Solution

Let 

$E_{xyz}$ be the simple event that $x, ~y$ and $z$ appear on first, second and third slot of spin machine.

Since, each slot has 4 possibilities, the total number of simple events is given by $4^3$ [ extended $mn$ rule ].

The number of ways of winning is the number of ways you can choose an object, given by $C^{4}_1=4.$

$\stackrel{\text{The probability of winning }}{}=\genfrac{}{}{0ex}{}{\text{Number of ways of winning}}{\text{Total number of simple events}}$

$=\; 4/4^3$

$=\; 1/16$

Answer

$1/16$