一口袋中装有$10$个小球，其中标有数字$1$，$2$，$3$，$4$，$5$的小球各两个，这些小球除数字外其余均相同.
$(1)$某人从中一次性摸出$4$个球，设事件$A$“摸出的$4$个球中至少有一个数字是$5$”，事件$B$“摸出的$4$个球中恰有两个数字相同”；分别求事件$A$和事件$B$的概率；
$(2)$现有一游戏，游戏规则是：游戏玩家每次有放回地从袋中随机摸出一球，若摸到$5$号球，则游戏结束；否则继续摸球，当摸到第$n\left(n\geqslant 2\right)$个球时，无论摸出的是几号球游戏都结束.设$X$表示摸球的次数$\left(1\leqslant X\leqslant n,X\in {\bf N}^{*}\right)$，求随机变量$X$的期望.

$(1)$ $P(A)=\\dfrac{2}{3},P(B)=\\dfrac{4}{7}$
$(2)$ $E(X)=5-4×{(\\dfrac{4}{5})}^{n-1}$

$(1)$从中一次性提出$4$个球有${\rm C}_{10}^{4}=210$种方法，
$n(A)=210-{\rm C}_{8}^{4}=140,n(B)={\rm C}_{5}^{1}\cdot {\rm C}_{4}^{2}\cdot {\rm C}_{2}^{1}\cdot {\rm C}_{2}^{1}=120$，
所以$P(A)=\dfrac{140}{210}=\dfrac{2}{3},P(B)=\dfrac{120}{210}=\dfrac{4}{7}$；
$(2)X$的取值可能为$1$，$2$，$3$，$\cdots $，$n$，
当$1\leqslant k\leqslant n-1$时，$P\left(X=k\right)={(\dfrac{4}{5})}^{k-1}\cdot \dfrac{1}{5}$，
当$k=n$时，$P\left(X=k\right)={(\dfrac{4}{5})}^{n-1}$，

所以$E(X)=1×{(\dfrac{4}{5})}^{0}\cdot \dfrac{1}{5}+2×{(\dfrac{4}{5})}^{1}\cdot \dfrac{1}{5}+3×{(\dfrac{4}{5})}^{2}\cdot \dfrac{1}{5}+⋯+\left(n-1\right)×{(\dfrac{4}{5})}^{n-2}\cdot \dfrac{1}{5}+n×{(\dfrac{4}{5})}^{n-1}$，
$=\dfrac{1}{5}[{(\dfrac{4}{5})}^{0}+2×{(\dfrac{4}{5})}^{1}+3×{(\dfrac{4}{5})}^{2}+⋯+\left(n-1\right)×{(\dfrac{4}{5})}^{n-2}]+n×{(\dfrac{4}{5})}^{n-1}$，
令$S={(\dfrac{4}{5})}^{0}+2×{(\dfrac{4}{5})}^{1}+3×{(\dfrac{4}{5})}^{2}+⋯+\left(n-1\right)×{(\dfrac{4}{5})}^{n-2}$，
则$\dfrac{4}{5}S={(\dfrac{4}{5})}^{1}+2×{(\dfrac{4}{5})}^{2}+⋯+\left(n-2\right)×{(\dfrac{4}{5})}^{n-2}+\left(n-1\right)×{(\dfrac{4}{5})}^{n-1}$，
相减得$\dfrac{1}{5}S={(\dfrac{4}{5})}^{0}+{(\dfrac{4}{5})}^{1}+{(\dfrac{4}{5})}^{2}+⋯+{(\dfrac{4}{5})}^{n-2}-\left(n-1\right)×{(\dfrac{4}{5})}^{n-1}=\dfrac{1-{(\dfrac{4}{5})}^{n-1}}{1-\dfrac{4}{5}}-\left(n-1\right)×{(\dfrac{4}{5})}^{n-1}$，
$=5-\left(n+4\right)×{(\dfrac{4}{5})}^{n-1}$，
所以$E(X)=5-4×{(\dfrac{4}{5})}^{n-1}$.