离散型随机变量及其分布列习题
前言
题目难度都比较大
典例剖析
(1). 求第一次检测出的是次品且第二次检测出的是正品的概率。
解法1:利用排列数公式和古典概型求解; \(P=\cfrac{A_2^1A_3^1}{A_5^2}=\cfrac{3}{10}\)
解法2:利用相互独立事件求解;
第一次检测出的是次品且第二次检测出的是正品的概率为\(P=\cfrac{2}{5}\times \cfrac{3}{4}=\cfrac{3}{10}\)
(2). 已知每检测一件产品需要费用100元,设\(X\)表示直到检测出2件次品或者检测出3件正品时所需要的检测费(单位:元),求\(X\)的分布列和数学期望。
解:先设检测过的产品数为\(x\),则由题目可知\(x=2,3,4\)
其中\(x=2\)时对应“次次”一种;
其中\(x=3\)时对应“正次次、次正次、正正正”三种;
其中\(x=4\)时对应“正次正次、正正次次、次正正次、次正正正、正正次正、正次正正”六种;
故\(X\)的所有可能取值为\(200,300,400\),(注意:由于是无放回的,故有顺序,故用排列而不是组合)
\(P(X=200)=\cfrac{A_2^2}{A_5^2}=\cfrac{1}{10}\),
\(P(X=300)=\cfrac{A_3^3+C_2^1\cdot C_3^1\cdot C_1^1+C_3^1\cdot A_2^2}{A_5^3}=\cfrac{3}{10}\),
\(P(X=400)=1-P(X=200)-P(X=300)=\cfrac{6}{10}\),[1]
故 \(X\) 的分布列为
\(X\) | \(200\) | \(300\) | \(400\) |
---|---|---|---|
\(P\) | \(\cfrac{1}{10}\) | \(\cfrac{3}{10}\) | \(\cfrac{6}{10}\) |
\(E(X)=200\times \cfrac{1}{10}+300\times \cfrac{3}{10}+400\times \cfrac{6}{10}=350\)
⑴. “星队”至少猜对 \(3\) 个成语的概率;
解析:先定义事件,记“星队”至少猜对3个成语 为事件\(E\),
“甲第一轮猜对”为事件\(A\),“乙第一轮猜对”为事件\(B\),
“甲第二轮猜对”为事件\(C\),“乙第二轮猜对”为事件\(D\),
且\(P(A)=P(C)=\cfrac{3}{4}\),\(P(B)=P(D)=\cfrac{2}{3}\),
则\(E=ABCD+\bar{A}BCD+A\bar{B}CD+AB\bar{C}D+ABC\bar{D}\),事件\(A、B、C、D\)相互独立,
事件\(ABCD、\bar{A}BCD、A\bar{B}CD、AB\bar{C}D、ABC\bar{D}\)互斥,故有
\(P(E)\)\(=\)\(P(ABCD+\bar{A}BCD+A\bar{B}CD+AB\bar{C}D+ABC\bar{D})\)
\(=\)\(P(ABCD)+P(\bar{A}BCD)+P(A\bar{B}CD)+P(AB\bar{C}D)+P(ABC\bar{D})\),
\(=\cfrac{3}{4}\times\cfrac{2}{3}\times\cfrac{3}{4}\times\cfrac{2}{3}\)\(+\)\(\cfrac{1}{4}\times\cfrac{2}{3}\times\cfrac{3}{4}\times\cfrac{2}{3}\)\(+\)\(\cfrac{3}{4}\times\cfrac{1}{3}\times\cfrac{3}{4}\times\cfrac{2}{3}\)\(+\)\(\cfrac{3}{4}\times\cfrac{2}{3}\times\cfrac{1}{4}\times\cfrac{2}{3}\)\(+\)\(\cfrac{3}{4}\times\cfrac{2}{3}\times\cfrac{3}{4}\times\cfrac{1}{3}\)\(=\)\(\cfrac{2}{3}\).
【反思】:能不能利用二项分布来求解?为什么?
分析:设甲猜对的成语个数为\(X\),则\(X\sim B\left(2,\cfrac{3}{4}\right)\),乙猜对成语的个数为\(Y\),
则\(Y\sim B\left(2,\cfrac{2}{3}\right)\),“星队至少猜对3个成语”为事件\(E\),则
\(P(E)=P(X=1)P(Y=2)+P(X=2)P(Y=1)+P(X=2)P(Y=2)\)
\(=P(\bar{A}BCD)+P(A\bar{B}CD)+P(AB\bar{C}D)+P(ABC\bar{D})+P(ABCD)\)
⑵. “星队”两轮得分之和\(X\)的分布列。
解析:由于“星队”每轮得分分别为0分、1分、3分,
则“星队”两轮得分之和 \(X\) 可能取值为0、1、2、3、4、6,
\(P(X=0)\)\(=\)\(P(\bar{A}\bar{B}\bar{C}\bar{D})\)\(=\)\(\cfrac{1}{4}\times\cfrac{1}{3}\times\cfrac{1}{4}\times\cfrac{1}{3}\)\(=\)\(\cfrac{1}{16\times 9}\)\(=\)\(\cfrac{1}{144}\)
\(P(X=1)\)\(=\)\(P(A\bar{B}\bar{C}\bar{D})\)\(+\)\(P(\bar{A}B\bar{C}\bar{D})\)\(+\)\(P(\bar{A}\bar{B}C\bar{D})\)\(+\)\(P(\bar{A}\bar{B}\bar{C}D)\)
\(=\)\(\cfrac{3}{4}\times\cfrac{1}{3}\times\cfrac{1}{4}\times\cfrac{1}{3}\)\(+\)\(\cfrac{1}{4}\times\cfrac{2}{3}\times\cfrac{3}{4}\times\cfrac{1}{3}\)\(+\)\(\cfrac{1}{4}\times\cfrac{1}{3}\times\cfrac{3}{4}\times\cfrac{1}{3}\)\(+\)\(\cfrac{1}{4}\times\cfrac{1}{3}\times\cfrac{1}{4}\times\cfrac{2}{3}\)
\(=\)\(\cfrac{10}{16\times 9}\)\(=\)\(\cfrac{5}{72}\)
\(P(X=2)\)\(=\)\(P(A\bar{B}C\bar{D})+P(A\bar{B}\bar{C}D)+P(\bar{A}B\bar{C}D)+P(\bar{A}BC\bar{D})\)
\(=\)\(\cfrac{3}{4}\times\cfrac{1}{3}\times\cfrac{3}{4}\times\cfrac{1}{3}\)\(+\)\(\cfrac{3}{4}\times\cfrac{1}{3}\times\cfrac{1}{4}\times\cfrac{2}{3}\)\(+\)\(\cfrac{1}{4}\times\cfrac{2}{3}\times\cfrac{1}{4}\times\cfrac{2}{3}\)\(+\)\(\cfrac{1}{4}\times\cfrac{2}{3}\times\cfrac{3}{4}\times\cfrac{1}{3}\)
\(=\)\(\cfrac{25}{16\times 9}\)\(=\)\(\cfrac{25}{144}\)
\(P(X=3)\)\(=\)\(P(\bar{A}\bar{B}CD)+P(AB\bar{C}\bar{D})\)
\(=\)\(\cfrac{1}{4}\times\cfrac{1}{3}\times\cfrac{3}{4}\times\cfrac{2}{3}\)\(+\)\(\cfrac{3}{4}\times\cfrac{2}{3}\times\cfrac{1}{4}\times\cfrac{1}{3}\)
\(=\)\(\cfrac{12}{16\times 9}\)\(=\)\(\cfrac{1}{12}\)
\(P(X=4)\)\(=\)\(P(\bar{A}BCD)+P(A\bar{B}CD)+P(AB\bar{C}D)+P(ABC\bar{D})\)
\(=\)\(\cfrac{1}{4}\times\cfrac{2}{3}\times\cfrac{3}{4}\times\cfrac{2}{3}\)\(+\)\(\cfrac{3}{4}\times\cfrac{1}{3}\times\cfrac{3}{4}\times\cfrac{2}{3}\)\(+\)\(\cfrac{3}{4}\times\cfrac{2}{3}\times\cfrac{1}{4}\times\cfrac{2}{3}\)\(+\)\(\cfrac{3}{4}\times\cfrac{2}{3}\times\cfrac{3}{4}\times\cfrac{1}{3}\)
\(=\)\(\cfrac{60}{16\times 9}\)\(=\)\(\cfrac{5}{12}\)
\(P(X=6)\)\(=\)\(P(ABCD)\)\(=\)\(\cfrac{3}{4}\times\cfrac{2}{3}\times\cfrac{3}{4}\times\cfrac{2}{3}\)
\(=\)\(\cfrac{36}{16\times 9}\)\(=\)\(\cfrac{1}{4}\)
所以,随机变量 \(X\) 的分布列为
\(X\) | \(0\) | \(1\) | \(2\) | \(3\) | \(4\) | \(6\) |
---|---|---|---|---|---|---|
\(P\) | \(\cfrac{1}{144}\) | \(\cfrac{5}{72}\) | \(\cfrac{25}{144}\) | \(\cfrac{1}{12}\) | \(\cfrac{5}{12}\) | \(\cfrac{1}{4}\) |
⑴. 甲至少投中\(1\)个球的概率;
解析:由题目可知,甲投球两次,每次投中的概率相等,
设甲投中球的次数为\(X\),则\(X\sim B\left(2,\cfrac{3}{4}\right)\)
故甲至少投中 \(1\) 个球的概率 \(P=P(X\ge 1)=C_2^1\times(\cfrac{3}{4})^1\times(\cfrac{1}{4})^1+C_2^2\times(\cfrac{3}{4})^2\times(\cfrac{1}{4})^0=\cfrac{15}{16}\).
⑵. 甲乙两人每轮投球得分之和\(Z\)的分布列。
解析: 再设乙投中球的次数为\(Y\),则\(Y\sim B\left(2,\cfrac{2}{3}\right)\),则\(Z\)可能的取值为\(0、1、2、3、4\),
\(P(Z=0)=P(X=0)\times P(Y=0)\)
\(P(Z=1)=P(X=0)\times P(Y=1)+P(X=1)\times P(Y=0)\)
\(P(Z=2)=P(X=0)\times P(Y=2)+P(X=1)\times P(Y=1)+P(X=2)\times P(Y=0)\)
\(P(Z=3)=P(X=1)\times P(Y=2)+P(X=2)\times P(Y=1)\)
\(P(Z=4)=P(X=2)\times P(Y=2)\)
反思1: 【分析】:“甲第一次投中”为事件\(A\),“乙第一次投中”为事件\(B\),“甲第二次投中”为事件\(C\),“乙第二次投中”为事件\(D\),
且\(P(A)=P(C)=\cfrac{3}{4}\),\(P(B)=P(D)=\cfrac{2}{3}\),
\(P(Z=2)=P(X=0)\times P(Y=2)+P(X=1)\times P(Y=1)+P(X=2)\times P(Y=0)\)
\(=P(\bar{A}B\bar{C}D)+P(\bar{A}\bar{B}CD+\bar{A}BC\bar{D}+A\bar{B}\bar{C}D+AB\bar{C}\bar{D})+P(A\bar{B}C\bar{D})\)
反思2:本问,为什么不能用二项分布来求解?
依上,设甲猜对的成语个数为\(\mu\),则\(\mu\sim B\left(2,\cfrac{3}{4}\right)\),
乙猜对成语的个数为\(\eta\),则\(\eta\sim B\left(2,\cfrac{2}{3}\right)\),则
\(P(X=2)\)\(=\)\(P(\mu=1)P(\eta=1)\)\(=\)\(P(\bar{A}C+A\bar{C})P(\bar{B}D+B\bar{D})\)
\(=\)\(P(AB\bar{C}\bar{D}+\bar{A}BC\bar{D}\)\(+\)\(A\bar{B}\bar{C}D+\bar{A}\bar{B}CD)\)
而由上可知,
\(P(X=2)\)\(=\)\(P(A\bar{B}C\bar{D})\)\(+\)\(P(A\bar{B}\bar{C}D)\)\(+\)\(P(\bar{A}B\bar{C}D)\)\(+\)\(P(\bar{A}BC\bar{D})\),
明显事件中有一样的,也有不一样的,主要是二项分布是单独刻画甲和乙的猜测结果,而例2中是把甲乙两个人绑成了“星队”这个整体。所以是不一样的。
详解:\(P(X=400)\)\(=\)\(\cfrac{3}{5}\times \cfrac{2}{4}\times \cfrac{2}{3}\times \cfrac{1}{2}\)\(+\)\(\cfrac{2}{5}\times \cfrac{3}{4}\times \cfrac{2}{3}\times \cfrac{1}{2}\)\(+\)\(\cfrac{3}{5}\times \cfrac{2}{4}\times \cfrac{2}{3}\times \cfrac{1}{2}\)\(+\)\(\cfrac{2}{5}\times \cfrac{3}{4}\times \cfrac{2}{3}\times \cfrac{1}{2}\)\(+\)\(\cfrac{3}{5}\times \cfrac{2}{4}\times \cfrac{2}{3}\times \cfrac{1}{2}\)\(+\)\(\cfrac{3}{5}\times \cfrac{2}{4}\times \cfrac{2}{3}\times \cfrac{1}{2}\)\(=\)\(\cfrac{6}{10}\) ↩︎