习题2
2( 甲、乙、丙3人进行独立射击( 每人的命中率依次为0(3( 0(4( 0(6( 设每人射击一次( 试求3人命中总数之概率分布律(
解 用X表示3人命中总数( 则X的取值为0( 1( 2( 3(
用A表示,甲命中”( B表示,乙命中”( C表示,命中”( 则
P(X(0)(P((A(B(C)(0(7(0(6(0(4(0(168(
P(X(1)(P(A(B(C)(P((AB(C)(P((A(BC)
(0(3(0(6(0(4(0(7(0(4(0(4(0(3(0(6(0(6(0(436(
P(X(2)(P(AB(C)(P(A(BC)(P((ABC)
(0(3(0(4(0(4(0(3(0(6(0(6(0(7(0(4(0(6(0(324(
P(X(3)(P(ABC)(0( 3(0( 4(0( 6(0(072(
X
0
1
2
3
pk
0(168
0(436
0(324
0(072
6( 设对某批产品的验收敛方案为( 从该批产品中随机地抽查5件产品( 若次品数小于等于1( 则该批产品通过验收敛( 否则不予通过( 若某批产品的次品率为0(05( 试求该批产品通过验收敛的概率(
解 用X表示5件产品中的次品数( 则X~B(5( 0(05)( 于该批产品通过验收敛的概率为
P(X(1)(P(X(0)(P(X(1)
(0(9774(
7( 某份试卷有10道选择题( 每题共有A( B( C( D四个答案供选择( 其中只有一个答案是正确的( 设某人对每道题均随机地选择答案( 试求该生10道题中恰好答对6道题的概率是多少？
解 用X表示10道题中答对的题目数( 则( 于是该生10道题中恰好答对6道题的概率是
(
12( 设随机变量X具有分布函数
(
试求( P(X((3)( ( ( (
解 P(X((3)(F((3)(0(
(
(
(
15( 设随机变量X具有概率密度
(
(1)求常数A(
(2)求X的分布函数(
(3)求X的取值落在区间内的概率(
解 (1)由( 得A(4.
(2)当x(0时( F(x)(0(
当0(x(1时( (
当x(1时( F(x)(1.
因此 (
(3) X的取值落在区间内的概率为
(
18( 设随机变量X~N(5( 4)( 试求( P(X(5)( P(3(X(6)( P(3(X(7)( P(|X|(1)以及常数C的范围( 使P(|X(5|(C)(0(99(
解 (
(((0.5)((((1)(((0.5)([1(((1)]
(0.6915([1(0.8413](0.5328,
(((1)((((1)(2((1)(1(2(0.8413(1(0.6826.
P(|x|(1)(1(P(|X|(1)(1(P((1(X(1)
(1((((2)((((3)(1([1(((2)]([1(((3)]
(((2)(((3)(1(0.9772(0.9987(1(0.9785.
(
要使P(|X(5|(C)(0.99( 只需( 即( 查表得( 故C(5.16.
20( 设某批鸡蛋每只的重量X(以克计)服从正态分布( X ~N(50( 25)(
(1)求从该批鸡蛋中任取一只( 其重量不足45克的概率(
(2)从该批鸡蛋中任取一只( 其重量介于40克到60克之间的概率(
(3)若从该批鸡蛋中任取五只( 试求恰有2只鸡蛋不足45克的概率(
(4)从该批鸡蛋中任取一只其重量超过60克的概率(
(5)求最小的n( 使从中任选n只鸡蛋( 其中至少有一只鸡蛋的重量超过60克的概率大于0(99(
解 (1)(
(2)
(2(0.9772(1(0.9544.
(3)设Y为5只鸡蛋中重量不足45克的鸡蛋数( 则Y~B(5( 0.1587)( 故所求概率为
(
(4)(
(5)设Z表示n只鸡蛋中重量大于60克的鸡蛋数( 则Z~B(n( 0.0228)(
因为
P(Y(1)(1(P(Y(1)(1((0.9772)n(
所以要使P(Y(1)(0.99( 只需
1((0.9772)n(0.99(
即 (0.9772)n(0.01(
解得 (
23( 设随机变量X具有概率分布律(
X
(3
(2
(1
0
1
2
3
4
5
pk
0.08
0.02
0.03
0.17
0.15
0.05
0.20
0.16
0.14
试求Y(X2的概率分布律(
解 Y的取值为0( 1( 2( 3( 4( 5( 其概率分布律为
P(Y(0)(P(X(0)(0.17(
P(Y(1)(P(X((1)(P(X(1)(0.03(0.15(0.18(
P(Y(2)(P(X((2)(P(X(2)(0.02(0.05(0.07(
P(Y(3)(P(X((3)(P(X(3)(0.08(0.20(0.28(
P(Y(4)(P(X(4)(0.16(
P(Y(5)(P(X(5)(0.14.
即
Y
0
1
2
3
4
5
pk
0.17
0.18
0.07
0.28
0.16
0.14
2( 甲、乙、丙3人进行独立射击( 每人的命中率依次为0(3( 0(4( 0(6( 设每人射击一次( 试求3人命中总数之概率分布律(
解 用X表示3人命中总数( 则X的取值为0( 1( 2( 3(
用A表示,甲命中”( B表示,乙命中”( C表示,命中”( 则
P(X(0)(P((A(B(C)(0(7(0(6(0(4(0(168(
P(X(1)(P(A(B(C)(P((AB(C)(P((A(BC)
(0(3(0(6(0(4(0(7(0(4(0(4(0(3(0(6(0(6(0(436(
P(X(2)(P(AB(C)(P(A(BC)(P((ABC)
(0(3(0(4(0(4(0(3(0(6(0(6(0(7(0(4(0(6(0(324(
P(X(3)(P(ABC)(0( 3(0( 4(0( 6(0(072(
X
0
1
2
3
pk
0(168
0(436
0(324
0(072
6( 设对某批产品的验收敛方案为( 从该批产品中随机地抽查5件产品( 若次品数小于等于1( 则该批产品通过验收敛( 否则不予通过( 若某批产品的次品率为0(05( 试求该批产品通过验收敛的概率(
解 用X表示5件产品中的次品数( 则X~B(5( 0(05)( 于该批产品通过验收敛的概率为
P(X(1)(P(X(0)(P(X(1)
(0(9774(
7( 某份试卷有10道选择题( 每题共有A( B( C( D四个答案供选择( 其中只有一个答案是正确的( 设某人对每道题均随机地选择答案( 试求该生10道题中恰好答对6道题的概率是多少？
解 用X表示10道题中答对的题目数( 则( 于是该生10道题中恰好答对6道题的概率是
(
12( 设随机变量X具有分布函数
(
试求( P(X((3)( ( ( (
解 P(X((3)(F((3)(0(
(
(
(
15( 设随机变量X具有概率密度
(
(1)求常数A(
(2)求X的分布函数(
(3)求X的取值落在区间内的概率(
解 (1)由( 得A(4.
(2)当x(0时( F(x)(0(
当0(x(1时( (
当x(1时( F(x)(1.
因此 (
(3) X的取值落在区间内的概率为
(
18( 设随机变量X~N(5( 4)( 试求( P(X(5)( P(3(X(6)( P(3(X(7)( P(|X|(1)以及常数C的范围( 使P(|X(5|(C)(0(99(
解 (
(((0.5)((((1)(((0.5)([1(((1)]
(0.6915([1(0.8413](0.5328,
(((1)((((1)(2((1)(1(2(0.8413(1(0.6826.
P(|x|(1)(1(P(|X|(1)(1(P((1(X(1)
(1((((2)((((3)(1([1(((2)]([1(((3)]
(((2)(((3)(1(0.9772(0.9987(1(0.9785.
(
要使P(|X(5|(C)(0.99( 只需( 即( 查表得( 故C(5.16.
20( 设某批鸡蛋每只的重量X(以克计)服从正态分布( X ~N(50( 25)(
(1)求从该批鸡蛋中任取一只( 其重量不足45克的概率(
(2)从该批鸡蛋中任取一只( 其重量介于40克到60克之间的概率(
(3)若从该批鸡蛋中任取五只( 试求恰有2只鸡蛋不足45克的概率(
(4)从该批鸡蛋中任取一只其重量超过60克的概率(
(5)求最小的n( 使从中任选n只鸡蛋( 其中至少有一只鸡蛋的重量超过60克的概率大于0(99(
解 (1)(
(2)
(2(0.9772(1(0.9544.
(3)设Y为5只鸡蛋中重量不足45克的鸡蛋数( 则Y~B(5( 0.1587)( 故所求概率为
(
(4)(
(5)设Z表示n只鸡蛋中重量大于60克的鸡蛋数( 则Z~B(n( 0.0228)(
因为
P(Y(1)(1(P(Y(1)(1((0.9772)n(
所以要使P(Y(1)(0.99( 只需
1((0.9772)n(0.99(
即 (0.9772)n(0.01(
解得 (
23( 设随机变量X具有概率分布律(
X
(3
(2
(1
0
1
2
3
4
5
pk
0.08
0.02
0.03
0.17
0.15
0.05
0.20
0.16
0.14
试求Y(X2的概率分布律(
解 Y的取值为0( 1( 2( 3( 4( 5( 其概率分布律为
P(Y(0)(P(X(0)(0.17(
P(Y(1)(P(X((1)(P(X(1)(0.03(0.15(0.18(
P(Y(2)(P(X((2)(P(X(2)(0.02(0.05(0.07(
P(Y(3)(P(X((3)(P(X(3)(0.08(0.20(0.28(
P(Y(4)(P(X(4)(0.16(
P(Y(5)(P(X(5)(0.14.
即
Y
0
1
2
3
4
5
pk
0.17
0.18
0.07
0.28
0.16
0.14
