BI TP L THUYT XC SUT CHNG 2NHM 4BI LM:
Bi 2.1:a)X : Khi lng 3 sn phmX = { 6,7,8,9}C 3 sn phm u xu: P (X
= 6) = = 2 sn phm xu, 1 sn phm tt : P (X = 7) = = 1 sn phm xu, 2 sn
phm tt : P (X = 8) = = C 3 sn phm u tt: P (X = 9) = = Ta c bng phn
phi xc sut ca X :X6789
P(X)
b)K vng : E (X) = 6 . + 7 . +8 . + 9. = 7,8Phng sai : D(X) =( 62
. + 72 . +82 . + 92. ) 7,82 0,56 lch chun: (X)Bi 2.2:a) X : S vin
phn mu ly cX = { 0,1,2,3,4}Trong 3 vin phn ly ra:_Khng c vin phn mu
no : P (X = 0) = = _C 1 vin phn mu: P (X = 1) = = _C 2 vin phn mu:
P (X = 2) = = _C 3 vin phn mu: P (X = 3) = = Ta c bng phn phi xc
sut ca X :X0123
P(X)
b) Hm phn phi xc sut ca X:F ( X) = c) K vng : E (X) = 1 . + 2 .
+ 3. = Phng sai : D(X) =( 1 . + 22 . + 3. ) 2 = lch chun: (X) = Bi
2.3:a)X : s ph phm trong 4 sn phm ly raX = { 0,1,2,3,4}C 4 sn phm u
tt : P (X =0) = =
1 ph phm, 3 sn phm tt : P (X = 1) = = 2 ph phm, 2 sn phm tt : P
(X = 8) = = 3 ph phm, 1 sn phm tt : P (X = 8) = = 4 ph phm: P (X =
9) = = Ta c bng phn phi xc sut ca X :X01234
P(X)
b)K vng : E (X) = 1 . Phng sai : D(X) =(1 . ) 12 lch chun: (X) =
Mod (X) = 1 : Med(X) =1
Bi 2.4:t gi thit trong trng hp khng hon li:Xc sut anh ta chn c 1
bi l: =0,2Xc sut anh ta chn c 1 bi trng l: =0,8Xc sut anh ta chn c
bi l rt t. Theo lut phn phi xc sut, nu anh ta chi cng nhiu ln th xc
sut chn c bi cng nh v chn c bi trng cng ln.Do vy, anh ta khng nn
chi tr ny nhiu ln.
Bi 2.5:a)Gi X l s trn thng ca i tuyn.X={0,1,2,3}Ta c bng phn phi
xc sut nh sau:X0123
P(X)0,1680,4360,3240,072
T , ta c Hm phn phi xc sut ca X:F(x) = b)Xc sut i tuyn khng thng
trn no: P(X=0)=0,6*0,7*0,4=0,168Xc sut i tuyn thng t nht mt trn:
P=1-0,168=0,832Bi 2.6:a)X l s mu nhm A chn c. Ta c bng phn phi xc
sut ca X:X0123
P(X)0,30,5
b)T , ta c hm phn phi xc sut ca X:(x) =c)K vng EX ==1,8Phng sai:
DX=E(X-EX)2=E(X2) (EX)2=0,56 lch chun: X:= =
Bi 2.7:a. Gi X l s ln bn trng. => ta c bng phn phi xc sut ca
X:X012
P0.360.480.16
b. Nu tham gia cuc thi ny th s tin m nhiu kh nng c nht l khi bn
trng 1 vin v trt mt vin tc l 10000.c. Gi Y l s tin ngi nhn c =>
ta c bng phn phi xc sut ca Y:Y-200001000040000
P0.360.480.16
tnh S tin trung bnh m 1 ngi tham gia cuc thi ta s dng K vng.Ta c
E(X) = -20000*0.36 + 10000*0.48 + 40000*0.16 = 4000Bi 2.8:a. Gi X l
s ln gp n ca ngi => ta c bng phn phi xc sut cu X:X01234
P
b. Ta c E(X) = 1* + 2* +3* + 4* = 1 => s ln trung bnh ngi gp
n l 1 ln. Bi 2.9:a. u tin ta lp bng ghi cc gi tr v xc sut tng
ngXY-112
-1-201
1023
2134
Bng 1: Gi tr ca tng X + Y
XY0.20.30.5
0.40.080.120.2
0.30.060.090.15
0.30.060.090.15
Bng 2: Xc sut tng ng.
XY-112
-11-1-2
1-112
2-224
Bng 3: Gi tr ca tch X*Y
X + Y:T bng 1 v bng 2 ta c:X+Y = { -2; 0; 1; 2; 3; 4 }P (X+Y =
-2) = 0.08;P (X+Y = 0) = 0.06 + 0.12 = 0.18;P (X+Y = 1) = 0.2 +
0.06 = 0.26;P (X+Y = 2) = 0.09;P (X+Y = 3) = 0.15 + 0.09 =0.24P
(X+Y = 4) = 0.15;Bng phn phi xc sut ca X + Y l:X+Y-201234
P0.080.180.260.090.240.15
X*Y:T bng 3 v bng 2 ta c:X*Y = { -2; -1; 1; 2; 4}P (XY = -2) =
0.2 + 0.06 = 0.26;P (XY = -1) = 0.12 + 0.06 = 0.18;P (XY = 1) =
0.08 + 0.09 = 0.17;P (XY = 2) = 0.15 + 0.09 = 0.24;P (XY = 4) =
0.15;=> Bng phn phi xc sut ca X*Y l:X*Y-2-1124
P0.260.180.170.240.15
b. E(X + Y) = -2*0.08 +0*0.18 + 1*0.26 + 2*0.09 + 3*0.24 +
4*0.15 = 1.6E[(X + Y)^2] = (-2)^2*0.08 + 0*0.18 + 1^2*0.26 +
2^2*0.09 + 3^2*0.24 + 4^2*0.15 = 6.124=> D(X + Y) = 6.124^2
1.6^2 = 34.94
=> = = Tng t ta c:E(X * Y) = 0.55E[(X * Y)^2] = 4.75=> D(X
* Y) = 4.75^2 0.55^2 = 22.26
=> = = Bi 2.10:a/V f(x) l hm mt sn xut, do :
1= =
m= b/
0 , x[0,2]
Ta c : f(x) = , x [0,2]
Cn trng cht trc khi n c 1 thng tui 0< X < 1
P( 0< X < 1) = = = c/Tui th trung bnh ca loi cn trng cng
chnh l k vng ca X.
E(X) = = = ( thng)Bi 2.11:
a/V f(x) l hm mt sn xut, do :
1= =
b/
0 , x [ 0,30]Ta c: f(x) =
, x [ 0,30]
Nhu cu hng khng vt qu 12 ngn sp trong 1 nm tc l 0 < X <
12
P( 0< X < 12) = = =
c/ Nhu cu trung bnh hng nm v loi hng cng l k vng ca X
E(X) = = = 10 ngn sn phmBi 2.12:a/0 nu x 0F(x) = ax2 nu 0 x
1
1 nu x > 1
V F(x) lin tc ti 1 nn ta c:
= F(1)
a -1 = 1
a=2Khi hm mt xc sut ca X s l:
0 nu x [0,1]f(x) = F (x) =
4x nu x [0,1]
b/Thi gian xp hng trung bnh cng l k vng ca X
E(X) = = 4 pht
Bi 2.13:a, F(x) l hm mt xc sut nn ta c:
= => k=b, E(X)=dx=1Var(X)= dx=
Bi 2.14:a, t u=x => du=dx dv=e-bxdx => v=-e-bx=>E(X)=
b, F(X)=vi x ta c F(x)= Vi x>0 ta c E(X)=
Hm phn phi xc sut : F(x)=c, Tui th TB ca mt loi mch in 5 nm
F(X)=5=> F(x)=P(0)= =1-
Bi 2.15:a, Ta c E(X Y) = E(X) E(Y)=2=> X, Y c lpb, E(X-Y)=
3-2=1E(Y-4)=-2Var(X)= = 1 =>= 1+1=2Bi 2.16 :a,Theo tnh cht hm mt
xc sut ta c:F(x)=Vi xta c F(x)=Vi 1 ta c F(x)== Vi x>1 ta c
F(x)=Vy hm phn phi xc sut l: F(x)=
b, E(X)=E(X2)=Var(X)=E(X2)-E((X))2 = lch chun x= F(x)= 3x2 x
F(x)=6x F(x)=0 => x=0F(max)=3 vi x=1Do : Mod= 1Bi 2.17:a) Da vo
bng ta c:P(X=1, Y=1)=O,10P(X=1).P(Y=1)=O,55.0,15=0,0825Ta
thy:P(X=1,Y=1)P(X=1).P(Y=1)X v Y khng c lp.b)Tham s c trng:- K
vng:E(X)=1. 0,55+2. 0,45=1,45E(Y)=1. 0,15+3. 0,55+7. 0,3=3,9- Phng
sai:E(X2)=12. 0,55+22.
0,45=2,35V(X)=E(X2)-[E(X)]2=2,35-(1,45)2=0,2475E(Y2)=12. 0,15+32.
0,55+72. 0,3=19,8V(Y)=E(Y2)-[E(Y)]2=19,8-(3,9)2=4,59- lch
chun:(X)=()=,2475(Y)=()=4,59-Mt:M0(X)=1M0(Y)=3Bi 2.18:a) Da vo bng
ta c:P(X=1,Y=2)=0,12P(X=1).P(Y=2)=0,3. 0,4=0,12Ta thy:P(X=1,
Y=2)=P(X=1).P(Y=2)=0,12X v Y c lp.b)K vng:E(X)=1. 0,3+2. 0,5+3.
0,2=1,9E(Y)=2. 0,4+5. 0,6=3,8- Phng
sai:E(X2)=12.0,3+22.0,5+32.0,2=4,1V(X)=4,1-(1,9)2=0,49E(Y2)=22.0,4+52.0,6=16,6V(Y)=E(Y2)-[E(Y)]2=16,6-(3,8)2=2,16-
lch chun:(X)=()=0,49=0,7(Y)=2,16-Mt:m0(X)=2m0(Y)=5Bi 2.19:Gi i,j l
s nt xut hin ln th nht v th hai = { (i,j) , 1i6 , 1j6 }X() = Y() =
{1,2,3,4,5,6}P(X=i, Y=i) = = P(X=i).P(Y=i) , vi mi i,jVy X v Y c
lp
Bi 2.20:1. Ta c: X() = Y() = {0,1,2,3} X~ B(3, ) Y~ B(3, ) X0
123
P1/83/83/81/8
Y0123
P1/83/83/81/8
(S dng phn phi nh thc tnh P)1. E(X) = E(Y) = 1,5 D(X) = D(Y) =
0,75E(XY) = 1,5R(X,Y) = = -1Vy X v Y tng quan nghch dng hm tuyn
tnhBi 2.21:a. X013
P0,20,60,2
Y-1012
P0,180,220,40,2
b.=E(X)= 0.0,2 + 1.0,6 + 3.0,2 = 1,2DX = Var(X) = = 0,96Tng t:
E(Y) = 0,62 DY = 0,9956c. Ta c: P(X=0,Y=1) = 0,1 P(X=0).P(Y=1) =
0,2.0,4 = 0,08Suy ra X v Y l 2 i lng ngu nhin ph thucE(XY) = =
1.1.0,3 + 1.2.0,1 + 3.(-1).0,08 + 3.2.0,1 = 0.86Hip phng sai:
Cov(X,Y) = E(XY) E(X)E(Y) = 0,116R(X,Y) = = 0,1345
a) Bc ln1Bc ln 2Bc ln 3
= 3/6XX
Xanh = 2/6 = 3/5X
Xanh = 1/5 =
Vng =
Vng = 1/5 =
Xanh =
Vng = 1/6 = 3/5X
Xanh = 2/5 =
Xanh =
Gi X = S bi xanh ly ra => X = Y = S bi vng ly ra => Y =
Bng phn phi xc sut ng thi ca X v Y l : YX01
01/10
11/51/10
21/201/20
b)
Hip phng sai :
H s tng quan l:
Ma trn tng quan:
Bi 2.26:b) Bng phn phi xc sut l ca trnh hc vn v ca la tui: YX(
25 35 )30( 35 55 )45( 55 100 )70PY
00,010,020,050,08
10,030,060,10,19
20,180,210,150,54
30,070,080,040,19
PX0,290,370,341
a) Bng phn phi xc sut ca trnh hc vn nhng ngi tui 30: X0123
Bi 2.27:a) YX10203040PY
100,20,040,0100,25
200,10,360,0900,55
3000,050,100,15
400000,050,05
PX0,30,450,110,051
b) Bng phn phi xc sut ca Y vi iu kin X=20:Y10203040
0
