Question

8．甲乙兩人相約打靶，甲射擊3次，每次射擊的命中率為$\frac{1}{2}$，乙射擊2次，每次射擊的命中率為$\frac{2}{3}$，記甲命中的次數(shù)為x，乙命中的次數(shù)為y
（1）求x+y的分布列和E（x+y）
（2）猜想兩個相互獨立的變量x，y的期望與x+y的期望間的關系，并證明你的猜想．
其中，x的分布列為：

x	x1	x2	…	xn
p	p1	p2		pn

y的分布列為：

y	y1	y2	…	ym
p	p${\;}_{1}^{′}$	p${\;}_{2}^{′}$	…	p${\;}_{m}^{′}$

Answer 1

分析 （1）由已知得x+y的可能取值為0，1，2，3，4，5，分別求出相應的概率，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），再利用離散型隨機變量的數(shù)學期望計算公式進行證明．

解答 解：（1）由已知得x+y的可能取值為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的分布列為：

x+y	0	1	2	3	4	5
p	$\frac{1}{72}$	$\frac{7}{72}$	$\frac{19}{72}$	$\frac{25}{72}$	$\frac{16}{72}$	$\frac{4}{72}$

E（x+y）=$0×\frac{1}{72}+1×\frac{7}{72}+2×\frac{19}{72}+3×\frac{25}{72}$+$4×\frac{16}{72}+5×\frac{4}{72}$=$\frac{204}{72}=\frac{17}{6}$．
（2）猜想：E（x+y）=E（x）+E（y）
證明：∵$P（ξ={x_i}+{y_j}）={p_i}{p_j}^/\;\;（1≤i≤n，1≤j≤m）$
∴E（x+y）=（x₁+y₁）×p₁p₁′+$（{x}_{1}+{y}_{2}）{p}_{1}{{p}_{2}}^{'}$+$（{x}_{1}+{y}_{3}）{p}_{1}{{p}_{3}}^{'}$…+（x₁+y_m）p₁p_m′
+$（{x}_{2}+{y}_{1}）{p}_{2}{{p}_{1}}^{'}$+$（{x}_{2}+{y}_{2}）{p}_{2}{{p}_{2}}^{'}+（{x}_{2}+{y}_{3}）{p}_{2}{{p}_{3}}^{'}$+…+（x₂+y_m）×p₂p_m′
+…+$（{x}_{n}+{y}_{1}）{p}_{n}{{p}_{1}}^{'}+（{x}_{n}+{y}_{2}）{p}_{n}{{p}_{2}}^{'}$+（x_n+y₃）p_np₃′+…+（x_n+y_m）p_np_m′
=${x}_{1}{p}_{1}{{p}_{1}}^{'}+{y}_{1}{p}_{1}{{p}_{1}}^{'}$+${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+…+x_np_n）+$（{y}_{1}{{p}_{1}}^{'}+{y}_{2}{{p}_{2}}^{'}+…+{y}_{m}{{p}_{m}}^{'}）$（p₁+p₂+…+p_n）
=E（x）+E（y）．

點評 本題考查離散型隨機變量的分布列、數(shù)學期望的求法及應用，是中檔題，解題時要認真審題，注意離散型隨機變量的數(shù)學期望的計算公式及性質的合理運用．