^3-2(-1)\\\\
=\frac{1}{3}*4*(-1)+\frac{5}{6}*27+2)\\\\
=\frac{-4}{3}+\frac{5}{2}*9+2\\\\
=\frac{-4}{3}+\frac{45}{2}+2\\\\
=\frac{2}{3}+\frac{45}{2}\\\\
=22+\frac{2}{3}+\frac{1}{2}\\\\
=22+\frac{4}{6}+\frac{3}{6}\\\\
=22+\frac{7}{6}\\\\
=23\frac{1}{6}\\$$

0.1666666666666667