2019年高考全国卷I理科数学21题的推广

其实只要概率和等比数列的基础知识就能做出这道压轴题QwQ

试卷随便在哪都能找到,就不放链接了QwQ

题意

\[P(A有用)=\alpha\] \[P(A无用)=1-\alpha\] \[P(B有用)=\beta\] \[P(B无用)=1-\beta\] \[a=P(X=-1)=P(A无用\cap B有用)=(1-\alpha)\cdot\beta\] \[b=P(X=0)=P((A无用\cap B无用)\cup(A有用\cap B有用))=\alpha\cdot\beta+(1-\alpha)\cdot(1-\beta)\] \[c=P(X=1)=P(A有用\cap B无用)=\alpha\cdot(1-\beta)\] \[进行n次试验,第i次获胜的概率p_i满足p_i=a\cdot p_{i-1}+b\cdot p_i+c\cdot p_{i+1}\] \[已知p_0=0,p_{2n}=1求p_n\]

推导

\[\because p_i=a\cdot p_{i-1}+b\cdot p_i+c\cdot p_{i+1}\] \[\therefore (1-b)\cdot p_i=a\cdot p_{i-1}+c\cdot p_{i+1}\] \[\because a+b+c=1\] \[\therefore 1-b=a+c\] \[\therefore (a+c)\cdot p_i=a\cdot p_{i-1}+c\cdot p_{i+1}\] \[变形得c\cdot(p_{i+1}-p_i)=a\cdot(p_i-p_{i-1})\] \[令x_i=p_i-p_{i-1}\] \[x数列等比,公比为k=\frac{a}{c}\] \[\begin{aligned} x_1+x_2+\dots+x_{2n}&=p_{2n}-p_0\\ &=x_1\cdot(1+k+k^2+\dots+k^{2n-1})\\ &=x_1\cdot\frac{k^{2n}-1}{k-1}=1 \end{aligned}\] \[\therefore x_1=\frac{k-1}{k^{2n}-1}\] \[\begin{aligned} p_n&=p_n-p_0\\ &=x_1+x_2+\dots+x_n\\ &=x_1\cdot(1+k+k^2+\dots+k^{n-1})\\ &=\frac{k-1}{k^{2n}-1}\cdot\frac{k^n-1}{k-1}\\ &=\frac{k^n-1}{k^{2n}-1} \end{aligned}\]