There are $\left(\begin{array}{c}100 \\ 3\end{array}\right)$ choices of three packages (in any ordering). There are $\left(\begin{array}{c}60 \\ 3\end{array}\right)$ choices of three packages without prizes. Hence $P(X=0)=\left(\begin{array}{c}60 \\ 3\end{array}\right) /\left(\begin{array}{c}100 \\ 3\end{array}\right) \approx 0.2116$. If a single prize is won this can happen in $\left(\begin{array}{c}40 \\ 1\end{array}\right) \cdot\left(\begin{array}{c}60 \\ 2\end{array}\right)$ ways. Hence $P(X=1)=\left(\begin{array}{c}40 \\ 1\end{array}\right) \cdot\left(\begin{array}{c}60 \\ 2\end{array}\right) /\left(\begin{array}{c}100 \\ 3\end{array}\right) \approx 0.4378$ and similarly $P(X=2)=\left(\begin{array}{c}40 \\ 2\end{array}\right)$.

$\left(\begin{array}{c}60 \\ 1\end{array}\right) /\left(\begin{array}{c}100 \\ 3\end{array}\right) \approx 0.2894$ and $P(X=3)=\left(\begin{array}{c}40 \\ 3\end{array}\right) /\left(\begin{array}{c}100 \\ 3\end{array}\right) \approx 0.0611$ (there is some small rounding error in the

given values).