8.(5 分)已知随机变量 $\xi_{\mathrm{i}}$ 满足 $\mathrm{P}\left(\xi_{\mathrm{i}}=1\right)=\mathrm{p}_{\mathrm{i}}, \mathrm{P}\left(\xi_{\mathrm{i}}=0\right)=1-\mathrm{p}_{\mathrm{i}}$ , $\mathrm{i}=1$ ,2.若 $0< \mathrm{p}_{1}<\mathrm{p}_{2}<\frac{1}{2}$ ,则( )
(5 分)已知随机变量 _ i 满足 P ( _ i =1…——2017 高考数学第 8 题答案解析
2017_浙江卷 (2017)
完整解析 · 逐步详解
【分析】由已知得 $0<\mathrm{p}_{1}<\mathrm{p}_{2}<\frac{1}{2}, \frac{1}{2}<1-\mathrm{p}_{2}<1-\mathrm{p}_{1}<1$ ,求出 $\mathrm{E}\left(\xi_{1}\right)=\mathrm{p}_{1}, \mathrm{E}$
$\left(\xi_{2}\right)=p_{2}$ ,从而求出 $D\left(\xi_{1}\right), D\left(\xi_{2}\right)$ ,由此能求出结果。
【解答】解:∵ 随机变量 $\xi_{i}$ 满足 $\mathrm{P}\left(\xi_{i}=1\right)=\mathrm{p}_{\mathrm{i}}, \mathrm{P}\left(\xi_{\mathrm{i}}=0\right)=1-\mathrm{p}_{\mathrm{i}}, \mathrm{i}=1,2, \ldots$ , $0<\mathrm{p}_{1}<\mathrm{p}_{2}<\frac{1}{2}$ ,
$\therefore \frac{1}{2}<1-\mathrm{p}_{2}<1-\mathrm{p}_{1}<1$ ,
$\mathrm{E}\left(\xi_{1}\right)=1 \times \mathrm{p}_{1}+0 \times\left(1-\mathrm{p}_{1}\right)=\mathrm{p}_{1}$ ,
$E\left(\xi_{2}\right)=1 \times p_{2}+0 \times\left(1-p_{2}\right)=p_{2}$ ,
$D\left(\xi_{1}\right)=\left(1-p_{1}\right)^{2} p_{1}+\left(0-p_{1}\right)^{2}\left(1-p_{1}\right)=p_{1}-p_{1}^{2}$,
$D\left(\xi_{2}\right)=\left(1-p_{2}\right)^{2} p_{2}+\left(0-p_{2}\right)^{2}\left(1-p_{2}\right)={p_{2}}-p_{2}{ }^{2}$,
$D\left(\xi_{1}\right)-D\left(\xi_{2}\right)=p_{1}-p_{1}^{2}-\left(p_{2}-p_{2}^{2}\right)=\left(p_{2}-p_{1}\right)\left(p_{1}+p_{2}-1\right)<0$,
$\therefore \mathrm{E}\left(\xi_{1}\right)<\mathrm{E}\left(\xi_{2}\right), \mathrm{D}\left(\xi_{1}\right)<\mathrm{D}\left(\xi_{2}\right)$.
故选:A.
【点评】本题考查离散型随机变量的数学期望和方差等基础知识,考查推理论证能力、运算求解能力、空间想象能力,考查数形结合思想、化归与转化思想,是中档题。