回復 29# anyway13 的帖子
\begin{align}
& P\left( X\_2 \right)=\frac{2\times 1}{6\times 5} \\
& P\left( X\_3 \right)=\frac{2\times 4\times 1\times C_{1}^{2}}{6\times 5\times 4} \\
& P\left( X\_4 \right)=\frac{2\times 4\times 3\times 1\times C_{1}^{3}}{6\times 5\times 4\times 3} \\
& P\left( X\_5 \right)=\frac{2\times 4\times 3\times 2\times 1\times C_{1}^{4}}{6\times 5\times 4\times 3\times 2} \\
& P\left( X\_6 \right)=\frac{2\times 4\times 3\times 2\times 1\times 1\times C_{1}^{5}}{6\times 5\times 4\times 3\times 2\times 1} \\
& \\
& E\left( X \right)=\frac{14}{3},E\left( {{X}^{2}} \right)=\frac{70}{3} \\
\end{align}