Chap1_Dai cuong ve Xac suat(handout).pdf

I CNG V XC SUT

ThS. Nguyn Th Nhn

MathematicsUniversity of [email protected]

Ngy 30 thng 3 nm 2015

1 / 31 Nhan N.T. I CNG V XC SUT

BIN C V XC SUT


Bin c v xc sut

Php th ngu nhin (random experiment)

L s thc hin mt nhm cc iu kin xc nh (lm th nghim) v tac th lp li nhiu ln. Kt qu ca php th ta khng xc nh trcc.

V d

Php th ngu nhin Kt qu

Tung ng xu Mt sp, mt nga

im thi kt thc mn hc {0, 1, 2, . . . , 10}Tung xc xc {1, 2, 3, 4, 5, 6}


Bin c ngu nhin

Tp hp tt c cc kt qu c th xy ra khi thc hin php th gil khng gian mu hay khng gian cc bin s s cp (samplespace).

Mi kt qu ca php th ngu nhin ( ) gi l bin s scp (simple event).

Mt tp con ca khng gian mu c nhiu bin c c gi l binc ngu nhin (event).K hiu A, B, C,. . .

Bin c lun xy ra khi thc hin php th l bin c chc chn.Khiu

Bin c lun khng xy ra khi thc hin php th l bin c bt kh(empty event). K hiu


Bin c ngu nhin

V d

Gieo xc xc 1 ln. Gi i ="Xut hin mt k chm" (k = 1, 6)

Khng gian cc bin c s cp: = {1, 2, . . . , 6} = {1, 2, . . . , 6}Bin c ngu nhin:A = {1, 3, 5} = "chm l"B = {2, 4, 6} = "chm chn"


Quan h gia cc bin c

Quan h ko theo

A ko theo B, k hiu A B, nu A xy ra th B xy ra. Hay A l binc thun li cho B.

V d

im thi kt thc hc phn Xc sut thng k ca SV Nguyen Van A.GiB={SV t im 8}C={SV u}Ta thy SV t im 5 th u. Do B xy ra th C xy ra, hayB C .


Quan h gia cc bin c

Quan h tng ng

A tng ng B, k hiu A = B, nu A xy ra th B xy ra, v ngc li.

V d

Tung 2 con xc xc. GiA={Tng s chm xut hin trn 2 xc xc l s chn.}B={S chm xut hin trn 2 xc xc l cng chn hoc cng l.}Khi : A = B


Cc php ton trn bin c

Tng

Bin c tng ca A v B, k hiu A + B (hay A B), l bin c xy ranu c t nht mt trong hai bin c A, B xy ra.



Tch

Bin c tch ca A v B, k hiu A.B (hay A B), l bin c xy ra nuA v B cng ng thi xy ra.



Hiu

Bin c hiu ca A v B, k hiu A \ B, l bin c xy ra nu A xy ranhng B khng xy ra.



Xung khc

Bin c A xung khc B (mutually exclusive) nu A v B khng ng thixy ra. K hiu A.B =

Dy cc bin c A1,A2, . . . ,An c gi l xung khc tng i mt nuAi .Aj = ,i 6= j



Bin c i lp

Bin c i lp (bin c b complement) ca A, k hiu A, l bin c xyra khi A khng xy ra v ngc li (A = \ A).{

A + A = A.A =



H y cc bin c

H y cc bin c (exhaustive) ca dy n cc bin c A1,A2, . . . ,Anc gi l mt h y cc bin c nu:{

Ai .Aj = ,i 6= j , i , j = 1, nA1,A2, . . . ,An =


Quan h gia cc bin c

V d

Kim tra cht lng 3 sn phm. t Ak l s kin sn phm th k tt(k = 1, 3). Hy biu din Ak qua cc s kin sau:

Tt c 3 sp u tt

Tt c 3 sp u xu

C t nht 1 sp tt

C t nht 1 sp xu

C ng 1 sp tt


Quan h gia cc bin c

Bi gii

A={Tt c cc sp u tt}=A1A2A3

B={Tt c cc sp u xu}=A1A2A3

C={C t nht 1 sp tt}=A1 + A2 + A3

D={C t nht 1 sp xu}=A1 + A2 + A3

E={C ng 1 sp tt}=A1A2A3 + A1A2A3 + A1A2A3


Khi nim v cc nh ngha v xc sut

Khi nim v xc sut

Xc sut ca bin c A l mt con s, s c trng cho kh nngxut hin ca bin c A trong php th tng ng. K hiu P(A)

P(A) cng ln (cng gn 1) th kh nng xut hin A cng nhiu.

P(A) cng nh (cng gn 0) th kh nng xut hin A cng t i.


Khi nim v cc nh ngha v xc sut

Tin Kolmogorov

Xc sut ca s kin A l 1 s khng m: P(A) 0Xc sut ca s kin chc chn bng 1: P() = 1

Xc sut ca tng hai s kin xung khc bng tng cc xc suttng ng:

P(A + B) = P(A) + P(B)

P( i=1

Ai)

=i=1

P(Ai )


nh ngha

Nu trong mt php th c tt c n bin c s cp ng kh nng, ngha

l P(1) = P(2) = . . . = P(n) =1

n, trong c m bin c thun li

cho bin c A th xc sut ca A:

P(A) =m

n

Tnh cht ca xc sut

Nu A B th P(A) P(B)Ta c B = A + B \ A v A.(B \ A) = nnP(B) = P(A) + P(B \ A)0 P(A) 1P(A) = 1 P(A)Do A + A = v A.A = nnP(A + A) = P(A) + P(A) = P() = 1

P() = 0


V d

Mt hp c 3 qu cu trng v 5 qu cu ging ht nhau v kchthc. Ly ngu nhin 3 qu cu t hp . Tnh xc sut c

3 qu

2 qu trng v 1 qu

Bi gii

Tng s qu cu l 8, mi cch ly ra 3 qu ng vi t hp chp 3 ca 8phn t. Do [] = C 38

A={ Ly c 3 qu }Bin c s cp thun li cho A l chn 3 trong s 5 qu cu C 35 .

Vy P(A) =C 35C 38

B={ Ly c 2 qu trng v 1 qu }.

Tng t, ta c P(B) =C 23C

15

C 38


Cc cng thc tnh xc sut c bn

Cng thc cng xc sut

Cc bin c ty

A, B ty : P(A+B)=P(A)+P(B)-P(AB)A1,A2, . . . ,An:

P( n

i=1

Ai)=

ni=1

P(Ai )

1i

V d

Trong s 300 sv nm I c 100 sv bit ting Anh, 90 sv bit ting Php,20 sv bit c 2 ngoi ng Anh-Php. Chn ngu nhin 1 sv. Tnh xcsut sv ny bit t nht 1 ngoi ng.

Bi gii

t:A={sv ny bit ting Anh}B={sv ny bit ting Php}C={sv ny bit t nht 1 ngoi ng}

P(C ) = P(A + B) = P(A) + P(B) P(AB) = 100300

+90

300 20

300=

17

30



Xc sut c iu kin

Cho 2 bin c A v B vi P(B) > 0. Xc sut ca bin c A vi iu kinbin c B xy ra l

P(A|B) = P(AB)P(B)

Cc tnh cht

0 P(A|B) 1P(B|B) = 1Nu A.C = th P[(A + C )|B] = P(A|B) + P(C |B)P(A|B) = 1 P(A|B)



Cng thc nhn xc sut

Vi cc bin c A v B ty

P(AB) = P(A|B)P(B) = P(B|A)P(A)

Dng qui np ta c th tng qut ha cho trng hp Ai (i = 1, n)l h n bin c:

P(A1A2 . . .An) = P(A1)P(A2|A1)P(A3|A1A2) . . .P(An|A1A2 . . .An1

Trong trng hp bin c c lp

P(AB) = P(A)P(B)

hay

P(n

i=1

Ai ) =n

i=1

P(Ai )


V d

Mt ngi c 3 con g mi, 2 con g trng nht chung mt chung. Mtngi n mua, ngi bn bt ngu nhin ra mt con, v ngi chpnhn mua.

Tnh xc sut ngi mua c con g mi?

Ngi th hai n mua, ngi bn li bt ngu nhin ra mt con.Tm xc sut ngi th hai mua c g trng, bit ngi th nhtmua c g mi.

Bi gii

t:Ai={Ngi th i mua c g mi}, i = 1, 2

P(A1) =3

5= 0.6

P(A2|A1) = 24

= 0.5


V d

Trong mt trm cp cu c 80% bnh nhn phng do nng v 20% bphng do ha cht. Loi phng do nng c 30% b bin chng, cn hacht th 50%.

Tnh xc sut khi bc s m tp h s th gp 1 bnh n ca bnhnhn b bin chng?

Bit gp bnh n ca bnh nhn b bin chng. Tnh xc sut donng gy nn?


Bi gii

t:A={Bnh nhn b bin chng}A1={Bnh nhn phng do nng}A2={Bnh nhn phng do ha cht}Ta thy A1,A2 lp thnh 1 nhm y .

Dng cng thc xc sut y

P(A) = P(A1)P(A|A1) + P(A2)P(A|A2)= 0.8 0.3 + 0.2 0.5 = 0.34 = 34%

Theo cng thc Bayes

P(A1|A) = P(A1)P(A|A1)P(A)

=0.8 0.3

0.34= 0.705 = 70.5%


V d

T l bnh B ti mt a phng bng 0.02. Dng mt phn ng gipchn on, nu ngi b bnh th phn ng dng tnh 95%, nu ngikhng b bnh th phn ng dng tnh 10%.

Tm xc sut dng tnh ca phn ng?

Mt ngi lm phn ng thy dng tnh, tm xc sut ngi lngi b bnh?

Tm xc sut chn on ng ca phn ng?


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