题目内容
8.甲乙两人相约打靶,甲射击3次,每次射击的命中率为$\frac{1}{2}$,乙射击2次,每次射击的命中率为$\frac{2}{3}$,记甲命中的次数为x,乙命中的次数为y(1)求x+y的分布列和E(x+y)
(2)猜想两个相互独立的变量x,y的期望与x+y的期望间的关系,并证明你的猜想.
其中,x的分布列为:
x | x1 | x2 | … | xn |
p | p1 | p2 | pn |
y | y1 | y2 | … | ym |
p | p${\;}_{1}^{′}$ | p${\;}_{2}^{′}$ | … | p${\;}_{m}^{′}$ |
分析 (1)由已知得x+y的可能取值为0,1,2,3,4,5,分别求出相应的概率,由此能求出x+y的分布列和E(x+y).
(2)先猜想E(x+y)=E(x)+E(y),再利用离散型随机变量的数学期望计算公式进行证明.
解答 解:(1)由已知得x+y的可能取值为0,1,2,3,4,5,
P(x+y=0)=${C}_{3}^{0}(\frac{1}{2})^{3}•{C}_{2}^{0}(\frac{1}{3})^{2}$=$\frac{1}{72}$,
P(x+y=1)=${C}_{3}^{1}(\frac{1}{2})(\frac{1}{2})^{2}{C}_{2}^{2}(\frac{1}{3})^{2}+{C}_{3}^{0}(\frac{1}{2})^{3}{C}_{2}^{1}(\frac{2}{3})(\frac{1}{3})$=$\frac{7}{72}$,
P(x+y=2)=${C}_{3}^{2}(\frac{1}{2})^{2}(\frac{1}{2}){C}_{2}^{0}(\frac{1}{3})^{2}+{C}_{3}^{0}(\frac{1}{2})^{3}{C}_{2}^{2}(\frac{2}{3})^{2}$+${C}_{3}^{1}(\frac{1}{2})(\frac{1}{2})^{2}{C}_{2}^{1}(\frac{2}{3})(\frac{1}{3})$=$\frac{19}{72}$,
P(x+y=3)=${C}_{3}^{3}(\frac{1}{2})^{3}{C}_{2}^{0}(\frac{1}{3})^{2}+{C}_{3}^{2}(\frac{1}{2})^{2}(\frac{1}{2}){C}_{2}^{1}(\frac{2}{3})(\frac{1}{3})$+${C}_{3}^{1}(\frac{1}{2})(\frac{1}{2})^{2}{C}_{2}^{2}(\frac{2}{3})^{2}$=$\frac{25}{72}$,
P(x+y=4)=${C}_{3}^{3}(\frac{1}{2})^{3}{C}_{2}^{1}(\frac{2}{3})(\frac{1}{3})+{C}_{3}^{2}(\frac{1}{2})^{2}(\frac{1}{2}){C}_{2}^{2}(\frac{2}{3})^{2}$=$\frac{16}{72}$,
P(x+y=5)=${C}_{3}^{3}(\frac{1}{2})^{3}{C}_{2}^{2}(\frac{2}{3})^{2}$=$\frac{4}{72}$,
∴x+y的分布列为:
x+y | 0 | 1 | 2 | 3 | 4 | 5 |
p | $\frac{1}{72}$ | $\frac{7}{72}$ | $\frac{19}{72}$ | $\frac{25}{72}$ | $\frac{16}{72}$ | $\frac{4}{72}$ |
(2)猜想:E(x+y)=E(x)+E(y)
证明:∵$P(ξ={x_i}+{y_j})={p_i}{p_j}^/\;\;(1≤i≤n,1≤j≤m)$
∴E(x+y)=(x1+y1)×p1p1′+$({x}_{1}+{y}_{2}){p}_{1}{{p}_{2}}^{'}$+$({x}_{1}+{y}_{3}){p}_{1}{{p}_{3}}^{'}$…+(x1+ym)p1pm′
+$({x}_{2}+{y}_{1}){p}_{2}{{p}_{1}}^{'}$+$({x}_{2}+{y}_{2}){p}_{2}{{p}_{2}}^{'}+({x}_{2}+{y}_{3}){p}_{2}{{p}_{3}}^{'}$+…+(x2+ym)×p2pm′
+…+$({x}_{n}+{y}_{1}){p}_{n}{{p}_{1}}^{'}+({x}_{n}+{y}_{2}){p}_{n}{{p}_{2}}^{'}$+(xn+y3)pnp3′+…+(xn+ym)pnpm′
=${x}_{1}{p}_{1}{{p}_{1}}^{'}+{y}_{1}{p}_{1}{{p}_{1}}^{'}$+${x}_{1}{p}_{1}{{p}_{2}}^{'}+{y}_{2}{p}_{1}{{p}_{2}}^{'}+…+$${x}_{2}{p}_{2}{{p}_{m}}^{'}+{{y}_{m}{p}_{1}{p}_{m}}^{'}$
+…+${x}_{2}{p}_{2}{{p}_{1}}^{'}+{y}_{1}{p}_{2}{{p}_{1}}^{'}+{x}_{2}{p}_{2}{{p}_{2}}^{'}+…+$${x}_{2}{p}_{2}{{p}_{m}}^{'}+{y}_{m}{p}_{2}{{p}_{{\;}^{\;}m}}^{'}$
+…+${x}_{n}{p}_{n}{{p}_{1}}^{'}$+${y}_{1}{p}_{n}{{p}_{1}}^{'}+{x}_{n}{p}_{n}{{p}_{2}}^{'}+{y}_{2}{p}_{n}{{p}_{2}}^{'}+…+$${x}_{n}{p}_{n}{{p}_{m}}^{'}+{y}_{m}{p}_{n}{{p}_{m}}^{'}$
=${x}_{1}{p}_{1}({{p}_{1}}^{'}+{{p}_{2}}^{'}+…+{{p}_{m}}^{'})$+${p}_{1}({y}_{1}{{p}_{1}}^{'}+{y}_{2}{{p}_{2}}^{'}+…+{y}_{m}{{p}_{m}}^{'})$
+${x}_{2}{p}_{2}({{p}_{1}}^{'}+{{p}_{2}}^{'}+…+{{p}_{m}}^{'})+{p}_{2}$(${y}_{1}{{p}_{1}}^{'}+{y}_{2}{{p}_{2}}^{'}+…+{y}_{m}{{p}_{m}}^{'}$)
+…+${x}_{n}{p}_{n}({{p}_{1}}^{'}+{{p}_{n}}^{'}+…+{{p}_{m}}^{'})+{p}_{2}$(${y}_{1}{{p}_{1}}^{'}+{y}_{2}{{p}_{2}}^{'}+…+{y}_{m}{{p}_{m}}^{'}$)
=(x1p1+x2p2+…+xnpn)+$({y}_{1}{{p}_{1}}^{'}+{y}_{2}{{p}_{2}}^{'}+…+{y}_{m}{{p}_{m}}^{'})$(p1+p2+…+pn)
=E(x)+E(y).
点评 本题考查离散型随机变量的分布列、数学期望的求法及应用,是中档题,解题时要认真审题,注意离散型随机变量的数学期望的计算公式及性质的合理运用.
A. | (0,1) | B. | (1,2) | C. | (2,3) | D. | (3,4) |
A. | -$\frac{1}{2}$ | B. | $\frac{1}{2}$ | C. | -$\frac{3}{4}$ | D. | $\frac{3}{4}$ |