已知由甲，乙两位男生和丙，丁两位女生组成的四人冲关小组，参加由某电视台举办的知识类答题闯关活动．活动共有四关，设男生闯过一至四关的概率依次是 $\dfrac{5}{6},\dfrac{4}{5},\dfrac{3}{4},\dfrac{2}{3}$，女生闯过一至四关的概率依次是 $\dfrac{4}{5},\dfrac{3}{4},\dfrac{2}{3},\dfrac{1}{2}$．
【难度】
【出处】
2018年全国高中数学联赛甘肃省预赛
【标注】
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求男生闯过四关的概率；标注答案$\dfrac{1}{3}$解析记男生四关都闯过为事件 $A$，则 $P(A)=\dfrac{5}{6}\times\dfrac{4}{5}\times\dfrac{3}{4}\times\dfrac{2}{3}=\dfrac{1}{3}$．
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设 $\xi$ 表示四人冲关小组闯过四关的人数，求随机变量 $\xi$ 的分布列和期望．标注答案$E(\xi)=\dfrac{16}{15}$解析记女生四关都闯过为事件 $B$，则 $P(B)=\dfrac{4}{5}\times\dfrac{3}{4}\times\dfrac{2}{3}\times\dfrac{1}{2}=\dfrac{1}{5}$．因为 $P(\xi=0)=(\dfrac{2}{3})^2(\dfrac{4}{5})^2=\dfrac{64}{225}$，$P(\xi=1)={\rm C}_2^1\dfrac{1}{3}\cdot\dfrac{2}{3}(\dfrac{4}{5})^2+{\rm C}_2^1\dfrac{1}{5}\cdot\dfrac{4}{5}(\dfrac{2}{3})^2=\dfrac{96}{225}$，$P(\xi=2)={\rm C}_2^2(\dfrac{1}{3})^2\dfrac{4}{5}+{\rm C}_2^2(\dfrac{1}{5})^2(\dfrac{2}{3})^2+{\rm C}_2^1\dfrac{1}{3}\cdot\dfrac{2}{3}\cdot{\rm C}_2^1\dfrac{1}{5}\cdot\dfrac{4}{5}=\dfrac{52}{225}$，$P(\xi=3)={\rm C}_2^1\dfrac{1}{3}\cdot\dfrac{2}{3}(\dfrac{1}{5})^2+{\rm C}_2^1\dfrac{1}{5}\cdot\dfrac{4}{5}(\dfrac{1}{3})^2=\dfrac{12}{225}$，$P(\xi=4)=(\dfrac{1}{3})^2(\dfrac{1}{5})^2=\dfrac{1}{225}$，所以 $\xi$ 的分布列如下：
$\begin{vmatrix}\hline\begin{array}{c|c|c|c|c|c}\xi & 0 & 1 & 2 &3 & 4 \\\hline P& \dfrac{64}{225} & \dfrac{96}{225} & \dfrac{52}{225} & \dfrac{12}{225}& \dfrac{1}{225} \\\hline
\end{array}\end{vmatrix}$
$E(\xi)=0\times\dfrac{64}{225}+1\times\dfrac{96}{225}+2\times\dfrac{52}{225}+3\times\dfrac{12}{225}+4\times\dfrac{1}{225}=\dfrac{240}{225}=\dfrac{16}{15}$．
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