概率论试题一与答案
《概率论》试卷一与答案
一.填空题
1. 设P (A ) =0. 4, P (A B ) =0. 7,若事件A 与B 互不相容,则P (B ) =A 与B 相互独立,则P (B ) = .
2. 已知在10件产品中有2件次品,在其中取两次,每次任取一件,作不放回抽样,则第二次取出的是次品的概率 .
1113. 三人独立地破译一密码,他们能单独破译出的概率分别为、、,则此密码被破译出的概率345
为 .
4. 设X ~N (3, 22), Φ(3. 5) =0. 9998, 则P (-4
5. 若随机变量X 服从泊松分布,且P {X =1}=P {X =2},则P {X =4}=
6.设随机变量X ~U (0, 5) , 则关于y 的二次方程y 2+4y +X =0有实根的概率⎧e -x , x ≥07. 设随机变量X 的概率密度函数为f (x ) =⎨,则随机变量Y =2X 的概率密度函数0, x
f Y (y ) = .
(写明分布名称及参数) .
(, Y ) 的分布函数9. 设X , Y 是两个独立同分布的随机变量,其分布函数为F (x ) ,则Z =m ax X 8. 设随机变量X , Y 分别服从N (1, 4) 和N (0, 9) , 且X 与Y 相互独立,令Z =2X -Y , 则Z 服从分布
F max (z ) = .
10. 设X 为非负随机变量,E (1211X -1) =2, D (X -1) =, ,则E (X ) =222
11.设随机变量X 的方差为25,根据契比雪夫不等式, 有P {|X -E (X ) |
2一.填空题答案 1.0.3, 0.5 2.0.2 3. 0.6 4.0.9996 5.e -2 6. 3
1- 7.f Y (y ) =e 2, y ≥0 8.N (2, 25) 9. F 2(z ) 10. 11. 2 y 二.将两信息分别编码0和1传递出去,接收站收到时0被误收作1的概率为0.02,而1被误收作
0的概率为0.01,信息0和信息1传送的频繁程度为2:1,求(1) 接收站收到信息0的概率;(2)
若接收站收到的信息是0,则原发信息是0的概率。
⎧a +bx 2, 0
1(2) X 的分布函数F (x ) ; (3) P {-1
《概率论》第 1 页 共 3 页
四.(本题10分)设随机变量X 与Y 独立同分布,X 的分布律为
X 12
P 33
记U =max{X , Y },V =min{X , Y },求(U , V ) 的联合分布律.
五.(本题16分)设二维随机变量(X , Y ) 的密度函数为
f (x , y ) =⎧⎨8xy , 0≤x ≤1, 0≤y ≤x
⎩0, 其他,
求 (1) X , Y 的边缘概率密度f X (x ), f Y (y ) ;
(2) P {Y ≥X 2};
(3) 当X =1
3时,Y 的条件密度函数f Y X (y |x =1
3) ;
(4) E (X ), E (Y ), E (XY ) 及Cov (X , Y ) .
六.(本题10分)某保险公司多年的统计资料表明,在索赔户中被盗索赔户中占20%,以在随意抽查的100个索赔户因被盗向保险公司索赔的户数.
(1)写出X 的概率分布律;
(2)利用中心极限定理,求被盗索赔户不少于14户且不多于30户的概率的近似值.
(答案)
二. 设A 为“接收站收到信息0”,B 为事件“原发信息是0”,
(1)P (A ) =P (B ) P (A B ) +P () P (A |=0. 98⨯21197
3+0. 01⨯3=300;„„„„4分
(2) P (B A ) =P (B ) P (A B ) 0. 98⨯2
P (A ) =3
0. 98⨯21=196
197. „„„„„„„„„6分
3+0. 01⨯3
⎧1
三.(1)由题意,得⎪⎨⎰0(a +bx 2) dx =1⇒⎧a =
⎪⎩⎰12) dx =0. 6⎨0. 6
⎩b =1. 2 „„„„„„ 4分
0x (a +bx
⎧0, x
(2) F (x ) =⎰x
-∞f (t ) dt =⎪⎨0. 6x +0. 4x 3, 0≤x
(3)P {-1
2=F (1
2) -F (-1) =F (1
2=0. 35 „„„„„„„„„„„„ 2分
《概率论》第 2 页 共 3 页 X 表示
四. (U , V ) 有三个可能值(1, 1), (2, 1), (2, 2) ,„„„„„„„„„„„„2分
4而P (U =1, V =1) =P (X =1, Y =1) =P (X =1) P (Y =1) =,„„„„„„„„„ 2分 9
P (U =2, V =1) =P (X =1, Y =2) +P (X =2, Y =1)
4 =P (X =1) P (Y =2) +P (X =2) P (Y =1) =„„„„„„„„„ 2分 9
1P (U =2, V =2) =P (X =2, Y =2) =P (X =2) P (Y =2) = „„„„„„„„„ 2分 9
即 12
14/90
24/91/9
x 3⎧+∞⎪⎰08xydy =4x , 0≤x ≤1五. (1) f X (x ) =⎰f (x , y ) dy =⎨ „„„„ 2分 -∞⎪0,其它⎩
⎧18xydx =4y -4y 3, 0≤y ≤1+∞⎪f Y (y ) =⎰f (x , y ) dx =⎨⎰y ;„„„„„„„„„„„2分 -∞⎪0其它⎩
1x 112(2) P {Y ≥X }=⎰⎰f (x , y ) dxdy =⎰dx ⎰28xydy =⎰(4x 3-4x 5) dx =;„„„„4分 0x 03y ≥x 2
⎧2y f (x , y ) ⎪2, 0≤y ≤x (3)当0
111当x =时, f Y |X (y |x =) =18y , 0≤y ≤ „„„„„„„„„„„„„„2分 333
1x 142(4)E (X ) =⎰dx ⎰8x y dy =⎰4x 4dx =, „„„„„„„„„„„„„„„„1分 0005
1x 18E (Y ) =⎰dx ⎰8xy 2dy =⎰(4y 2-4y 4) dy = „„„„„„„„„„„„„1分 00015
1x 4E (XY ) =⎰dx ⎰8x 2y 2dy = „„„„„„„„„„„„„„„„„„„1分 009
4Cov (X , Y ) =E (X ) E (Y ) -E (XY ) =- „„„„„„„„„„„„„„„1分 225
六. (1)由题意,X ~b (100, 0. 2) ,
k 则P {X =k }=C 1000. 2k 0. 8100-k , k =0, 1, , 100 „„„„„„„„„„„„„„ 4分
(2) E (X ) =np =20, D (X ) =np (1-p ) =16 „„„„„„„„„„„„„„„ 2分 根据棣莫佛—拉普拉斯定理,
14-20X -2030-20P {14≤X ≤30}=P ≤≤ „„„„„„„„ 2分 X -20=P {-1. 5≤≤2. 5} ≈Φ(2. 5) -Φ(-1. 5) =0. 927 „„„„„„„„„„„ 2分
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