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8/2/2019 gtxstk_dhnn1

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B GIO DC V O TOTRNG I HC NNG NGHIP I

**********************

Ths.L C VNH

GIO TRNH

XC SUT THNG K

H NI - 2006


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Trngi hc Nng nghip H Ni Gio trnh Gio trnh Xc sut thng k..1

Chng 1 : Php th. Skin

Nhng kin thc v gii tch t hp sinh vin c hc trong chng trnh phthng. Tuy nhin gip ngi hc d dng tip thu kin thc ca nhng chng k tip

chng ti gii thiu li mt cch c h thng nhng kin thc ny. Php th ngu nhinv s kin ngu nhin l bc khi u ngi hc lm quen vi mn hc Xc sut.Trong chng ny chng ti trnhby nhng kin thc ti thiu v s kin ngu nhin,cc php ton v cc s kin ngu nhin, h y cc s kin ng thi ch ra cchphn chia mt s kin ngu nhin theo mt h y . Nhng kin thc ny l cn thit ngi hc c th tip thu tt nhng chng tip theo.

I. Gii tch thp

1.Qui tc nhn: Trong thc t nhiu khi hon thnh mt cng vic, ngi taphi thchin mt dy lin tip k hnh ng.

Hnh ng th nht: c 1 trong n1 cch thc hinHnh ng th hai: c 1 trong n2 cch thc hin

. . .. . . . . .. .. . . . . . . .. . . . . . . . . .. .. .. . . .

Hnh ng th k: c 1 trong nk cch thc hin

Gi n l s cch hon thnh cng vic ni trn, ta c:

n = n1n2..nk

Qui tc trn gi l qui tc nhn.

V d: i t thnh ph A ti thnh ph Cphi qua thnh ph B. C mt trong bnphng tin i t A ti B l: ng b, ng st, ng khng v ng thu. C

mt trong hai phng tin i t B ti C l ng b v ng thu. Hi c bao nhiucch i t A ti C?

thc hin vic i t A ti C taphi thc hin mt dy lin tip hai hnh ng.

Hnh ng th nht: chn phng tin i t A ti C c n1= 4 cch

Hnh ng th hai: chn phng tin i t B ti C c n2 = 2 cch

Vy theo qui tc nhn, s cch i t A ti C l n= 4.2 = 8 cch

2.Qui tc cng: hon thnh cng vic ngi ta c th chn mt trong k phng n.

Phng n th nht: c 1 trong n1 cch thc hin

Phng n th hai: c 1 trong n2 cch thc hin

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

Phng n th k: c 1 trong nk cch thc hin

Gi n l s cch hon thnh cng vic ni trn, ta c:

n = n1 + n2 +. . . ..+ nk


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Qui tc trn gi l qui tc cng

V d: Mt t sinh vin gm hai sinh vin H Ni, ba sinh vin Nam nh v ba sinhvin Thanh Ho. Cn chn hai sinh vin cng tnh tham gia i thanh nin xung kch.Hi c bao nhiu cch chn.

Phng n th nht: Chn hai sinh vin H Ni c n1= 1 cch

Phng n th hai: Chn hai sinh vin Nam nh c n2= 3 cchPhng n th ba: Chn hai sinh vin Thanh Ho c n3= 3 cch

Theo qui tc cng ta c s cch chn hai sinh vin theo yu cu:

n = 1 + 3 + 3 = 7 cch

3.Hon vTrc khi a ra khi nim mt hon v ca n phn t ta xt v d sau:.

V d: C ba hc sinh A,B,C c sp xp ngi cng mtbn hc. Hi c bao nhiucch sp xp?

C mt trong cc cch sp xp sau:ABC, ACB, BAC, BCA, CAB, CBA.

Nhn thy rng: i ch bt k hai hc sinh no cho nhau ta c mt cch sp xpkhc. T mt cch sp xp ban u, bng cch i ch lin tip hai hc sinh cho nhau tac th a v cc cch sp xp cn li. Mi mt cch sp xp nh trn cn c gi lmt hon v ca ba phn t A, B, C. Tng qut vi tp hp gm n phn t ta c nhngha sau:

3.1nh ngha: Mt hon v ca n phn t l mt cch sp xp c th t n phn t .

3.2 S hon v ca n phn t: Vi mt tp gm n phn t cho. S tt c cc hon vca n phn t k hiu l Pn.Ta cn xy dng cng thc tnh Pn.

to ra mt hon v ca n phn t taphi thc hin mt dy lin tip n hnh ng.Hnh ng th nht: Chn 1 phn t xp u c n cch chn

Hnh ng th hai: Chn 1 phn t xp th 2 c n-1 cch chn

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

Hnh ng cui: Chn phn t cn li xp cui c 1 cch chn

Theo qui tc nhn, s cch to ra 1 hon v ca n phn t l

Pn = n.(n-1) ....2.1= n!

4. Chnh hp khng lp

4.1 nh ngha: Mt chnh hp khng lp chp k ca n phn t l mt cch sp xp cth t gm k phn t khc nhau ly t n phn t cho.

V d: C 5 ch s 1, 2, 3, 4, 5. Hy lp tt c cc s gm 2 ch s khc nhau

Cc s l: 12, 13, 14, 15, 21, 23, 24, 25, 31, 32, 34, 35, 41, 42, 43, 45, 51, 52, 53, 54.

Mi mt s trn chnh l mt cch sp xp c th t gm hai phn t khc nhau ly tnm phn t l nm ch s cho. Vy mi s l chnh hp khng lp chp hai ca nmphn t.


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4.2 S cc chnh hp khng lp: S cc chnh hp khng lp chp k ca n phn t k hiul knA . Ta xy dng cng thc tnh

knA .

to ra mt chnh hp khng lp chp k ca n phn t ta phi thc hin mt dy lintip k hnh ng.

Hnh ng th nht: chn 1 trong n phn t xp u: c n cch

Hnh ng th hai: chn 1 trong n-1 phn t xp th 2: c n -1 cch

. . . . . . . .. . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . .

Hnh ng th k: chn 1 trong n-k+1 phn t xp cui: c n-k+1 cch

Theo qui tc nhn: S cch to ra mt chnh hp khng lp chp k ca n phn t l :knA = n(n-1).. ....(n-k+1)

d nhta s dng cng thc sau:

)!kn(

!n

1.2)......kn(

1.2).......kn().1kn)...(1n.(n)1kn)....(1n.(nA kn

=

+=+=

5. Chnh hp lp: hiu th no l mt chnh hp lp ta xt v d sau:V d: Hy lp cc s gm 2 ch s t 4 ch s: 1, 2, 3, 4.

Cc s l: 11, 12, 13, 14, 21, 22, 23, 24, 31, 32, 33, 34, 41, 42, 43, 44.

Mi s trong cc s ni trn l mt cch sp xp c th t gm hai ch s, mi ch sc th c mt n hai ln ly t bn ch s cho. Mi cch sp xp nh vy cn gi lmt chnh hp lp chp hai ca bn phn t. Tng qut ho ta c nh ngha sau:

5.1 nh ngha: Mt chnh hp lp chp k ca n phn t l mt cch sp xp c th tgm k phn t m mi phn t ly t n phn t cho c th c mt nhiu ln.

5.2 S cc chnh hp lp chp k:

S cc chnh hp lp chp k ca n phn t c k hiu l knA . Ta s a ra cng thc

tnh knA .

to ra mt chnh hp lp chp k ca n phn t taphi thc hin mt dy lin tip khnh ng.

Hnh ng th nht: chn 1 trong n phn t xp u c n cch

Hnh ng th hai: chn 1 trong n phn t xp th 2 c n cch

. . . . . . . . . . . . . . . . .. . . . . . . . . .. . . . . . . . . . . . . . . . . . .

Hnh ng th k: chn 1 trong n phn t xp th k c n cch

Theo qui tc nhn ta c: knA= nk

6.T hp: Cc khi nim trn lun n trt t ca tp hp ta ang quan st. Tuynhin trong thc t c nhiu khi ta ch cn quan tm ti cc phn t ca tp con ca mttp hp m khng cn n cch sp xp tp con theo mt trt t no. T y tac khi nim v t hp nh sau

6.1nh ngha: Mt t hp chp k ca n phn t l mt tp con gm k phn t ly t nphn t cho.


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V d: Cho tp hp gm bn phn t {a,b,c,d}. Hi c bao nhiu tp con gm haiphn t?

Cc tp con l {a,b},{a,c},{a,d},{b,c},{b,d},{c,d}

Vy tp hp gm bn phn t {a,b,c,d} c su tp con va nu.

6.2: Sthp chp kca n phn t c k hiu l knC

Bng cch i ch cc phn t cho nhau, mt t hp chp k ca n phn t c thto rak! chnh hp khng lp chp k ca n phn t.

C knC t hp chp k ca n phn t to raknA chnh hp khng lp chp k ca n phn t.

Vy ta c :)!kn(!k

!n

!k

AC

knk

n

==

7.T hp lp:7.1nh ngha: Mt t hp lp chp k ca n phn t l mt nhm khng phn bit th tgm k phn t, mi phn t c th c mt n k ln ly t n phn t cho.

V d: Cho tp {a,b,c} gm 3 phn t

Cc t hp lp ca tp hp trn l {a,a},{a,b},{a,c},{b,b},{b,c},{c,c}

7.2 S cc thp lp chp kca n phn t k hiu l:. knC

Vic to ra mt t hp lp chp k ca n phn t tng ng vi vic xp k qu cuging nhau vo n ngn ko t lin nhau, hai ngn lin tip cng chung mt vch ngn.Cc vch ngn tr vch ngn u v cui c th x dch v i ch cho nhau. Mi cchsp xp k qu cu ging nhau vo n ngn l mt cch b tr n+k-1 phn t ( gm k qucu v n-1 vch ngn) theo th t t phi sang tri. Cch b tr khng i khi cc qu cui ch cho nhau hoc cc vch ngn i ch cho nhau. Cch b tr thay i khi cc qucu v cc vch ngn i ch cho nhau. Ta c (n+k-1)! cch b tr n+k-1 phn t (gm kqu cu v n-1 vch ngn). S cch i ch k qu cu l k! , s cch i ch n-1 vchngn l (n-1)! . Vy ta c s cc t hp lp chp k ca n phn t l:

k

kn

k

n Cnk

knC 1)!1(!

)!1(+=

+=

V d: Ti mt tri ging g c ba loi g ging A, B, C. Mt khch hng vo nhmua 10 con. Hi c bao nhiu cch mua ( gi s rng s lng cc ging g A, B, C miloi ca tri u ln hn 10).

Ta thy mi mt cch mua 10 con g chnh l mt t hp lp chp 10 ca 3 phn t. Vy

s cch mua l: 103C= 1012C = 66

8. Nh thc Newton

Ta c: 2012111

2020

2222 baCbaCbaCbab2a)ba( ++=++=+

3033

2123

1213

0303

32233 baCbaCbaCbaCbab3ba3a)ba( +++=+++=+

Mrng ra:n0n

nkknk

n11n1

n0n0

nn baC................baC........baCbaC)ba( +++++=+

Cng thc trn gi l cng thc nh thc Newton.

Ta chng minh cng thc nh thc Newton theo qui np..


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Vi n = 2 ta c cng thc ng.

Gi s cng thc ng vi n = m tc l:m0m

m11m1

m0m0

mm baC.......baCbaC)ba( +++=+

Ta s chng minh:1m01m

1m1m1

1m01m0

1m1m

baC.........baCbaC)ba(++

++

+

+

+

+++=+ Tht vy:

)ba)(baC...baC...baC()ba()ba()ba( m0mmkkmk

m0n0

mm1m +++++=++=+ + =>

1011101101 )(...)(...)()( ++++ +++++++=+ mmmm

m

kkmk

m

k

m

m

mm

mbaCCbaCCbaCCba

Mt khc: k 1mkm

1km CCC +

=+ suy ra:1m01m

1m1m1

1m01m0

1m1m baC.........baCbaC)ba( ++++

+

+

+ +++=+ .

Theo nguyn l qui np cng thc nh thc Newton c chng minh.

V d: Tm h s ca x12 trong khai trin: 202

)1

(x

x +

Ta c:20

2020

k220k20

20020

20

x

1C.......xC........xC)

x

1x( ++++=+ .

Xt 20 - 2 k = 12

=> k = 4 Vy h s ca x12 l: 4745420 =C

II.Php th, skin

1.Php thngu nhin v khng ngu nhin

Mt php th c th coi l mt th nghim, mt quan st cc hin tng t nhin, cchin tng x hi v cc vn kthut vi cng mt hiu kin no .

Trong cc loiphp th c nhngphp th m khi bt u tin hnh thc hin ta bitc kt qu s xy ra sau khi th nhun nc iu kinbnh thng (di p sut 1atmotphe) th n 100oC nc s si, hoc cho dung dch NaOH khng dvo dung dchHCl cng khng d ta thu c mui n NaCl v nc H2O.

Nhngphp th m khi bt u tin hnh th ta bit c nhng kt qu no s xy rasau khi th c gi l ccphp th khng ngu nhin.

Tuy nhin c rt nhiu loiphp th m ngay khi bt u tin hnh php th ta khngth bit c nhng kt qu no s xy ra sau khi th chng hn nh khi gieo 100 ht

u ging, s ht ny mm sau mt thi gian gieo c th l t 0 n 100 hoc khi cho p10 qu trng th s trng g c th nra g con l t 0 n 10 con. Nhngphp th loiny gi l nhngphp th ngu nhin.

Trong gio trnh ny chng ta ch quan tm ti nhngphp th ngu nhin, l nhngphp th m khi bt u tin hnh th ta cha th bit nhng kt qu no s xy ra. n gin t y tr i khi ni tiphp th taphi hiu y l php th ngu nhin


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2. Skin:Cc kt qu c th c ca mtphp th ng vi mt b cc iu kin xc nh no gil cc s kin ngu nhin hoc n gin gi l cc s kin hoc cc bin c.

Ta thng ly cc ch ci A, B, C, D. . .. . . hoc Ai, Bj, Ck, Dn.. . . ch cc s kin.

V d 1: Tung mt con xc xc cn i v ng cht c th c cc s kin sau:

A: S kin xut hin mt chnB: S kin xut hin mt l

Ai: S kin xut hin mt c i chm.

V d 2: Trong mt gi ng hoa qu c cha 1 qu cam, 1 qu qut, 1 qu o v 1qu l. Chn ngu nhin ra 2 qu c th c cc s kin sau:

A: Hai qu c chn gm 1 cam 1 qut

B: Hai qu c chn gm 1 cam 1 o

C: Hai qu c chn gm 1 cam 1 l

D: Hai qu c chn gm 1 qut 1 l

E: Hai qu c chn gm 1 qut 1 oG: Hai qu c chn gm 1 o 1 l

3. Skin tt yu v skin khng th cS kin tt yu hoc s kin chc chn l s kin nht thitphi xy ra sau khiphp thc thc hin. Ta k hiu s kin ny l ..

S kin khng th c hoc s kin bt kh hoc s kin rng l s kin khng bao gixy ra sau khi th. Ta k hiu s kin ny l .

V d: ng ti H Ni nm mt hn

S kin ri xung a gii Vit Nam l s kin tt yuS kin ri xung i Ty Dng l s kin bt kh.

4. Quan h gia cc skin, hai skin bng nhauS kin A c gi l ko theo s kin B nu A xy ra th B cng xy ra v k hiu

A B ( hoc A B).

Nu A ko theo B v B ko theo A th ta ni A bng B v vit A = B. Trong xc sut hais kin bng nhau c coi l mt

V d: Mt hc sinh thi ht mt mn hc

A l s kin hc sinh (t im t 5 ti 10)

B l s kin hc sinh trungbnh hoc kh (t im t 5 ti 8)

C l s kin hc sinh kh hoc gii

G l s kin hc sinh gii (t im 9, 10)

K l s kin hc sinh d kh (t im 7, 8)

TB l s kin hc sinh trungbnh (t im 5, 6)

Ai l s kin hc sinh t i im (i = 0, 1, . . . .,9, 10).

Ta c: ...TBA;KA;BA;GA;AA;AC;AB;AG 57796


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5.Cc php tnh v skin5.1Php hp: Hp ca 2 s kin A v B l s kin C, s kin C xy ra khi A xy ra hocB xy ra.

K hiu: CBA = v c l A hp B bng C

Ta c th m t hp ca 2 s kin A v B bng hnh v sau:

Hnh 1

Da vo hnh v trn c th thy C xy ra khi:

A xy ra v B khng xy ra. B xy ra v A khng xy ra. C A v B cng xy ra.

V vy c thni hp ca hai s kin A v B l mt s kin C xy ra khi t nht 1 trong 2s kin A, B xy ra.

V d: Mt sinh vin thi ht mt mn hc

Gi : A l s kin sinh vin khngphi thi li (im thi t 5 n 10)

B l s kin sinh vin t im trungbnh kh (im thi t 5 n 8)

C l s kin sinh vin t im kh gii ( im thi t 7 n 10)

Ta c: A = CB .5.2Php giao: Giao ca 2 s kin A v B l s kin D, s kin D xy ra khi c A v Bcng xy ra.

K hiu: DBA = hoc AB = D v c l A giao B bng D hoc A nhn B bng D

Hnh v sau m t giao ca 2 s kin A v B

Hnh 2

V d: Quay li v d mc 5.1

Gi K l s kin sinh vin t im kh (im thi t 7 n 8)

Ta c:K = CB

Nu =BA ta ni A v B l 2 s kin xung khc vi nhau. Khi A xung khc vi B thhp ca 2 s kin A v B c k hiu l A + B v c l A cng B.


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5.3Php tr. Skin i lp: Hiu ca s kin A tr s kin B l s kin E, s kin Exy ra khi A xy ra v B khng xy ra.

K hiu: A\B= E v c l A tr B bng E

Ta cng c th m t hiu ca s kin A tr s kin B bng hnh v sau:

Hnh 3

D nhn thy rng: Nu A B = th A \ B = A

S kin : A\ Gi l s kin i lp ca s kin A v k hiu l__

A .

T nh ngha s kin i lp ca s kin A ta thy:

* A v__A . xung khc vi nhau

* Nu A khng xy ra th__

A xy ra v ngc li

Hai s kin i lp nhau xung khc vi nhau mnh m theo kiu c anh th khng cti nhng khng c anh th phi c ti.

V d: Mt t hc sinh gm 3 hc sinh nam 3 hc sinh n. Chn ngu nhin 2 ngi.

Gi : A l s kin 2 hc sinh c chn l cng gii

B l s kin 2 hc sinh c chn u l nam

C l s kin 2 hc sinh c chn u l n

D l s kin 2 hc sinh c chn c mt nam mt nTa c A \ B = C, D = A .

Hnh sau m t s kin i lp ca s kin A

Hnh 45.4 Tnh cht

1/ AA;A

2/ AA;A;A;AA ====

3/ Nu CB;BA th CA

4/ BAAB;ABBA ==


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5/ C)AB()BC(A;C)BA()CB(A ==

6/ )CA)(BA()BC(A;ACAB)CB(A ==

7/__

BAB\A =

8/

____________________

BAAB;BABA == Vic chng minh cc tnh cht trn kh d dng xin dnh cho bn c. Chng ti chchng minh tnh cht 8 phn 1 nhl mt v d minh ho cho vic chng minh cc skin bng nhau:

Ta chng minh:___________

BABA =

Gi s_______

BA xy ra theo nh ngha ca s kin i lp => BA khng xy ra, theonh ngha ca hp hai s kin => A khng xy ra v B khng xy ra, li theo nh ngha

ca s kin i lp =>A xy ra v__

B xy ra, theo nh ngha caphp giao hai s kin

=>____

BA . xy ra.

Vy ta c:___________

BABA (1)

Ngc li gi s____

BA xy ra, theo nh ngha caphp giao, =>__

A xy ra v__

B xy ra,li theo nh ngha ca s kin i lp => A khng xy ra v B khng xy ra, theo nhngha ca hp hai s kin => BA . khng xy ra, theo nh ngha ca s kin i lp

=>_______

BA xy ra. Vy ta cng c:____________

BABA (2)

T (1) v (2) =>___________

BABA =

6. Skin c th phn chia c, skin scp cbn6.1 Skin c thphn chia c

S kin A c gi l c th phn chia c nu tn ti hai s kin B , C ,

BC = v A = B + C. Khi ta ni A phn chia c thnh hai s kin B v C.

V d: Trong mt con xc xc cn i v ng cht.

Gi A l s kin xut hin mt c s chm chia ht cho 3.

Gi Ai l s kin xut hin mt i chm

S kin A c th phn chia c v tn ti A3; A

6=

63AA; v A = A

3+ A

6.

6.2 Skin scp cbn: S kin khc rng v khng th phn chia c gi l s kinscp cbn.

V d: Quay li v d mc 6.1. Cc s kin A1, A2, A3, A4, A5, A6 l cc s kin scp cbn.

Ta nhn thy rng cc s kin scp cbn l cc s kin m sau mtphp th ch cmt trong cc s kin ny xy ra.


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7. H y cc skin7.1 H y cc skin: H cc s kin A1, A2,. ... An gi l mt h y cc s kinnu:

1/ Ai vi mi i = 1, 2 . . . . n

2/ =jiAA vi mi i khc j

3/ A1+ A2+.. . . . . .+ An =

V d:em hai c th th h F1 mang gen Aa, Aa lai vi nhau. Cc c th con thh F2 c th c 1 trong 4 kiu gien AA, Aa, aA v aa. Chn 1 c th con trong cc c thni trn.

Gi: A l s kin c th con l ng hp t (mang gen AA hoc aa)

B l s kin c th con l d hp t (mang gen Aa hoc aA)

C l s kin c th con c mang gen tri (AA, Aa, aA)

A1 l s kin c th con ch mang gen tri (AA)

A2 l s kin c th con ch mang gen ln (aa)Ta c: A, B l mt h y cc s kin

C, A2 cng l mt h y cc s kin

B, A1, A2 cng l mt h y cc s kin

Nh vy: vi mtphp th cho c th c nhiu h y cc s kin khc nhau.

7.2 Phn chia mt skin theo h y .

Gi s A1, A2, . ...An l mt h y cc s kin. A l mt s kin khc rng no . Tac:

A= ni1n21 AA.....AA....AA)A..........AA(AA ++=+++=

Khi ta ni A c phn chia gin tip nhh y cc s kin: A1, A2 , A3 ,..., An.Nh bit vi mi php th c th lp ra nhiu h y cc s kin v vy mi skin khc rng A cng c th phn chia theo nhiu cch khc nhau. Mc ch ca vicphn chia s kin A ra mt s s kin n gin hn nhm nh gi kh nng xy ra cas kin A nh cc s kin n gin ny.

8. i s v - i s cc skin

Xt l mt tp hp khc rng m ta gi l s kin chc chn. C l mt h cc tp conno ca .Mi tp con A ca , AC gi l mt s kin. H C c gi l

i s cc s kin nu:

1/ C .

2/ Nu A C th __

A C

3/ Nu A1, A2. . . . . . An. . .l cc s kin thuc C th

=n

1nA C


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H C c gi l i s cc s kin nu yu cu 1, 2 nu trn tho mn v hp ca mts hu hn cc s kin thuc C cng l mt s kin thuc C. Ta nhn thy rng nu C l

i s cc s kin th C cng l mt i s cc s kin.

V d: Tung ng thi 2 ng tin, cc s kin scp cbn l:

SS, SN, NS, NN. Xt = SS + SN +NS +NN.

Tp tt c cc tp hp con ca l mt i s cc s kin.v cng l mt i s ccs kin

Bi tp chng 1

1. Mt on gen gm 2 gen X, 2 gen Y, 2 gen Z, 2 gen T lin kt vi nhau theo mt hngdc.


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a. Hi c bao nhiu cch lin kt 8 gen ni trn?

b. Hi c bao nhiu cch lin kt 2 gen X ng lin nhau?

c. Hi c bao nhiu cch lin kt c 3 gen XYZ ng lin nhau theo th t trn.

2. C 10 ngi xp theo mt hng dc

a. C bao nhiu cch sp xp 2 ngi A v B ng lin nhau?

b. C bao nhiu cch sp xp 2 ngi A v B ng cch nhau ng 3 ngi?

3. C th lp c bao nhiu s gm 10 ch s khc nhau sao cho:

a.Khng c 2 ch s chn no ng lin nhau

b. Khng c 2 ch s l no ng lin nhau

c. Cc ch s chn ng lin nhau

d. Cc ch s l ng lin nhau

4. Cho 6 ch s: 1, 2, 3, 4, 5, 6

a. C th lp c bao nhiu s gm 8 ch s sao cho ch s 1 v ch s 2 mi ch sc mt ng 2 ln, cc ch s cn li c mt ng 1 ln.

b. C th lp c bao nhiu s chn gm 8 ch s trong ch s 2 c mt ng 3 ln,cc ch s cn li c mt ng mt ln.

c. C th lp c bao nhiu s l gm 8 ch s trong ch s 1 c mt ng 3 ln,cc ch s cn li c mt ng 1 ln.

5*. Trong mt k thi tin hc quc t ti mt khu vc gm 6 phng thi nh s t 1 n

6dnh cho ba on Vit nam , M v Nga mi on gm 4 th sinh. Miphng thi c 2my tnh (khng nh s) dnh cho 2 th sinh. Vic xp 2 th sinh vo mi phng thi theonguyn tc hai th sinh cng mt quc tch khng c xp cng mtphng. Hi c baonhiu cch sp xp cc th sinh ca ba on vo 6 phng?

6*.Dc theo hai bn ng vo mt trng trong hc ngi ta d nh trng mi bn 3cybng, 3 cy phng v 3 cy bng lng.

a. Hi c bao nhiu cch trng cc cy cng loi trng i din nhau?

b. Hi c bao nhiu cch trng khng c hai cy cng loi no trng i din nhau?

7*. Vng chung kt gii v chbng chu u gm 16 i trong c i ch nh vi v ch bn nm trc.a. C bao nhiu cch chia 16 i vo bnbng A, B, C, D.

b, C bao nhiu cch chia 16 i vo bnbng A, B, C, D sao cho i ch nh v iv ch bn nm trc khng cngbng.

c. Giibi ton trn trong trng hp khng ti vai tr ca ccbng.


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8. Mt n g gm 4 con g mi v 6 con g trng. Trong 4 con g mi c 2 con muvng, 2 con mu en. Trong 6 con g trng c 3 con mu vng v 3 con mu en. Chnngu nhin 2 con g

a. C bao nhiu cch chn c 1 con trng 1 con mi

b. C bao nhiu cch chn c 2 con mu vng

c. C bao nhiu cch chn c1 con trng 1 con mi cng mu

9. Mt t sinh vin gm 6 nam 4 n. Trong 6 nam c 2 sinh vin H Ni v 4 sinh vintnh H Ty. Trong 4 n c 2 n sinh H Ni v 2 n sinh Thi Bnh. Chn ngu nhin ra3 ngi

a. C bao nhiu cch chn ra 3 sinh vin nam?

b. C bao nhiu cch chn ra 2 sinh vin nam 1 sinh vin n?

c. C bao nhiu cch chn ra 3 sinh vin gm 3 tnh?

10. Cho a gic u gm 2n cnha. Hi c th lp c bao nhiu hnh ch nht c 4 nh l 4 nh ca a gic u ny?

b. Hi a gic u ni trn c bao nhiu ng cho?

11. Cho tp A = { }10,9,8,7,6,5,4,3,2,1,0

a. A c bao nhiu tp con c t nht 2 ch s nh hn 6

b. A c bao nhiu tp con c t nht 2 ch s ln hn 6

12. C 4 vin bi ging nhau cb vo 3 ci hp. Hi c bao nhiu cchb?

13*. C 4 hnh khch i tu ti nh ga A i ti B. Mt on tu gm 4 toa chunbri ga A i ti B.

a. C bao nhiu cch ln tu ca 4 hnh khch trn.

b. C bao nhiu cch ln tu ca 4 hnh khch trn sao cho mi ngi ln mt toa.

c. C bao nhiu cch 4 hnh khch trn ln hai toa mi toa 2 ngi.

14. Trong khai trin 502)

x

2x( .

a. Tm s hng khng cha x

b. Tm h s ca x20

c. Tm h s ca x-40

15. Chng minh cc ng nht thc:

a. nnnkn

1n

0n 2C......C.............CC =+++++

b. 1nnnkn

2n

1n 2nnC.....kC........C2C

=+++++


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c.1n

12C

1n

1.....C

1k

1......C

2

1C

1nnn

kn

1n

0n

+

=

+++

++++

+

d. 1n2n21k2

n23

n21

n2n2n2

k2n2

2n2

0n2 C......C....CCC....C.....CC

+ +++++=+++++

16. Cho p, q > 0, p + q = 1. Tm s hng ln nht trong dy s sau:0nn

nkknk

n1n1

nn00

n qpC;..........;qpC.......;..........;pqC;qpC

17. Xp 3 ngi theo mt hng dc. Nu cc s kin scp cbn

18. T 4 ngi A, B, C, D ly ngu nhin 2 ngi. Nu tp cc s kin scp cbn.

19. Hai c th sinh vt c cng kiu gen Aa Bb em lai vi nhau. Hy nu cc kiu genc th c ca cc c th con.

20. T hai nhm hc sinh, nhm th nht gm 4 hc sinh nam A, B, C, D nhm th haigm 4 hc sinh n X, Y, Z, T. Chn mi nhm ra 2 hc sinh.

a. Ch ra tp cc s kin scp cbn ng viphp th trn

b. Ch ra hai h y cc s kin.

21. Tung mt ln 3 ng tin.

a.Hy ch ra cc s kin scp cbn.

b.Hy ch ra mt h y cc s kin ch gm hai s kin

22. Tung ng thi hai con xc xc.

a. C bao nhiu s kin scp cbn

b. Hy ch ra mt h y cc s kin gm 11 s kin

23. Mt a gic u gm 2n cnh (n > 2). Chn ngu nhin bn nh.

a. C bao nhiu s kin scp cbn?

b. C bao nhiu s kin bn nh c chn lp thnh hnh ch nht?

Khi n = 3 Chn ngu nhin 3 nh ca mt lc gic u.

c. C bao nhiu s kin scp cbn?

d. C bao nhiu s kin ba nh c chn lp thnh tam gic u?

25.Chng minh cc tnh cht v ccphp ton ca cc s kin.


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Chng 2 : Xc sut

Vic a ra nhng s o thch hp nh gi kh nng khch quan xy ra ca mi skin c trnhby trong phn u ca chng ny. Cc dng nh ngha xc sut t cc

nh ngha c in ti nh ngha xc sut theo h tin gip ngi hc hnh dung cs pht trin v tnh phongph, a dng ca mn xc sut. Cc tnh cht cc nh l vxc sut c trnhby mc ti thiu ngi hc khi cm thy nng n khi tip thuchng. Nhng v d a ra gip ngi hc thy c nhng p dng thc thc t camn xc sut v qua cc v d ny ngi hc c th hiu cch lm cc bi ton xc sut.

I. Cc nh ngha caxc sut

1. M u: Khi tin hnh mtphp th, c th c mt trong nhiu s kin s xy ra, mis kin l mt c tnh nh tnh, vic ch ra s o kh nng xy ra ca mi mt skin l iu cn thit. Ta c th hiu xc sut ca mi s kin l s o kh nng xy raca s kin . Vic gn cho mi s kin mt s o kh nng xy ra ca n phi mbo tnh khch quan, tnh hp l v tnh phi mu thun. Trong mc ny chng ta s ara cc nh ngha ca xc sut. Mi dng c nhng u v nhc im nht nh. Tuyvy, qua cc dng nh ngha ny c th hnh dung ra s pht trin ca mn xc sut, mtmn hc c ngun gc xut pht t nhng sngbc nhng nhs t hon thin trongqu trnhpht trin nn mn xc sut khng nhng c y cc yu t cbn ca mtngnh khoa hc chnh xc m cn l mt trong nhng ngnh ca Ton hc c th h trcho tt c cc lnh vc khoa hc khc t khoa hc t nhin n khoa hc kthut v k cnhng ngnh tng nh xa l vi Ton hc l cc ngnh khoa hc x hi.

2. nh ngha xc sut theo quan nim ng kh nng.2.1Php th ng kh nng: Mtphp th ng kh nng l mtphp th m cc ktqu trc tip (cn gi l s kin scp) ng vi php th ny c kh nng xut hin nhnhau sau khi th. Chng hn khi ta gieo mt con xc xc cn i v ng cht th vicxut hin mt trong cc mt c s chm t 1 n 6 l c kh nng nh nhau hoc khichn ngu nhin hai trong nm ngi A, B, C, D, E th vic chn c AB hoc CD . . .DE l c kh nng xut hin nh nhau.

2.2nh nghaxc sut theo quan nim ng kh nng:

Xt mtphp th ng kh nng. Gi s sauphp th ny c mt trong n s kin scp

c th xy ra v c mt trong nA s kin scp xy ra ko theo A xy ra. Ta thy lyn

nA

lm s o khch quan xy ra s kin A l hp l. V vy ta c nh ngha sau:

nh ngha: Xc sut ca s kin A l s P(A) =n

nA

* n l s kt qu ng kh nng sauphp th

* nA l s kt qu xy ra ko theo A xy ra hoc s kt qu thun li cho s kin A hays kt qu hp thnh s kin A


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Vic tnh xc sut da trn nh ngha trnphi thc hin theo trnh t sau:* Xtphp th ang quan st c phi l php th ng kh nng khng

* Nuphp th l ng kh nng th phi tm s s kin ng kh nng n

* tnh xc sut ca s kin A taphi tm s kt qu ko theo A sau s dng nhngha

P(A) =n

nA

2.3 Cc v d

V d 2.1: Gieo hai ng tin cn i v ng cht. Tnh xc sut c hai cng xuthin mt quc huy.

Gi A l s kin c hai ng tin cng xut hin mt quc huy.

Ta c: S s kin ng kh nng: n = 4

S s kin ko theo A: nA = 1 .Vy P (A) =4

1

V d 2.2: Mt n g c bn con g ri gm hai mi hai trng v su con g tamhong gm hai trng bn mi. Chn ngu nhin hai con g

Gi A l s kin hai con g c chn u l trng

B l s kin hai con g c chn gm mt trng mt mi

C l s kin hai con g c chn l g mi ri

Hy tnh xc sut ca cc s kin A, B, C

Ta c: S s kin y kh nng l 210C = 45

S s kin ko theo A l 24C = 6

S s kin ko theo B l 16

1

4

CC = 24

S s kin ko theo C l 22C = 1

Vy: P(A)=15

2

45

6= , P(B) =

15

8

45

24= , P(C) =

45

1

V d 2.3: C ba gen X, Y, Z v ba gen x, y, z xp ngu nhin theo mt dy dc. Tnhxc sut cc gen x, y, z xp lin nhau.

Gi A l s kin cn tnh xc sut

S s kin ng kh nng: n = 6! = 720

S s kin ko theo A: nA = 3!4! = 144. Vy: P(A) =

5

1

720

144=

V d 2.4: Hai c th b v m cng c kiu gen AaBb. Tnh xc sut c th con ckiu gen ging kiu gen ca b m. Ta c bng lin kt gen sau:


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M

B

AB Ab aB ab

AB AABB AABb AaBB AaBb

Ab AABb AAbb AaBb AabbaB AaBB AaBb aaBB aaBb

ab AaBb Aabb aaBb aabb

Da vobng trn ta c: S s kin ng kh nng n = 16

S s kin ko theo A: nA = 4. Vy P(A) =4

1

16

4=

3- nh ngha xc sut theo tn sut

nh ngha xc sut theo quan nim ng kh nng c u im l ch ra cch tnh xcsut ca mt s kin r rng v n gin. Tuy nhin nh ngha ny ch p dng c viloiphp th ng kh nng v s kt qu sauphp th l hu hn. Trong thc t thnggp nhng loiphp th khng c tnh cht trn, khcphc hn ch ny ta c th nhngha xc sut theo quan im thng k.3.1 Tn sutca skin: Gi s ta tin hnh nphp th vi cng mt h iu kin thyc nA ln xut hin s kin A. S nAc gi l tn s xut hin s kin A v t s:

n

n)A(f An = gi l tn sut xut hin s kin A.

Ta nhn thy rng khi n thay i nA thay i v th fn(A) cng thay i. Ngay c khi tinhnh dy nphp th khc vi cng mt iu kin th tn s v tn sut ca n ln th nycng c th khc tn s v tn sut ca n ln th trc. Tuy nhin tn sut c tnh n nhngha l khi s php th n kh ln tn sut bin i rt nh xung quanh mt gi tr xcnh. minh chng cho nhn xt trn ta xt mt v d kinh in v xc nh tn s vtn sut vic xut hin mt sp (mt khng c ch) ca mt ng tin do Buffon vPearson thc hin

Ngi lm th nghim S ln tung 1 ng tin Tn s mt sp Tn sut mt sp

Buffon 4040 2040 0.5080

Pearson 12000 6010 0.5010

Pearson 24000 12012 0.5005

Ta nhn thy rng khi s ln tung tin n tng ln, tn sut xut hin mt sp n nh dnv gi tr 0,5 c ly lm xc xut xut hin mt sp khi tung mt ng tin cn i vng cht.

3.2nh ngha: Xc sut ca mt s kin l tr s n nh ca tn sut khi s php thtng ln v hn.

Vic khng nh tn sut ca mt s kin n nh (hay tin ti) mt gi tr xc nh khis php th tng ln v hn c m bo bi nh l Bernoulli s c pht biu v


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chng minh trong chng sau. Tuy nh ngha xc sut bng tn sut khng ch ra gi trc th xc sut ca s kin nhng trong thc t khi s ln th n l ln ta thng ly tnxut fn(A) thay cho xc sut ca s kin A. Vo cui th k 19 nh ton hc Laplace theodi ccbn thng k v dn s trong vng 10 nm ca London, Peterbua, Berlin v nc

Php ng ta tm ra tn sut sinh con trai ca ba vng trn v c nc Php l

43

22. Khi

xem xt t l sinh con trai ca Paris ng tm c tn sut49

25, tn sut ny nh hn

43

22.

Ngc nhin v s khc nhau , Laplace iu tra thm v tm ra hai iu th v sau:

Mtl: Vo thi by gi cc tr em ra khng ghi tn cha trong giy khai sinh thd sinh Marseille, Bordeaux hay bt c ni no trn t Php u c trongbn thngk tr sinh Paris.

Hai l: Phn ln nhng a tr ni trn u l con gi.

Sau khi loi nhng a tr khng sinh Paris ra khi danh sch ny th t l tr trai

Paris trv con s

43

22.

Qua v d nu trn chng ti mun cc nh nng hc tng lai khi quan st hoc thnghim thy c mt s liu no khc vi s liu bit th cnphi tm nguyn do skhc bit ny xutpht t u, rt c th qua ta c th pht hin c nhng iu bchphc v cho chuyn mn.

4. nh ngha xc sut bng hnh hcVi nhngphp th ng kh nng m s kt qu sau mtphp th l v hn th vic sdng nh ngha xc sut mc 2 tnh xc sut ca mt s kin l khng thc hinc. khcphc hn ch ny ngi ta a ra nh ngha xc sut bng hnh hc.

4.1 o ca mt min: Gi s D l mt min hnh hc no chng hn D l mt onthng, mt hnh phng hay mt khi khng gian. S o di, din tch, th tch tngng c gi l o ca min D v k hiu l m(D)4.2.nh ngha :

Xt mtphp th vi v hn kt qu ng kh nng, gi s c th thit lp s tng ngmt - mt mi kt qu vi mt im thuc min G c o l m(G) . Mi kt qu kotheo s kin A tng ng vi mi im thuc min D G c o m(D).

Xc sut ca s kin A l s P(A) =)G(m)D(m

V d 1: Mt ng dy cp quang ni H Ni vi thnh ph H Ch Minh di 1800

km gp s c kthut lm tc nghn vic thng tin lin lc. S c kthut c th xy ra bt c mt v tr no trn ng cp quang trn vi cng mt kh nng. Tnh xc sut s c kthut xy ra cch H Ni khng qu 300km.Min G y l ng cp quang ni H Ni- thnh ph H Ch Minh c m(G) = 1800.Min D tng ng vi s kin cn tnh xc sut l on cp quang t H ni ti v trcch H Ni 300 km, m(D) = 300.


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Vy xc sut cn tnh P =6

1

1800

300= .

V d 2: Hai ngi A, B hn gp nhau ti mt a im trong qung thi gian t 12gi n 13 githeo qui c, ngi n trc i ngi n sau khng qu 15pht. Tnhxc sut hai ngi gp c nhau. Bit rng mi ngi c th n im hn vo bt

c thi im no trong qung thi gian ni trn.Gi x l thi im A n ch hn, y l thi im B n ch hn, 0 60y,x

Vic hai ngi n ch hn tng ng vi im M(x, y) thuc hnh vung OABC ccnh di 60 n v di. Hai ngi gp c nhau

+ 15xy15x15yx M(x, y) thuc hnh ODEBGH.

Hnh 1

Ta c min G l hnh vung OABC, min D l hnh ODEBGH.

m(G) = 602 , m(D)= 602- 452.

Vy xc sut cn tnh P =

16

7

16

91

60

4560

)G(m

)D(m2

22

==

=

Mt s bi ton thc t nhqu trnh th phn, qu trnh th tinh .... c th p dng nhbi ton gp g ni trn.

5. H tin KolmogoropMc d ra i t th k 17 nhng do ngun gc xutpht v nhng khi nim c nura c tnh m t thiu nhng lun c khoa hc nn c mt qung thi gian di t th k 17n trc nhng nm 30 ca th k 20 xc sut khng c coi l mt ngnh ton hcchnh thng. Mi ti nm 1933 khi nh ton hc Nga A.N Kolmogorop xy dng h tin cho l thuyt xc sut th xc sut mi c cng nhn l mt ngnh ton hc chnh

thng

snh ngang

hng v

i nhi

u

ngnh

ton

hc

khc nh

s

hc,

hnh

hc,

i s

,gi

itch...

Tuy c chp nhn mun mng nhng xc sut c mt trong hu ht cc lnh vckhoa hc t khoa hc t nhin , khoa hc kthut dn khoa hc x hi. V l mt giotrnh dnh cho cc ngnh khng chuyn v ton chng ti ch c nh trnhby slch tin v l thuyt xc sut do A.N Kolmogorop a ra

Xt C l mt - i s cc s kin . Xc sut P l mt hm xc nh trn C tho mn :

1/ P(A) 0 A C


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2/ 1)(P =

3/ Nu A1, A2 , ... ,An,.. . . . . . . ... xung khc tng i, An C , n =1,2,... th

=

=

=1i

i1i

i )A(P)A(P

B ba ( , A, P ) c gi l khng gian xc sut.

IICc tnh cht v cc nh l

1. Cc tnh cht. n gin, ta ch s dng nh ngha theo quan im ng kh nng chng minhcc tnh cht s nu trong mc ny. Tuy nhin cc tnh cht cng ng vi mi dngnh ngha xc sut khc.

1/ 1)A(P0 v 1

n

n)A(P

n

n0nn0 AA ==

2/ 1)(,0)( == PP v nn,0n == suy ra iu cn chng minh.

3/ Nu =BA th P(A+B) = P(A) + P(B)

Gi nA l s s kin ko theo A, nB l s s kin ko theo B do A xung khc vi B nn ss kin ko theo A + B l

)B(P)A(Pn

n

n

n

n

nn

n

n)BA(Pnnn BABABABABA +=+=

+==++= ++

4/ )AB(P)B(P)A(P)BA(P +=

Gi nA l s s kin ko theo A, nB l s s kin ko theo B, nAB l s s kin ko theo

AB, BAn l s s kin ko theo BA . Ta c

)AB(P)B(P)A(Pn

n

n

n

n

n)BA(P

n

nnn

n

n)BA(Pnnnn

ABBA

ABBABAABBABA

+=+=

+==+=

H qu 1: )A(P1)A(P = . Tht vy ta c

=+=+=+ 1)A(P)A(P)(P)AA(PAA iu cn chng minh.

H qu 2: Nu A1, A2 , .. .An xung khc tng i th ==

=n

i

i

n

i

i APAP

11

)()(

ap dng nhiu ln tnh cht 1.3 ta c h qu trn.

5/ Nu )B(P)A(PBA

V )B(Pn

n

n

n)A(PnnBA BABA ==

2. Xc sut c iu kin


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Xt hai s kin A v B trong mtphp th c tin hnh ng vi mt b iu kin no. Vic xut hin s kin ny i khi nh hng n xc sut xut hin ca s kin kiav ngc li .

Chng hn trong mt hp c 3 bi trng v 2 bi , rt ln lt 2 bi. Ln u rt c bitrng hay khng r rng nh hng n xc sut xut hin bi trng ln th hai.

2.1.nh ngha: Xc sut ca s kin A vi gi thit s kin B xy ra l xc sut ciu kin ca A vi iu kin B.

Ta k hiu xc sut ny l P(A/B) hoc PB(A)

V d 2.1: Quay li v d va nu trn. Gi B l s kin ln u rt c bi trng , A

l s kin ln sau cng rt c bi trng. Ta c P(A/B)=2

1

4

2= cn P(A/

4

3)B = . R rng

vic xut hin hay khng xut hin B nh hng ti xc sut xut hin A.

V d 2.2: Tnh trng hoa vng gen A l tnh trng tri, hoa trng gen a l tnh trngln. Hai cy u hoa vng d hp t ( cng mang gen Aa) em lai vi nhau cc c th conc cc kiu gen AA, Aa, aA, aa v cng mt kh nng. Chn mt c th con th thy c

th ny c hoa mu vng. Tnh xc sut c th l ng hp tGi B l s kin c th con c hoa mu vng, A l s kin c th con c gen ng hp t.

Ta c: P(A/B) =3

1

2.2 Cng thcxc sutc iu kin

)B(P

)AB(P)B/A(P =

Tht vy gi nB l s s kin ko theo B( do gi thit B xy ra nn nB 0 , gi nAB ls kin ko theo AB

Ta c)B(P

)AB(P

n

nn

n

n

n)B/A(P

B

AB

B

AB ===

3. Cng thc nhn xc sut

T )B/A(P)B(P)AB(P)B(P

)AB(P)B/A(P == (1)

Thay i vai tr ca A v B cho ta c P(AB) = P(A)P(B/A)Mrng ta c: P(A1A2...An) =P(A1)P(A2/A1)...P(An/A1A2...An-1) (2)

Cng thc trn gi l cng thc nhn xc sut. p dng lin tip cng thc (1) nhiu lnta c cng thc (2)

V d 3.1: C 6 cy u hoa vng v 2 cy u hoa trng ly ln lt 2 cy u. Tnhxc sut c 2 cy u ly ra l cy u hoa vng.

Gi A l s kin c 2 cy ly ra l u hoa vng

A1 l s kin cy ly ra ln u mu vng


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A2 l s kin cy ly ra ln hai mu vngTa c: A = A1A2 t suy ra

28

15

56

30

7

5.

8

6)A/A(P)A(P)AA(P)A(P 12121 =====

S dng nh ngha xc sut theo quan im ng kh nng ta cng c kt qu trn.

V d 3.2: Mt ging la mi ti mt tri lai to ging trc khi a ra sn xut i trphi tin hnh lin tip ba ln kim nh do ba trung tm kho cu ging cp mt, cphai, cp ba tin hnh. Nu ging la c chp nhn trung tm cp di th cchuyn ln trung tm cp trn kim nh tip. Qua thng k cho thy ging ca tritrn c trung tm cp mt chp nhn vi xc sut 0,7. Sau khi chuyn ln trung tmcp hai n c chp nhn vi xc sut 0,8. Nu c chuyn ln trung tm cp ba nc chp nhn vi xc sut 0,9. Tnh xc sut ging la c a ra sn xut i tr.

Gi: A l s kin ging la c a ra sn xut i tr.

Ai l s kin ging la c chp nhn trung tm cp i.

Ta c: A = A1A2A3

P(A) = P(A1A2A3) = P(A1)P(A2/A1)P(A3/A1A2) =0,7.0,8.0,9 = 0,486

4. Cc skin c lp.4.1 Hai skin c lp: S kin A c gi l c lp vi s kin B nu:

P(A/B) = P(A)

T nh ngha trn ta c

* Nu A c lp vi B th P(AB)=P(A)P(B)Tht vy P(AB)=P(B)P(A/B) =P(B)P(A)

* Nu A c lp vi B th B cng c lp vi A

Do P(AB)=P(A)P(B/A) =P(B)P(A)P(B/A)=P(B). Do vy B cng c lp vi A.

* A c lp vi B P(AB)= P(B)P(A)

4.2. H c lp tng i v c lp hon ton

H: A1,A2,...,Anc gi l c lp tng i nu Aic lp AjijH: A1,A2,...,Anc gi l c lp hon ton nuP( { } { }A,...,A,AA,...A,A)A(P)A...AA/A n21jjjijjji k21k21 =

T nh ngha trn ta thy h c lp hon ton th c lp tng i nhng iu ngc lini chung khng ng.4.3. Cc v d

V d 4.1: Mt mng cp nc nhhnh v

Hnh 2


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Nc c cp t E n F qua ba trm bm tng p A, B, C. Cc trm bm lm vic clp vi nhau. Xc sut cc trm bm A,B,C c s c sau mt thi gian lm vic lnlt l: 0,1; 0,1; 0,05. Tnh xc sut vng F mt nc

Gi: F l s kin vng F mt nc

A l s kin trm A c s c

B l s kin trm B c s cC l s kin trm C c s c

Ta c: F = ( ) ( )[ ]CBAP)F(PCBA =

= P(AB)+P(A)-P(ABC) = P(A)P(B)+P(C)-P(A)P(B)P(C)

= 0,01 + 0,05 - 0,005 = 0,055V d 4.2: C hai lng g ging. Lng th nht c 2 g trng, 4 g mi. Lng th hai

c 4 g trng, 2 g mi. Ly ngu nhin t mi lng ra 1 con. Tnh xc sut 2 con gly ra u l g mi

Gi : A1 l s kin con g ly ra lng mt l g mi

A2 l s kin con g ly ra lng hai l g miTa c: P(A1A2) = P(A1)P(A2) =

9

2

6

2.

6

4=

5. Dy php th c lp: Trong thc t nhiu khi ta gp nhngphp th hp gm mtdy lin tip ccphp th nh nhau c lp i lp li n ln v n s xut hin camt s kin A no trong n ln th ny. Chng hn khi gieo mt ng tin cn i vng cht n ln hoc tung mt con xc xc cn i v ng cht n ln th nhngphp ththuc loi ny chnh l dyphp th c lp.

5.1. Lc Bernoulli. Tin hnh mt dy n php th m php th sau c lp vi ccphp th trc , xc sut xut hin s kin A miphp th l nh nhau v bng p

(p 0, p 1). Dy nphp th c lp loi ny cn c gi l mt lc Bernoulli.5.2. Cng thc Bernoull: Trong mt lc Bernoulli s kin A c th xut hin t 0n n ln. Gi Bk l s kin A xut hin ng k ln trong lc Bernoulli. ta xy dngcng thc tnh P(Bk)

Gi Ai l s kin A xut hin ln th i trong n ln th

Ta c Bk = A1A2...Ak n1knkn1n1k A...AA...A...A...A ++ ++ . Mi s kin ca tng cc s

kin trn gm tch ca n s kin trong A xut hin k ln v A xut hin n-k ln. Mitch trn tng ng vi vic chn ra kphp th (A xut hin) t nphp th cho, theol thuyt t hp c tt c knC tch nh vy.

Do nphp th l c lp P(Ai) = p, P qp1)A( j == nn P(A1A2...Ak (P...)A...A n1k ==+ n1knkn1 A...AA...A + ) = p

kqn-k

Suy ra: P(Bk) =knC p

kqn-k

y l cng thc Bernoulli cho ta bit xc sut A xut hin k ln trong mt lc BernoulliGi: Pn(k) l xc sut s kin A xut hin k ln trong mt lc Bernoulli v


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Pn(k1, k2) l xc sut A xut hin trong khong t k1 n k2 ln (k1< k2)

Ta c: Pn(k) = P(Bk) =knC p

kqn-k

Pn(k1,k2) =knk

k

kk

kn

k

kkn qpC)k(P

2

1

2

1

==

=

V d 5.1: Xc sut mt qu trng g em p nra g con l 0,8. em p 5 qutrng. Tnh xc sut c 3 qu nra g con?

Ta c mt lc Bernoulli vi n = 5, p = 0,8. Xc sut cn tnh l

2048,02,08,0C)3(P 23355 ==

V d 5.2: T l u hoa vng ng hp t gen AA, hoa vng d hp t gen Aa v hoatrng gen aa l 1 : 2 : 1. Chn10 ht u em gieo

1/Tnh xc sut c 4 cy u hoa vng l ng hp t

2/Tnh xc sut c 5 cy u hoa vng

Nu ch xt ti cc cy u hoa vng ng hp t trong s cy u ta c lc

Bernoullie vip1 =

4

3q,

4

11 = . Vy xc sut cn tnh l

6441010 )4

3.()

4

1(C)4(P =

Trong trng hp th 2 ta c p2 =4

1q,

4

31 = v xc sut cn tnh

5551010 )4

1.()

4

3(C)5(P =

5.3. Sln xut hin chc nht: Xt mt lc Bernoullie vi s ln th n v xc sutxut hin s kin A, P(A) = p .

k0c gi l s ln xut hin chc nht hoc s ln xut hin c kh nng nht nu:

Pn(k0) Pn(k) k = 0, 1..., n. tm k0 ta ch cn xt dy s Pn(0), Pn(1),...Pn(k),...Pn(n)xem s no ln nht th k ng vi s chnh l k0 cn tm. Tuy nhin vic tnh tt ccc s trong dy s trn s mt nhiu thi gian. V vy ta a ra thut ton tm s ln

xut hin chc nht t nhn xt sau. Trong dy s u1, u2,... un nuk

1k

u

u + cn ln hn 1 th

dy s cn tng n khi nok

1k

u

u + nh hn 1 th dy s bt u gim. S k0 m t dy

chuyn t tng sang gim l s cn tm. p dng nhn xt trn ta xt

q

p.

1k

kn

qpC

qpC

)k(P

)1k(Pknkk

n

1kn1k1kn

n

n

+

==

+

++

Pn(k+1)>Pn(k) kqnp)qp(kqnpqkqkpnp >+>+>

Do np - q l mt hng s nn khi k cn nh hn np - q dy cn tng ti khi k vt qua

np q th dy bt u gim.


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Nu np - q khngphi l s nguyn s ln xut hin chc nht l: k0 = [np-q]+1

Ch : Phn nguyn ca s thc x l s nguyn ln nht nh thua hoc bng x, k hiul[ ]x

Nu np - q l s nguyn th s ln xut hin chc nht l k0 = np - q v k0+1.

Khi ta c Pn(k0) = Pn(k0+1) n,0k)k(Pn = V d 5.3: Xc sut mi cy sng sau thi gian trng l 0,8. Trng 1000 cy, Tm scy c kh nng sng cao nht

Ta c n =1000, p = 0,8, q = 0,2 np - q = 799,8

Vy s cy c kh nng sng cao nht k0 = 800

6. Cng thc xc sut ton phnXt A1, A2,..., An l mt h y cc s kin, A l mt s kin no .

Ta c:

A= A n21n21 AA...AAAA)A...AA(A +++=+++= )AA...AAAA(P)A(P n21 +++=

S dng cng thc cng v nhn xc sut ta c

P(A) = P(A1)P(A/A1)+ P(A2)P(A/A2)+...+ P(An)P(A/An)

Cng thc trn c gi l cng thc xc sut ton phn

V d 6.1: Mt kho hng c 10 kin hng trong c 4 kin do my A sn xut,

3 kin do my B sn xut v 3 kin cn li do my C sn xut. T l sn phm loi hai docc my sn xut ln lt l 0,02; 0,03; 0,05. Ly ngu nhin t kho ra mt kin hng rit ly ra mt sn phm. Tnh xc sut sn phm ly ra l sn phm loi hai

Gi: A l s kin sn phm ly ra l sn phm loi hai

Ai l s kin sn phm ly ra do my i sn xut

Khi A1, A2, A3 l mt h y A = 321 AAAAAA ++

Theo cng thc xc sut ton phn ta c

P(A) = P(A1)P(A/A1)+ P(A2)P(A/A2)+ P(A3)P(A/A3)

= 032,005,0.10

303,0.

10

302,0.

10

4=++

V d 6.2: Mt loi sinh vt c cc kiu gen AA, Aa, aa theo t l: 1 : 2 : 1.

Nu c th b (m) c kiu gen AA lai vi cc th m (b) c kiu gen AA th cc c thcon u c kiu gen AA.

Nu c th b (m) c kiu gen AA lai vi cc th m (b) c kiu gen Aa th c th conc kiu gen AA, Aa theo t l 1 : 1.

Nu c th b (m) c kiu gen AA lai vi cc th m (b) c kiu gen aa th c th conchc cc kiu Aa. Chn mt c th con t c th m c kiu gen AA.

1/Tnh xc sut c th con c kiu gen AA

2/Tnh xc sut c th con c kiu gen Aa


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Gi: A l s kin c th con c kiu gen AA

B l s kin c th con c kiu gen Aa

A1 l s kin c th b c kiu gen AA

A2 l s kin c th b c kiu gen Aa

A3 l s kin c th b c kiu gen aa

Theo u bi :

0)/(;2

1)/(;1)/(;

4

1)(;

4

2)(;

4

1)( 321321 ====== AAPAAPAAPAPAPAP

1)/(;2

1)/(;0)/( 321 === ABPABPABP

H: A1, A2, A3 l h y .

A = 321 AAAAAA ++

suy ra P(A) = P(A1)P(A/A1)+ P(A2)P(A/A2)+ P(A3)P(A/A3)

= 2

1

4

1

4

10.4

1

2

1.4

21.4

1=+=++

B = 321 BABABA ++

suy ra P(B) = P(A1)P(B/A1)+ P(A2)P(B/A2)+ P(A3)P(B/A3)

=2

1

4

1

4

11.

4

1

2

1.

4

20.

4

1=+=++

7. Cng thc BayesGi s A1, A2,...Ai. . . . An l mt h y cc s kin. A l mt s kin no

c P(A) 0Theo cng thc xc sut ton phn ta c:

P(A) = P(A1)P(A/A1)+ P(A2)P(A/A2)+... +P(Ai)P(A/Ai)+...+P(An)P(A/An)

Xt Aj l mt s kin no trong h cc s kin cho

Ta c P(Aj/A)=( )

=

=n

1iii

jjj

A/AP)A(P

)A/A(P)A(P

)A(P

)AA(P

Cng thc trn c gi l cng thc Bayes. Cc xc sut P(Aj/A) gi l cc xc sut hunghim phn bit vi cc xc sut tin nghim P(Ai)

V d 7.1: Cp tr sinh i c th l sinh i tht ( do cng mt trng sinh ra) trongtrng hp ny chng lun cng gii. Trng hp cp sinh i do hai trng sinh ra gi lgi sinh i. Nu cp sinh i do hai trng sinh ra th xc sut chng cng gii l 1/2.Bit xc sut cp sinh i do cng mt trng sinh ra l p. Mt cp tr sinh i ra ibit chng cng gii. Tnh xc sut chng l sinh i tht.

Gi: A l s kin cp tr sinh i l cng gii

A1 l s kin cp tr sinh i l sinh i tht

A2 l s kin cp tr sinh i l gi sinh i


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A1, A2 l h y , P(A1) = p; P(A2) =1-p; P(A/A1) = 1; P(A/A2) =2

1

Theo cng thc Bayes ta c xc sut cn tnh l:

p1

p2

2

1)p1(1.p

1.p

)A/A(P)A(P)A/A(P)A(P

)A/A(P)A(P)A/A(P

2211

111

+=

+

=+

=

V d 7.2: T l ngi n khm ti mt bnh vin mc bnh A l 60%, trong snhng ngi mc bnh A c 50% mc c bnh B, cn trong s nhng ngi khng mcbnh A c 70% mc bnh B.

1/Khm cho mt ngi th thy ngi mc bnh B. Tnh xc sut ngic khm cng mc bnh A.

2/ Nu ngi c khm khng mc bnh B tm xc sut ngi khng mcbnh A.

Gi : A l s kin ngi chn i khm mc bnh A

B l s kin ngi chn i khm mc bnh BTa c A v A l mt h y , P(A) = 0,6; P( A ) =0,4

V vy ta c: B = BA +B A

Xc sut cn tnh phn 1 l P(A/B) =)B(P

)A/B(P)A(P

Xc sut cn tnh phn 2 l)B(P

)A/B(P)A(P)B/A(P =

Ta c: P(B/A) = 0,5; P( )A/B = 0,3.

Suy ra: P(B) = P(A)P(B/A) + P( A )P(B/A )= 0,6.0,5 + 0,4.0,7 = 0,58 P( B ) = 0,42

Vy: P(A/B) =7

2

42,0

3,0.4,0)B/A(P,

29

15

58,0

30,0===

V d 7.3: gy t bin cho mt tnh trng ngi ta tm cch tc ng ln hai genA, B bngphng x. Xc sut t bin ca tnh trng do gen A l 0,4; do gen B l 0,5 vdo c hai gen l 0,9.

1/Tnh xc sut c t bin tnh trng bit rngphng x c th tc ngln gen A vi xc sut 0,7 v ln gen B vi xc sut 0,6.

2/Tnh trng c du hiu t bin. Xc nh vai tr ng gp ca tng gen

Gi : C l s kin c t bin tnh trng ang xtA l s kinphng x tc dng ln gen AB l s kinphng x tc dng ln gen B

C1 l s kinphng x ch tc ng ln gen A

C2 l s kinphng x ch tc dng ln gen B

C3 l s kinphng x tc dng ln c 2 gen


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C4 l s kinphng x khng tc dng ln gen no

Khi h C1, C2, C3, C4 l mt h y

C1 = A B , C2 = BA , C3 = AB, C4 = A B . Mt khc A, B l c lp nn

P(C1) = P(A)P( B ) = 0,28, P(C2) = P( A )P(B) = 0,18

P(C3) = P(A)P(B) = 0,42; P(C4) = P( A )P( B ) = 0,12Mt khc P(C/C1) = 0,4; P(C/C2) = 0,5; P(C/C3) = 0,9 v P(C/C4) = 0

Theo cng thc xc sut ton phn ta c

P(C) = 0,28.0,4 +0,18.0,5 +0,42.0,9 +0,12.0 = 0,58

Vai tr ng gp ca ring gen A cho s t bin l

1931,0580

112

)(

)/()()/( 111 ==

CP

CCPCPCCP

Vai tr ng gp ca ring gen B cho s t bin l

1552,0580

90

)(

)/()(

)/(22

2 == CP

CCPCP

CCP

Vai tr ng gp ca c hai gen cho s t bin l

6517,0580

378

)(

)/()()/( 333 ==

CP

CCPCPCCP


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Bi tp chng II

1.Tung ng thi 3 ng tin. Tnh xc sut c 3 ng tin cng xut hin mt sp.

2. Mt t hc sinh c 4 nam, 4 n. Chn ngu nhin 3 hc sinh.a.Tnh xc sut trong 3 ngi c chn c 2 nam, 1 n.

b. Tnh xc sut trong 3 ngi c chn u l n.

3. C 6 tm th nh s t 1 n 6. Rt ln lt 3 tm ri t t tri quaphi

a.Tnh xc sut s lp c l s chn

b. Tnh xc sut s lp c chia ht cho 3

c. Tnh xc sut s lp c chia ht cho 5

4. C n ngi xp theo mt hng dc (n >5)

a.Tnh xc sut 2 ngi A, B ng lin nhau

b.Tnh xc sut 2 ngi A, B ng cch nhau ng 3 ngi

5. Mt hc sinh c 5 quyn sch Ton, 3 quyn sch Vn v 2 quyn sch Ngoi ng.Hc sinh ny xp ngu nhin cc quyn sch ny trn mt ngn ca gi sch.

a. Tnh xc sut 5 quyn sch Ton ng lin nhau

b. Tnh xc sut khng c 2 quyn sch Ton no xp lin nhau

6. Chn ngu nhin 10 hc sinh, tnh xc sut khng c 2 hc sinh no c cng sinh

nht.

7. Cho mt l hng c n sn phm trong c m ph phm. Ly ngu nhin k sn phm

(k < n, k < m ). Tnh xc sut trong k sn phm ly ra c l ph phm (l < k).

8*. Mt on tu vo ga gm c 4 toa, trn sn ga c 8 hnh khch i ln tu. Cc hnhkhch ny c ln mt trong bn toa trn mt cch ngu nhin

a. Tnh xc sut mi toa c ng 2 hnh khch mi ln

b. Tnh xc sut mi toa c ng 4 hnh khch mi ln

c. Tnh xc sut mt toa c ng 5 hnh khch mi ln 3 toa cn li mi toa c 1 hnhkhch ln.

9. Ti mt tri ln ging c 4 con ln ni thuc cc loi A, B , C, D cho phi ging vi 4ln c cng thuc 4 loi trn mt cch ngu nhin .

a. Tnh xc sut cc cp ln cng loi phi ging vi nhau

b. Tnh xc sut khng c cp no cng loi phi ging vi nhau


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10. Trong 10 ht u ging c 4 ht u hoa vng thun chng, 3 ht u hoa vng khngthun chng v 3 ht u hoa trng. Chn ngu nhin 3 ht u

a. Tnh xc sut 3 ht u c chn gm 3 loi khc nhau

b. Tnh xc sut 3 ht u c chn l u cho hoa mu vng

c. Tnh xc sut 3 ht u c chn c t nht mt ht cho hoa mu trng

11. Mt on thng c chiu di 2l cb ngu nhin thnh 3 on. Tnh xc sut 3on ny lp thnh mt tam gic.

12. Do thiu kinh nghim mt nhn vin th tinh nhn to cho b ch chun on cb s rng trng trong khong 0h sng n 24h cng ngy. Bit rng trng v tinh trngc th sng trong t cung khng qu t gi(t < 12).

a. K thut vin tin hnh th tinh nhn to vo lc 12h. Tnh xc sut vic th tinhthnh cng.

b. Kthut vin tin hnh vic th tinh nhn to mt cch ngu nhin trong qung thigian t 10h n 14h. Tnh xc sut vic th tinh thnh cng.

13. Lai g lng mu nu vi g lng mu trng g con th h F1 c lng mu nu, muxm v mu trng theo t l: 1 : 2 : 1. Chn ngu nhin 5 qu trng th h F1em pv c 5 qu trng u n. Tnh xc sut :

a. C ng 3 g con c lng mu nu.

b. C 2 g con c lng mu nu v 3 g con c lng mu xm.

c. C 1 g con c lng mu nu, 2 g con c lng mu xm, 2 g con c lng mu trng.

14. Bit t l ngi c nhm mu O, A, B v AB trong cng ng tng ng l:

34%, 37%, 21%, 8%. Ngi c nhm mu O, A, B ch c th nhn mu ca ngi cngnhm vi mnh hoc nhn t ngi c nhm mu O, cn ngi c nhm mu AB c thnhn mu t bt c mt ngi c nhm mu no. C mt ngi cn tip mu v mtngi cho mu. Vic truyn mu c thc hin.

a. Tnh xc sut ngi nhn mu c nhm mu A

b. Tnh xc sut ngi nhn mu c nhm mu B

15. Mt cng ty c haiphng chc nng. Phng A gm 3 nhn vin nam, 2 nhn vin n.

Phng B gm 3 nhn vin nam, 3 nhn vin n. kim tra nng lc lm vic ca miphng, gim c cng ty quyt nh chn mi phng 2 nhn vin kim tra chuynmn. Bit rng mi nhn vin phng A c th vt qua k kim tra vi xc sut 0,8 ivi nam v 0,7 i vi n. Mi nhn vin phng B c th vt qua k kim tra vi xcsut 0,7 i vi nam v 0,8 i vi n.

a. Tnh xc sut 4 nhn vin c chn u l nam.

b. Tnh xc sut 4 nhn vin c chn u qua k kim tra


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c. Kh nng vt qua k kim tra ca phng no cao hn?

16. Mt nhm bnh nhn gm 6 ngi trong c 4 ngi mc bnh A v 5 ngi mcbnh B.

a. Tm s bnh nhn mc c hai loi bnh

b. Chn ngu nhin 2 trong s 6 bnh nhn ni trn. Tnh xc sut 2 ngi mc chai loi bnh.

c. Ngi ta nh s dng mt loi bit dc X iu tr cho nhm bnh nhn trn. Xcsut mt bnh nhn ch mc mt loi bnh khi s dng bit dc X khi bnh l 0,8.Xc sut mt bnh nhn mc c hai loi bnh khi s dng bit dc X khi bnh l0,6. Chn ngu nhin hai bnh nhn trong 6 bnh nhn ni trn ri cho dng bit dc X.Tnh xc sut c hai bnh nhn khi bnh.

17. Baphng th nghim c giao nhim v to ging la mi. Baphng lm vic clp, xc sut thnh cng tng ng l 0,4; 0,3; 0,2.

a. Tnh xc sut c ng mtphng thnh cng.b. Tnh xc sut c t nht mtphng thnh cng.

c. Trong mt nm nuphng no thnh cng trong vic to ra ging la mi th c coil hon thnh nhim v. Nu thtbi c lm thm mt ln na v nu ln ny thnhcng th cng c coi l hon thnh nhim v. Tnh xc sut c ba phng cng honthnh nhim v.

18. Mt mng cung cp in nhhnh v

Hnh 3

in c cung cp t E ti khu tiu dng F qua nm trm bin p A, B, C, D, G. Cctrm bin p ny lm vic c lp, xc sut mi trm bin p A, B, C c s c kthutsau mt thi gian hot ng l 0,1. Xc sut trn vi hai trm D, G l 0,05.

a. Tnh xc sut F mt in.

b. Bit Fb mt in.Tnh xc sut c 2 trm D, G c s c.

19. Cho A, B l hai s kin c P(A) = 0,45; P(B) = 0,30; P(AB) = 0.60. Hy tnh ccxc sut sau:

a. P( )BA ; b. P(B/A) ; c. P(AB) ; d. P(A/B)

20. Cho P(A) = 3/14; P(B) = 1/6; P(C) = 1/3; P(AC) = 1/7; P(B/C) = 5/21.

Tnh: a. P(A/C) ; b. P(C/A) ; c. P(BC) ; d. P(C/B)


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21. Nu mt cnbo xut hin bbin Philippin th cnbo s b vo Vit Namvi t l p1. Kinh nghim cho bit xc sut mt cnbo xut hin vng bin nytrong thng Tm l p2.

a. Tnh xc sut mt cn bo s xut hin bbin Philippin v s b vo Vit

Nam trong thng Tm nm nay.b. Nu cnbo hnh thnh vng bin Philippin m c lm nh i bng kthut phunho cht khibo qua vng bin Trng sa th kh nng n b vo Vit Nam s gimi 1/4. Tnh xc sut phn a trong trng hp ny. Bit rng cc cn bo xut hin vng bin Philippin khi b vo t lin lun i qua qun o Trng sa.

22. Mt nh phn tch th trng chng khon xem xt trin vng ca cc chng khonca nhiu cng ty ang pht hnh. Mt nm sau 25% s chng khon t ra tt hn nhiuso vi trungbnh ca th trng, 25% s chng khon t ra xu hn nhiu so vi trungbnh ca th trng v 50% bng trungbnh ca th trng. Trong s nhng chng khon

trnn tt c 40% c nh phn tch nh gi l mua tt, 20% s chng khon l trungbnh cng c nh gi l mua tt v 10% s chng khon trnn xu cng c nhgi l mua tt.

a. Tnh xc sut mt chng khon c nh gi l mua tt s tr thnh tt.

b. Tnh xc sut mt chng khon c nh gi l mua tt s tr thnh xu.

23. Mt i l ti H Ni kinh doanh ung do ba cng ty A, B, C sn xut theo t l

2 : 3 : 5. T l ung c ga tng ng ba cng ty trn l 70%, 60% v 50%.

a. Chn ngu nhin mt kin hng ti kho ca i l. Tnh xc sut kin ung cchn l ung c ga.

b. Bit kin hng c chn l ung c ga. Tnh xc sut kin hng do cng tyA sn xut.

24. Trong mt kho s lng ru loi A v loi B l nh nhau. Ngi ta chn ngu nhint trong kho ra mt chai ru v a cho 5 ngi snh ru nm th xem y l loiru no. Gi s xc sut on ng ca mi ngi l 0,7. C 3 ngi kt lun l ruloi A, 2 ngi kt lun l ru loi B. Tnh xc sut chai ru trn l ru loi A.

25. Mt trung tm phn phi ging cy trng nhn cy ging t 3 cs khc nhau theo tl: 2 : 3 : 5. T l cy ging xu tng ng l 5%, 3% v 2%.

a. Chn ngu nhin mt cy ging ca trung tm. Tnh xc sut cy ging c chnl cy xu.

b. Bit cy ging c chn l cy ging xu. Kh nng cy ging thuc cs no lcao nht? Ti sao?

26. Cho lai g lng xm thun chng (tnh trng tri) vi g lng trng thun chng (tnhtrng ln) th h F1 tt c g con u c lng mu xm , th h F2 g c lng mu


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xm v mu trng theo t l 3 : 1. Bit t l g lng xm thun chng , g lng xm khngthun chng, g lng trng thun chng trong mt n g l 1 : 2 : 1. Mt qu trng gca mt g m lng xm khng thun chng sp nra mt ch g con.

a. Tnh xc sut g con nra c lng mu trng.

b. Bit g con nra c lng mu xm. Tnh xc sut g b c lng mu trng.

27. T l ngi c k sinh trng st rt trong mu ca mi ngi dn vng cao l 0,2.

a. Chn ngu nhin 4 ngi. Tnh xc sut trong 4 ngi c chn c 3 ngi trongmu c k sinh trng st rt.

b. Ly mu ca 100 ngi em th. Tnh xc sut c t nht mt ngi c k sinhtrng st rt trong mu.

28. C hai t hc sinh. T th nht c 4 nam 5 n, t th hai c 5 nam 6 n. Chon ngunhin ra mi t 3 hc sinh ri ghp mi hc sinh t ny vi mi hc sinh ca t kia lmmt nhm hc tp

a.Tnh xc sut cc nhm hc tp u cng gii.b.Tnh xc sut cc nhm hc tp u khc gii

29. Nhn ngy quc t ph n, sinh vin A vo ca hng hoa ti cng trng mua ngunhin 3 bng hoa tng cho 3bn n mi ngi 1 bng. Sinh vin B cng vo cahng hoa ny v cng mua ngu nhin 3 bng hoa tng cho 3bn n ni trn mingi 1 bng. Ca hng hoa ch bn 3 loi hoa l hngbch, hng vng v hng nhung.a. Tnh xc sut mibn n c tng 2 bng hoa cng mu.b. Tnh xc sut mibn n c tng 2 bng hoa gm 2 mu khc nhau.

30. Mt hp u ging gm 2 ht u trng v 3 ht u . Mt hp khc gm 3 ht utrng v 2 ht u . T l ny mm l 0,8 i vi mi ht u trng, l 0,7 i vi miht u . Ly ngu nhin t mi hp ra 2 ht em gieo.a. Tnh xc sut c 4 ht u ny mm.b. Bit 4 ht em gieo u ny mm. Tnh xc sut 4 ht ny u l u .

31. Lai hai ging hoa mu hng v mu thun chng, cc cy con F1 c th cho hoamu hng, mu hoc mu cnh sen vi t l 1: 1: 2. Chn ngu nhin 5 ht hoa F1emgieo. Tnh xc sut :a. C ng 3 cy cho hoa mu .b. C 2 cy hoa mu , 3 cy mu hng.c. C 1 cy mu , 1 cy mu hng v 3 cy mu cnh sen.

32. Di tc ng caphng x cc nhim sc th ca mt t bob gy lm hai mnhtrong ch c mt mnh cha tm ng. Cc mnh gy theo thi gian s t ghp li vinhau mt cch ngu nhin v t bo s sng st nu mi cp mnh ghp vi nhau chcha mt tm ng. Tm xc sut t bo sng st, bit rng t bo c n nhim scth b gy.


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33. (Bi ton Buffon) Trn mt phng c mt di cc ng thng song song cch unhau mt khong 2a. Gieo ngu nhin mt ci kim c chiu di 2l (l < a). Tnh xc sut ci kim ct mt trong nhng ng thng trn.

34. Hai tu thu cp vo mt cng tr hng mt cch c lp trong vng 24h. Bit

rng thi gia bc d hng ca tu th nht l 2h, ca tu th hai l 3h Tnh xc sut mt trong hai tu trnphi ch cp bn.

35. C n vin bib ngu nhin vo m ci hp (n > m).a. Tnh xc sut c ng 1 hp khng cha vin bi no.b. Tnh xc sut c 1 hp cha c n vin bi.c. Tnh xc sut mi hp c t nht 1 vin bi.

36. (Bi ton Banach) Mt nh ton hc c 2 bao dim, mi bao c n que. ng ta mibn ti 1 bao. Khi s dng nh bc hc rt ngu nhin 1 bao ri rt ra 1 que dng.Tm xc sut khi ngpht hin 1 bao ht dim th bao kia cn k que.

37. C k thng ht ging gm k loi khc nhau c gi n mt trung tmbo qunging. Trung tm ny c kphng c nh s t 1 n k miphngbo qun mt loiht ging. Do ngiph trch kthut ca trung tm vng mt, nhn vinbo v nhxp tm mi thng ht ging vo mtphng.a. Tnh xc sut khng thng no ng v tr.b. Tnh xc sut cc thng u ng v tr.

38*. Trn mt toa tu c 30 hnh khch. n ga tip theo mi hnh khch c th xung

tu vi xc sut 0,3. Ti ga ny mi hnh khch mi c th ln toa tu trn vi xc sut0,5. Tnh xc sut khi ra khi ga toa tu vn cn 30 hnh khch.

39. Mt ca hngbn mt loi sn phm trong 30% do nh my A sn xut, 40% donh my B sn xut, 30% do nh my C sn xut.T l sn phm loi mt ca ba nh my trn ln lt l: 0,9 ; 0,8 , 0,9.a. Mua ngu nhin mt sn phm ti ca hng. Tm xc sut sn phm mua c lloi mt.b. Bit sn phm mua cl loi mt. Tnh xc sut sn phm do nh my A snsut.

40. Ti mt vng dn c, t l ngi nghin ht thuc l l 0,2. Bit rng t l vim hngtrong s ngi nghin thuc l l 0,7 v vi ngi khng nghin l 0,2. Khm ngunhin 1 ngi th thy ngi b vim hng. Tnh xc sut ngi nghin thuc l.

41. C 8 ngi rt thm chn cc cn h trong mt chung c t tng 8 n tng 15,mi tng c 8 cn h.a. Tnh xc sut c 8 ngi trn u nhn c cc cn h trong cng mt tng.b. Tnh xc sut 8 ngi trn nhn c 8 cn h trn 8 tng khc nhau.


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Chng 3 Bin ngu nhin

nh ngha chnh xc mang tnh ton hc thun tu v bin ngu nhin vt khi yu cuca gio trnh. nh ngha c trnhby y mang tnh m t, tuy nhin n cng gipcho ngi hc hiu c th no l bin ngu nhin, bin ngu nhin ri rc, bin nhinlin tc. Cc khi nim khc nhbng phn phi xc sut hm phn phi cng nhhmmt xc sut u c trnhby vi nhng kin thc n gin nht. Cc s c trngquan trng nht ca bin ngu nhin nhk vng, phng sai, lch chun c trnhby k hn cc s c trng khc.Cc bin ngu nhin ri rc v lin tc thng gptrong thc t cng nh cc s c trng ca chng c gii thiu kh k. Khi nimvctngu nhin c gi thiu mt cch slc. Cc v d lin quan ti cc kin thcl thuyt cng nhcc ng dng thc t gip ngi hc hiu v c hng th hn i vimn hc. Lut s ln, mt s nh l v lut s ln v mt s nh l gii hn c giithiu slc trong chng ny.

I Bin ngu nhin

Khi tin hnh mtphp th ngu nhin, cc kt qu caphp th thng l cc c tnhnh tnh. Tuy nhin trong nhiuphp th mi mt kt qu ca php th thng c gntng ng vi mt gi tr nh lng no . Chng hn khi chi cc tr chi n tin mikt qu ca mt ln chi c gn tng ng vi mt s tin ( c tnh nh lng) mngi chi c hay mt hoc khi nhm bn mtpht n vo bia, mi kt qu ca vicbn tng ng vi im s ( c tnh nh lng) m x th t c.

1.nh ngha:

Bin ngu nhin (thc) l bin nhn gi tr l cc s thcph thuc vo kt qu caphp

th ngu nhin .Ta thng dng cc ch ci hoa X, Y, Z... ch cc bin ngu nhin v cc ch cithng x, y, z...hoc xi , yj.... . ch cc gi tr c th m bin ngu nhin nhn.

2. Cc v d:V d 1: Tung ng thi hai con xc xc. Gi X l tng s chm hai mt trn, X l

mt bin ngu nhin v c th nhn mt trong cc gi tr t 2 n 12

V d 2: Mt ngi nhm bn vo bia cho ti khi trng bia th ngng. Gi Y l s ncn dng. Y l mt bin ngu nhin nhn cc gi tr:

1, 2, ..., n, ...

V d 3: Thp sng lin tc mt bng n in cho ti khi dy tc ca bng n bchy. Gi Z l thi gianbng n sng. Z l mt bin ngu nhinQua ba v d trn ta thy c hai loi bin ngu nhin:

Loi th nht l loi bin ngu nhin ch nhn mt s hu hn hay v hn m c ccgi tr.

*Mt tp c gi l v hn m c nu tn ti mtphp tng ng mt - mt ti tpcc s t nhin N.


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Loi th hai l loi bin ngu nhin m n c th nhn cc gi tr trong mt khong hocmt s khong thc no . Loi bin ngu nhin th nht gi l bin ngu nhin ri rc.Loi bin ngu nhin th hai gi l bin ngu nhin lin tc.

Vic a ra mt nh ngha thun tu ton hc v bin ngu nhin khng c trnhby y. Ngi c mun bit c th tham kho cc ti liu dn ra cui gio trnh ny.

3. Bng phn phi xc sut ca mt bin ngu nhin ri rc.3.1.nh ngha: Bng phn phi xc sut ca mt bin ngu nhin ri rc X l mtbnggm hai dngDng trn ghi cc gi tr c th c ca bin ngu nhin X, dng di ghi cc xc suttng ng. Nu X nhn 1 s hu hn cc gi tr th bng phn phi xc sut ca X l:

X x1 x2 ... xi ... xn

P p1 p2 ... pi ... pn

Nu X nhn 1 s v hn m cc gi tr th bng phn phi xc sut ca X l

X x1 x2 ... xi ... xn ...

P p1 p2 ... pi ... pn ...

pi = P(X= xi) l xc sut X nhn gi tr l xi.

Do X nhn v ch nhn mt trong cc gi tr xi nn ta c =

n

1iip = 1 i vibng th nht

v

=1iip = 1 i vibng th hai.

3.2. Cc v d

V d 1: Mt ngi chi tr chi n tin bng cch tung ng thi 2 ng tin cn iv ng cht. Nu c hai xut hin mt sp anh ta c 100 ng, nu c hai xut hinmt nga anh ta mt 40 ng cn xut hin mt sp mt nga anh ta mt 30 ng. Gi Xl s tin anh ta nhn c sau mt vn chi. Lpbng phn phi xc sut ca X

Nhn thy X c th nhn cc gi tr - 40, - 30, 100 tng ng vi vic mt 40 ng , mt30 ng v c 100 ng.

Ta c P(X = - 40) =4

1)100X(P,

2

1)30X(P,

4

1====

Vybng phn phi xc sut ca X l

X - 40 - 30 100

P4

1

2

1

4

1

V d2 : Mt phng th nghim c cp ba triu ng tin hnh th nghim tmmt chng vi rt trong gia cm. Mt ln th nghim chiph mt triu ng. Nupht hinra chng vi rt ny th ngng th nghim. Nu khngpht hin ra th lm th nghim choti khi pht hin ra chng vi rt trn hoc ht kinh ph th dng. Gi Y l s tin m


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phng th nghim trn tit kim c. Lp bng phn phi xc sut ca Y bit cc thnghim l c lp vi nhau v xc sut tm ra chng vi rt mi ln th nghim l 0,3.

Ta thy Y c th nhn mt trong ba gi tr 0, 1, 2. Vi xc sut tng ng

P(Y= 0 ) = 0,72 = 0,49; P( Y = 1 ) = 0,7.0,3 = 0,21; P( Y = 2 ) = 0,3.

Vybng phn phi xc sut ca Y l

Y 0 1 2

P 0,49 0,21 0,3

V d 3: Mt ngi nhm bn vo mt mc tiu cho ti khi trng ch th dng. Ccln bn c lp, xc sut trng ch ca mi ln bn l p

(0 < p < 1). Gi Z l s nphi dng. Lpbng phn phi xc sut ca Z

Nhn thy Z c th nhn cc gi tr 1, 2, ..., n,...

P(Z = n) = qn-1p (q = 1 - p). Vybng phn phi xc sut ca Z l

Z 1 2 ... i ... n ...

P p qp qi-1p qn-1p ...

4. Hm phn phi xc sut4.1. nh ngha: Hm phn phi xc sut ca bin ngu nhin X l hm s F(x) hocFX(x) cho bi F(x) = P( X


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1/ Chng minh rng F(x) l hm phn phi xc sut ca bin ngu nhin X no 2/Gi X l bin ngu nhin c hm phn phi F(x).

Tnh: P(4

X0

2

, F(x) = 1 cng khng

gim.T nhng kt qu nu trn ta thy F(x) khng gim trn R. Theo cc yu cu mthm s vi bin s thc tr thnh hm phn phi xc sut ca mt bin ngu nhin no, hm F(x) tho mn cc yu c ny. Vy n l hm phn phi xc sut ca mt binngu nhin.Gi X l bin ngu nhin c hm phn phi xc sut F(x) nu trn, theo tnh

cht 4 ta c:

P(4

x0


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Tnh cht 7.3: =


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C F(x) =

>


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v nu

=1nnn |x|p hi t th k vng ton ca X l s M(X) hoc E(X) cho bi

E(X) =

=1nnn xp

Nu bin ngu nhin X c hm mt xc sut f(x) v nu +

dx)x(f|x| hi t th k

vng ton ca X l s E(X) = +

dx)x(xf

T cc nh ngha trn ta nhn thy:

* nh ngha ch ra cch tnh k vng ton ca bin ngu nhin .

*Cc bin ngu nhin ri rc nhn mt s hu hn cc gi tr lun c k vng ton.

* Cc bin ngu nhin ri rc nhn mt s v hn m c hoc khng m c ccgi tr c th khng c gi tr k vng .

* K vng ca bin ngu nhin X l gi tr c trng cho v tr (trng tm hoc trungtm) ca bin ngu nhin .

* K vng cn oc gi l trung bnh s hc ca bin ngu nhin.

1.2 Cc v d

V d 1: Mt nhm 10 ngi trong ba ngi cao 1,62 m, hai ngi cao 1,66m, haingi cao 1,70m v ba ngi cao 1,74m. Chn ngu nhin mt ngi trong nhm ngitrn. Gi X l chiu cao ca ngi c chn. Tnh E(X)

Ta c bng phn phi xc sut ca X

X 1,62 1,66 1,70 1,74

P 0,3 0,2 0,2 0,3Vy E(X) = 0,3.1,62 + 0,2.1,66 + 0,2.1,70 + 0,3.1,74 = 1,68m

V d 2: Sau mt nm bn hng, mt ca hng kinh doanh hoa ti ti H ni nhnthy s lng hoa Xbn ra trong ngy theo t l (xc sut ) sau:

X 9 10 11 12 13 14 15

P 0,05 0,10 0,15 0,25 0,20 0,15 0,10

Mt lng hoa ti mua vo 60.000 ngbn ra 100000 ng, nu trong ngybn khnght s cn li b vt b. S lng hoa cn mua vo l bao nhiu li nhun trungbnh thuc l cao nht?

thc hinbi ton trn ta lpbng sau:Hng u cabng ghi s lng hoa Y dnh mua vo trong ngy.

Ct u cabng ghi s lng hoa X c th bn ra trong ngy.

Ct cui ghi xc sutbn c s lng hoa tng ng.

giao gia dng i v ct j l tin li (trm nghn) thu c khi mua vo j lngbn ra ilng.


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Y

X

9 10 11 12 13 14 15 P

910

1112

13

14

15

360 300 240 180 120 60 0

360 400 340 280 220 160 100

360 400 440 380 320 260 200360 400 440 480 420 360 300

360 400 440 480 520 460 400

360 400 440 480 520 560 500

360 400 440 480 520 560 600

0,05

0,10

0,150,25

0,20

0,15

0,10

Vic quyt nh mua cc lng hoa hng ngy c th thc hin theo cc phng n sau:

Phng n 1: Mua vo 9 lng, tin li trungbnh l:

E1 = 360

Phng n 2: Mua vo 10 lng, tin li trung bnh l:

E2 = 300. 0,05 + 400. 0,95 = 395Phng n 3: Mua vo 11 lng, tin li trungbnh l:

E3 = 240. 0,05 + 340. 0,10 + 440. 0,85 = 420

Phng n 4: Mua vo 12 lng, li nhun l:

E4 = 180. 0,05 + 280. 0,1 + 380. 0,15 + 480. 0,70 = 430

Phng n 5:Mua vo 13 lng, li nhun l:

E5 = 120. 0,05 + 220. 0,1 + 320. 0,15 + 420. 0,25 + 520. 0,45 = 415

Phng n 6: Mua vo 14 lng, li nhun trungbnh l:

E6 = 60. 0,05 + 160. 0,1 + 260. 0,15 +360. 0,25 + 460. 0,20 + 560. 0,25 = 380

Phng n 7: Mua vo 15 lng, li nhun trungbnh l:E7 = 0. 0,05 + 100. 0,1 +200. 0,15 + 300. 0,25 + 400. 0,2 + 500. 0,15 + 600. 0,1 = 330

T cc kt qu trn ta thy khi ca hng mua vo 12 lng hoa th li nhun trungbnh lcao nht.

V d 3: Mt x th nhm bn vo mt mc tiu cho ti khi trng mc tiu th dng.Cc ln bn c lp, xc sut trng mc tiu ca mi ln bn l 0,8. Gi X l lng nphi dng. Tnh k vng ca X

X c bng phn phi xc sut sau

X 1 2 ... n ...

P 0,8 0,2.0,8 ... 0,2n-1

0,8 ...

Do X ch nhn cc gi tr nguyn dng nn nu chui

=

1n

1n 8,0.2,0.n hi t th gi tr

chnh l k vng ton ca X

Xt

=

=1

)(n

nxxf vi x )1,0( . Do chui hi t u trong min ang xt nn


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=

=1n

1nnx)x(f

Mt khc

=

= =

+=

==

1n22

1n

1n

n

)x1(

1

)x1(

xx1)

x1

x(nxx

x-1

x

=

====1

221 25,1

8,08,0)(

8,01)2,0('2,0.

n

nXEfn

V d 4:i bin ngu nhin X c hm mt )x1(

1)x(f

2+=

( phn phi Cauchy ). Tnh E(X)

Ta c: dxx1

x2dx)x(f|x|

2+

+

+

=

Do tch phn ny phn k nn X khng c k vng.

V d 5: Bin ngu nhin X c hm mt

=

][0,xnu

][0,xu

xsin2

1n0

)x(f

Ta c E(X) = +

dx)x(xf =

=

0 2xdxsinx

2

1

1.3 Tnh cht

1/K vng ca hng s bng chnh n

Tht vy ta c th coi hng s l bin ngu nhin ch nhn gi tr C vi xc sut bng 1

nn E(C) = 1.C = C2/ Hng s c th a ra ngoi du k vng

Xt: Y= kX, nu X l bin ngu nhin ri rc vi P(X = xi) = pi th

P(Y = kxi) = pi. Vy E(Y) = == )X(kExpk)kx(p iiii 3/K vng ca mt tng bng tng cc k vng

Ta chng minh trong trng hp X, Y l cc bin ngu nhin ri rc.

Gi: Z = X + Y vi zij = xi + yj , P(Z = zij ) = pij

Ta c:

E(Z) = +=+=+= ijjijijijiijjiijijij pypxypxp)yx(pzp = +=+ )Y(E)X(Eypxp jjii

=

=

======1i

ijjj1j

ijii p)yY(Pp;p)xX(Pp

H qu 1: E(aX + b) = aE(X) + b


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H qu 2: E( ==

=n

1iii

n

1iii )X(Ea)Xa

p dng nhiu ln tnh cht 2 v tnh cht 3 ta c hai h qu trn

4/ Nu X c lp vi Y th E(XY) = E(X). E(Y)

t: Z = XY, jiij yxz = . Gi s P(X = xi) = ip , P(Y= yj) = jp jiij p.p)zZ(P == == jjiiijji yp.x.pzp.p)Z(E = = )Y(E).X(Eypx.p jjii

5/ Nu Y = )X( vi l mt hm s xc nh no . Nu X l bin ngu nhin

ri rc vi P(X= xi) = pi th P[Y = ii p)]x( = .

Vy E(Y) = )x(p ii Nu X c hm mt f(x) th Y c hm mt f(x) vy

E(Y) =

+

dx)x(f)x(

2. Phng saiXt bin ngu nhin X c k vng E(X)

2.1.nh ngha: Phng sai ca bin ngu nhin X l s k hiu lD(X) hoc VarX v D(X) = E[X-E(X)]2

T nh ngha trn ta thy:

* Nu X c phng sai th Xphi c k vng

* Phng sai cn c gi l lchbnh phng trungbnh ca X i vi k vng can.* Phng sai cng nh th X cng tp trung xung quanh k vng E(X)

Vy phng sai l i lng c trng cho mc phn tn ca bin ngu nhinquanhgi tr trungbnh l thuyt ca n.Xt: D(X) = E[X-E(X)]2 = E{X2-2XE(X) + [E(X)]2}

= E(X2)-2E(X)E(X) + E[E(X)]2 = E(X2)-[E(X)]2

t [E(X)]2 = E2(X) ta c D(X) = E(X2) - E2(X)2.2 Cc v d

V d 1: Bin ngu nhin X c bng phn phi xc sut

X 0 1 2

P 0,3 0,4 0,3

Tnh D(X).

C: D(X) = E(X2) - E2(X)

E(X) = 0.0,3 +1.0,4 + 2.0,3 = 1

E(X2) = 0.0,3 + 12.0,4 +22.0,3 = 1,6. Vy D(X) = 1,6 - 12 = 0,6


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V d 2: Mt x th nhm bn vo mt mc tiu cho ti khi trng ch th dng. Ccln bn l c lp, xc sut trng ch mi ln bn l 0,8. Gi X l lng n cndng.Tnh D(X)

Nhn thy X l bin ngu nhin nhn cc gi tr t nhin dng vi P(X = k) = 0,2k-1.0,8

E(X) =

=

=

= 1

1

1

1

2,0.8,08,0.2,0. k

k

k

k

kk

Xt chui:

=

=

==1k

1k

1k

k kx)x(fx (x)f(0,1)xvi

Mt khc f(x) =22 )x1(

1

)x1(

xx1)x(f

x1

x

=

+=

21

1

8,0

1)2,0('2,0. ==

=

fkk

k

25,18,0

1)( == XE

E(X2) =

=

=

=1k 1k

1k21k2 2,0.k8,08,0.2,0.k

=

=

+=1k 1k

1k1k 2,0.k8,02,0).1k(k8,0

=

+=1

2

8,0

12,0).1(2,0.8,0

k

kkk

T

=

=

=

===1

2

1

1

1

)1()('')(k

k

k

k

k

k xkkxfkxxfx (x)f;(0,1)xvi

=

=

1k

2kx)1k(k)x(f

==

=

= 2332

2,0).1(8,0

2)2,0(''

)1(

2)(''

)1(

1)( kkkf

xxf

xxf

E(X2) =223 8,0

2,1

8,0

1

8,0

4,0

8,0

1

8,0

22,0.8,0 =+=+

D(X) =222

22

8,0

2,0

8,0

1

8,0

2,1)()( == XEXE 0,3125

V d 3: Cho bin ngu nhin X c hm mt xc sut

=

][0,xnu

][0,xu

xsin21

n0)x(f Tnh D(X).

Ta c: E(X) = +

dx)x(xf =

=

0 2xdxsinx

2

1


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E(X2) = +

dx)x(fx 2 =

=0

22 1

2xdxsinx

2

1

Vy: D(X) = 144

12

222

=

2.3 Cc tnh chtca phng sai1/ Phng sai ca hng s bng 0

Tht vy: D(C) = E(C2) - E2(C) =C2 - C2 = 0

2/ Hng s a ra ngoi du phng saiphibnh phng ln

V: D(kX) = E(k2X2) - [E(kX)2] = k2E(X2) - k2E2(X)

= k2[E(X2) - E2(X)] = k2D(X)

3/ Nu X c lp vi Y th D(X+Y) = D(X) + D(Y)

Ta c: D(X + Y) = E[(X+Y)2] - [E(X +Y)]2

= E( X2 +2XY + Y2) - [E(X) + E(Y)]2

= E(X2) + 2E(XY) + E(Y2) - E2(X) - 2E(X)E(Y) - E2(Y)= E(X2) + E(Y2) - E2(X) - E2(Y) = D(X) +D(Y)

H qu 1: D(aX +b) = a2D(X)

H qu 2: Nu X1, X2,....,Xnc lp th D( = )X(Dk)Xk i2iii Hai h qu trn suy trc tip t ba tnh cht va nu. Minh ho cho s hu ch ca h qu2 ta xt v d sau.

V d: Gieo ng thi 10 con xc xc cn i v ng cht

Gi X l tng s chm cc mt trn. Tnh D(X)

Gi Xi l s chm mt trn ca con xc xc th i

X = =

10

1iiX Do cc Xi l c lp vi nhau nn D(X) =

=

10

1ii )X(D

P(Xi =1) = P(Xi =2) =...= P(Xi =6) =6

1

E(Xi) =6

1( 1 + 2 + ... + 6) =3,5

E( 2iX ) = 6

1(1 + 4 + ... +36) =

12

35

36

105

36

441546)

6

21(

6

91)(;

6

91 2 ==

==iXD

Vy: D(X) = 12

350

36

1050= 6

175=

3. lch chunVic dng phng sai o mc phn tn ca bin ngu nhin quanh gi tr k vngca n s trnn khng thch hp i vi bin ngu nhin c th nguyn (c n v o ikm) bi nu X c th nguyn bc nht th phng sai D(X) li c th nguyn bc hai.


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khc phc nhc im ny ngi ta a ra mt gi tr cng c trng cho mc phn tn ca X nhng c cng th nguyn vi X l lch chun

3.1.nh ngha: lch chun ca bin ngu nhin X l s)X(D)X( =

* lch chun cng th nguyn vi X nu X l bin c th nguyn* Bin ngu nhin X c lch chun khi v ch khi X c phng sai3.2. Tnh cht

1/ 0)X(

2/ )X(|k|)kX( =

3/ Nu X c lp vi Y th :

)Y()X()Y(D)X(D)YX( ++=

Vic chng minh cc tnh cht trn kh d dng nn dnh cho ngi c

4. Mt s gi tr c trng khcNgoi cc gi tr c trng l k vng, phng sai hoc lch chun ngi ta cn a ramt s gi tr c trng khc4.1Mmen: Mmen cp k ca bin ngu nhin X i vi a l s

k(a) = E(X - a)k

* Nu a = 0 th k(0) gi l m men gc cp k ca X

* Nu a = E(X) th k(a) gi l m men trung tm cp k ca X v k hiu l k

Ta c (0) = E(X), 2 = D(X), 1 = 0 vi mi bin ngu nhin

Mt khc khi X c phn phi i xng qua k vng E(X) th k = 0 vi k l s t nhin l.

V vy ta c th s dng 3 xt xem phn phi xc sut ca X c i xng hay khng.Tuy nhin, nu X l i lng c th nguyn th 3 c th nguyn bc ba so vi X, v vy

ngi ta dng H3 = 33

lm s o cho tnh cht i xng hay khng i xng ca X v H3

gi l h s bt i xng ca X. Ngi ta dng h s 3H44

4

= lm h s nhn ca

phn phi xc sut ca X, 2 l phng sai ca X.4.2Mode: Nu X l bin ngu nhin ri rc th mode ca i lngngu nhin X l s k hiu l ModX tho mn P(X = modX) P(X = xi) xi

Nu X l bin ngu nhin lin tc th ModX l im m ti hm mt f(x) t gi trcc i.

V d 1: Cho X c bng phn phi xc sut

X 1 2 3 4 5

P 0,3 0,2 0,1 0,3 0,1

ModX = 1 hoc ModX = 4. X c 2 Mod


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V d 2: Cho X c hm mt f(x) cho bi th sau:

Hnh 3

ModX l honh im cc i ca hm mt f(x)4.3. Phn v:

Phn v mc 1 - ca bin ngu nhin X l s X xc nh bi:P(X < X ) 1 - )( XXP

Nu = 0,5 th X0,5 gi l trung v ca X

Hnh 4

III. Mt squi lut xc sut ri rc thng gp

1. Phn phi nh thc1.1nh nghaBin ngu nhin X c gi l c phn phi nh thc nu X c bng phn phi xc sutsau:

X 0 1 . . . k . . . n

P n00n qpC

1n1n qpC

knkkn qpC 0nnn qpC


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Vi 0 < p < 1, q = 1 - p

Khi X c phn phi nh thc ta k hiu X B(n , p), n, p gi l cc tham s ca phn phiPhn phi nh thc vi n = 1 cn gi l phn phi Bernoulli

T nh ngha trn ta thy: S ln xut hin s kin A trong mt lc Bernulli c phnphi nh thc.1.2K vng vphng sai

X B(n,p) ta c E(X) = =

n

0k

k knkkn qpC = p

=

n

1k

k kn1kkn qpC

Xt: ( ==

)qxCn

1k

knkkn

=

n

1k

k kn1kkn qxC

Mt khc nn

0k

knkkn )xq(qxC +=

=

nn 1nn

0k

knkkn )xq(n)qxC(

=

+=

=

n

1k

k kn1kkn qxC = n(q + x)n-1

Thay x = p ta c: E(X) = p=

n

1k

k kn1kkn qpC = p. n(q + p)n-1 = np

Phng sai: D(X) = E(X2) - E2(X)

E(X2) = =

n

0k

2k knkkn qpC = p2

=

n

1k

k (k-1) kn2kkn qpC +

=

n

0k

k knkkn qpC

Xt: (q + x)n = =+

=

2nn

0k

knkkn )xq)(1n(nqxC

=

n

1k

k (k-1) kn2kkn qxC

Thay x = p ta c: n(n - 1) = =

n

1kk (k-1) kn2kkn qpC .

Mt khc =

n

0k

k knkkn qpC = E(X) = np. vy ta c E(X2) = n( n - 1)p2 + np

D(X) = n2p2 - np2 + np - (np)2 = np(1 - p) = npq

V d 1: Xc sut mt cy sng sau mt thi gian trng l 0,8. Trng 1000 cy. GiX l s cy sng sau mt thi gian trng. Tnh E(X), D(X).

Ta c X B(1000; 0,8) vy E(X) = 1000.0,8 =800,

D(X) =1000.0,8.0,2 =160

2. Phn phi siu bi2.1 nh ngha: Xt mt m ng gm N c th trong c M c th c c tnh A.Chn ngu nhin n c th ( n M, n N - M). Gi X l s c th c c tnh A trong n cth c chn. X l bin ngu nhin c phn phi siu bi.

Nhn thy X c th nhn 1 trong cc gi tr t 0, 1, ... ,n


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P(X = k) =nN

knMN

kM

C

CC Nh vy quy lut siu biph thuc vo ba tham s N, M, n.

2.2 Bng phn phi xc sutca quy lut siu bi, k vng, phng sai ca phn phisiu bi.

T cng thc xc sut: P(X=khng = nN

knMN

kM

CCC

ta c bng phn phi xc sut ca phn

phi siu bi l:

X 0 1 . . . k . . . n

P nN

nMN

0M

C

CC nN

1nMN

1M

C

CC ...nN

knMN

kM

C

CC ...nN

0MN

nM

C

CC

T bng phn phi cc xc sut ca phn phi siu bi xt t s ca

P( X = k+1) v P(X = k) ta c:

Nu 2N )1n)(1N( +++ khngphi l s nguyn th ModX = [ 2N )1n)(1N( +

++ ]

Cn nu2N

)1n)(1N(

+

++= k0 l s nguyn th ModX l k0 hoc k0 -1

p dng cng thc tnh k vng v phng sai ca X ta c:

E(X) =N

nMv D(X) = )

1

11(.

N

n

N

MN

N

nM

Nhn xt 1: Nu trong m ng gm N c th trong c M c th c c tnh A,chn ngu nhin ln lt n c th c hon li. Gi Y l s c th c c tnh A trong n cth th

Y B(n,p) vi p =NM

Theo cng thc tnh k vng v phng sai ca phn phi nh thc

ta c:

E(Y) = np =N

nM, D(Y) =

N

MN.

N

nM

T y ta c E(X) =E(Y), D(X) < D(Y). Nh vy phn phi siu bi v phn phi nhthc (tng ng vi vic ly mu khng hon li v c hon li t mt m ng s cp ti trong phn thng k) c cng k vng. Tuy vy phng sai ca phn phi siu binh hn phng sai ca phn phi nh thc. S khc bit ny cng cao nu n cng ln.Kt qu nyph hp vi trc quan l mu khng lp li t phn tn hn mu c lp li.

Tuy nhin nu s lng cc c th trong m ng N l rt ln so vi n th s sai khcgia phn phi siu bi v phn phi nh thc l khng ng k. iu ny c ngha l nuX v Y l phn phi siu bi v phn phi nh thc tng ng ni trn khi N ln so vin th:


54/156


P(X = k) P(Y = k) k = 0, 1 ..., n

Nhn xt 2:iu kin MNn c th b qua, khi X c th c th khng nhnmt s gi tru hoc mt s gi tr cui cn cng thc tnh xc sut, k vng, phngsai ca X vn tnh nh c.

2.1 Cc v dV d 1: T mt n g gm 10 con trong c 5 con mc bnh A. Chn ngu nhin

ra 3 con. Gi X l s g mc bnh A trong 3 con g c chn, X l bin ngu nhin cphn phi siu bi

V d 2: Mt t hc sinh gm 5 nam 4 n, chn ngu nhin 3 ngi, gi Y l s nsinh trong 3 ngi c chn. Y l bin ngu nhin c phn phi siu bi.

3. Phn phi hnh hc.3.1nh ngha: Bin ngu nhin X c phn phi hnh hc vi tham s thc p > 0 nu Xc bng phn phi xc sut sau:

X 1 2 3 .k.n

P p p(1-p) p(1-p)2

p(1-p)k-1

p(1-p)n-1

.3.2 Cc sc trng:

t q = 1 p => E(X) =

=

=

=1

1

1

1 ...k

k

k

kpkppqk

Xt chui:

=

=

==1k

1k

1k

k kx)x(fx (x)f(0,1)xvi

Mt khc f(x) =22 )x1(

1)x1(xx1

)x(fx1

x

=

+=

21

1 1)('.p

qfqkk

k ==

=

pXE 1)( =

E(X2) =

=

=

=1 1

1212 ...k k

kk pkppqk

=

=

+=1 1

11 .).1(k k

kk qkpqkkp

=

+=1

2 1).1(.k

k

pqkkqp

T

=

=

==== 1

2

1

1

1 )1()('')( kk

k

k

k

k

xkkxfkxxfx (x)f;(0,1)xvi

=

=1k

2kx)1k(k)x(f

==

=

= 2332

).1(2

)('')1(

2)(''

)1(

1)( kqkk

pqf

xxf

xxf


55/156


E(X2) =223

21212

p

pq

pp

q

pppq

+=+=+

D(X) =2222

22 1212)()(p

q

p

pq

pp

pqXEXE =

+=

+=

V d 3: Tin hnh lin tip cc th nghim tm nguyn t vi lng trong cc mu t choti khi th nghim thnhcng th dng. Cc th nghim thc hin c lp, Xc sut thngcng mi th nghim l 0,6. Gi X l s ln phi thc hin th nghim.

1/ Ch ra qui lut xc sut ca X.

2/ Tnh E(X), D(X).

T gi thuyt suy ra: X nhn cc gi tr t nhin dng, 14,0.6,0)( == nnXP . Vy cphn phi hnh hc vi tham s p = 0,6.

1111,16.0

4.0)(;6666,1

6,0

11)(

22====

p

qXD

pXE

4. Phn phi Poisson4.1nh ngha: Bin ngu nhin X c phn phi Poisson vi tham s thc > 0 nu Xc bng phn phi xc sut sau:

X 0 1 2 ... n ...

P e !1

e

!2e

2 ...!n

en ...

Ta k hiu X P c l X c phn phi Poisson vi tham s

4.2 Cc s c trng

Xt: k1k)kX(P

)1kX(P ==

+= T y ta c

Nu khng l s nguyn th ModX = [ ]

Nu l s nguyn th ModX = hoc ModX = -1

E(X) = ==

=

=

=

e.e)!1n(.e

!n.ne

1n

1n

1n

n

Tng t trn ta cng c D(X) = . Nh vy phn phi Poisson rt c bit l c k vngv phng sai u bng tham s

Nhn xt: Gi X(t) l s ln xy ra s kin A trong qung thi gian [0,t). Hin tng

ngu nhin trn c gi l mt qu trnh Poisson nu n tho mn 3 yu cu sau:* Nu t < t th qui lut ca bin X(t) - X(t) ch ph thuc vo ttt = ch khngphthuc vo t. Yu cu ny c gi l yu cu thun nht theo thi gian.

* Nu t < 11 ttt


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0t

)t(Plim;

t

)t(Plim 2

0t

1

0t=

=

Vic nhn in thoi ti mt trm in thoi l mt qu trnh Poisson. S ln cc cucin thoi gi n trm in thoi ny l bin ngu nhin c phn phi Poisson. Qu trnhphn r cc nguyn t phng x cng l mt qu trnh Poisson v s nguyn t b phn

hu trong mt qung thi gian cng l mt bin ngu nhin c phn phi Poisson.

5. Phn phi a thc5.1nh ngha: Xt mt dy nphp th c lp , sau miphp th c th

c mt trong k s kin A1, A2, ... , Ak xy ra. Xc sut P(Ai) = pi

P(Ai) = pi , 1pk

1ii =

=

. Gi Xi l s ln xut hin s kin Ai trong n ln th. Lut phn

phi ca cc bin X1, X2, ... ,Xkc gi l lut a thc.

Ta thy X1 + X2 + ... + Xk = n , khc vi cc qui lut nu trn qui lut a thc l qui

lut ca k - 1 bin ngu nhin. Bin th kph thuc vo cc bin cn li bi:Xk = n - (X1 + X2 + ... + Xk-1)

Nu k = 2 lut a thc tr thnh lut nh thc5.2 Cng thcxc sutca luta thc

Gi s X1, X2, ... ,Xk l cc bin ngu nhin c lut a thc

Ta i tnh xc sut P(X1 = n1, X2 = n2, ... ,Xk = nk) vi nnk

1ii =

=

Gi ijA l s kin Ai xut hin ln th thj, i = n,1j,k,1 = nn mt trong nhng kt

qu ca dy nphp th tho mn Xi = ni c xc sut xut hin l: k21n

kn

2n

1 p....pp

C tt c!n!...n!n

!n

k21

.Vy xc sut cn tnh l:

P(X1 = n1 , X2 = n2 , , Xk = nk) =!n!...n!n

!n

k21

. k21 nkn

2n

1 p....pp

IV Cc phn phi lin tc

1. Phn phi u1.1.nh ngha: Bin ngu nhin X c hm mt

=b][a,xnu

a-b

1

b][a,xnu0)x(f

c gi l bin ngu nhin c phn phi u trn on [a,b]

Nu X c phn phi u trn on [a,b] th hm phn phi xc sut ca X l


57/156


>

XXPlim,0 nn

= 1. Khi ta k hiu:

Xn XP .1.3 Hi t theo quy lut : Dy cc bin ngu nhin { }nX gi l hi t theo quy lut ti

bin X nu dy hm phn phi{ })x(Fn ca dy { }nX hi t ti hm phn phi F(x) ca

X.Ngi ta chng minh c rng nu mt dy bin ngu nhin hi t hu chc chn ti Xth n cng hi t theo xc sut ti X. Mt dy bin ngu nhinhi t theo xc sut ti Xth cng hi t theo quy lut ti X.

2 Lut s ln v cc nh l v lut s ln.

2.1 Squa v lut s ln: Dy cc bin ngu nhin { }nX gi l tun theo lut s ln

nu ==

==n

1ii

n

1ii

P EXn

1a;X

n

1X;0aX .

gtxstk_dhnn1

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