一个柜子里有n双不同型号的鞋子，现从其中随意拿取2r只(2rleqslant n)，求下列事件的概率：(1)没有一双配对；(2)恰有一双配对；(3)恰有两双配对；(4)恰有r双配对.

【解析】

由于$2r\leqslant n$，$\therefore r\lt n$,

总的事件数为${C}_{2n}^{2r}$,

$\left(1\right)$ 设没有一双配对的概率为${P}_{1}$，没有一双配对，等价于从$n$只鞋子中选择$2r$只鞋子，

$\therefore {P}_{1}=\dfrac{{C}_{n}^{2r}}{{C}_{2n}^{2r}}=\dfrac{\frac{n!}{\left(2r\right)!\left(n-2r\right)!}}{\frac{\left(2n\right)!}{\left(2r\right)!\left(2n-2r\right)!}}=\frac{n!}{\left(2r\right)!\left(n-2r\right)!}\cdot \dfrac{\left(2r\right)!\left(2n-2r\right)!}{\left(2n\right)!}$

$=\frac{{2}^{n-r}n!\left(n-r\right)!}{\left(n-2r\right)!\left(2n\right)!}$;

$\left(2\right)$ 设恰有一双配对的概率为${P}_{2}$，恰有一双配对，等价于从$n$双鞋子中任选一双，再从$n-1$只鞋子中选择$2r-2$只鞋子，

$\therefore {P}_{2}=\dfrac{{C}_{n}^{1}\cdot {C}_{n-1}^{2r-2}}{{C}_{2n}^{2r}}=\dfrac{n\cdot {\displaystyle \dfrac{\left(n-1\right)!}{\left(2r-2\right)!\left(n-1-2r+2\right)!}}}{{\displaystyle \dfrac{\left(2n\right)!}{\left(2r\right)!}}}$

$=n\cdot \dfrac{\left(n-1\right)!}{\left(2r-2\right)!\left(n-1-2r+2\right)!}\cdot \dfrac{\left(2r\right)!}{{\displaystyle \left(2n\right)!}}$

$=\dfrac{\left(2r\right)!\cdot \left(2r-1\right)}{\left(2n\right)!}$;

$\left(3\right)$ 设恰有两双配对的概率为${P}_{3}$，恰有两双配对，等价于从$n$双鞋子中任选两双，再从$n-2$只鞋子中选择$2r-4$只鞋子，

$\therefore {P}_{3}=\dfrac{{C}_{n}^{2}\cdot {C}_{n-2}^{2r-4}}{{C}_{2n}^{2r}}=\dfrac{\dfrac{n\left(n-1\right)}{2}\cdot {\displaystyle \dfrac{\left(n-2\right)!}{\left(2r-4\right)!\left(n-2-2r+4\right)!}}}{\dfrac{\left(2n\right)!}{\left(2r\right)!}}$

$=\left(r-1\right)\left(2r-3\right)\cdot \dfrac{{\displaystyle \left(2r\right)!}}{\left(2n\right)!}$;

$\left(4\right)$ 设恰有$r$双配对的概率为${P}_{4}$，恰有$r$双配对，等价于从$n$只鞋子中选择$r$只鞋子，

$\therefore {P}_{4}=\dfrac{{C}_{n}^{r}}{{C}_{2n}^{2r}}=\dfrac{n!}{r!\left(n-r\right)!}\cdot \dfrac{\left(2r\right)!\left(2n-2r\right)!}{\left(2n\right)!}$

$=\dfrac{\left(2r\right)!\left(2n-2r\right)!n!}{\left(2n\right)!r!\left(n-r\right)!}$.