Xacsuatthongke

8/8/2019 Xacsuatthongke

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Chng 1

GII TCH T HP

1.1. Quy tc nhn

Cc tnh cht sau ca php m s l nn tng ca tt c cng vic ca chng ta.

Tnh cht 1 (Quy tc nhn)Gi s c 2 cng vic c thc hin. Nu cng vic1 c th thc hin mt trongm cch khc nhau vng vi mi cch thc hin cng vic1, cng vic2 c n cch thc hin khc nhau th cm.n cch khcnhau khi thc hin hai hai cng vic.

Proof: Tnh cht c bn c th c chng minh bng cch lit k tt c cc cch thc hin c thca hai cng vic nh sau:

(1, 1), (1, 2), . . . , (1, n)

(2, 1), (2, 2), . . . , (2, n)...

(m, 1), (m, 2), . . . , (m, n)

trong , chng ta ni cch thc hin l (i, j) nu cng vic 1 thc hin theo cch th i trong m cchc th v cng vic 2 thc hin cch th j trong n cch. V th tp tt c cc cch c th thc hinbng mn.

V d 1.1.1 Mt cng ng nh c 10 ph n, mi ngi c 3 ngi con. Chn mt ngi ph nv mt a con ca h. Hi c bao nhiu cch chn ?

GiiTa xem vic chn ngi ph n nh l cng vic 1 v vic chn con ca h l cng vic 2. Khi

t tnh cht c bn ta c 10.3 = 30 cch chn khc nhau.Khi chng ta c nhiu hn hai cng vic c thc hnh, tnh cht c bn c th c tng qut

ho nh sau:

Tnh cht 2 (Quy tc nhn tng qut)Gi s c k cng vic c thc hin. Nu cng vic1 c th thc hin trongn1 cch khc nhau v ng

vi mi cch thc hin cng vic1, cng vic2 c n2 cch thc hin khc nhau; ng vi mi cch thchin hai cng vic u, cn3 cch khc nhau thc hin cng vic3, v.. . v .. th c n1.n2.n3. . . . nk cchkhc nhau thc hink cng vic .

V d 1.1.2 Mt hi ngh hc tp mt trng i hc bao gm 3 sinh vin nm th nht, 4 sinhvin nm th 2, 5 sinh vin nm th 3 v 2 sinh vin nm cui. Mt tiu ban gm 4 ngi trong 4kho khc nhau. Hi c th lp c bao nhiu tiu ban khc nhau?

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2 Chng 1. GII TCH T HP

GiiVic chn mt tiu ban nh l vic thc hin 4 cng vic khc nhau. Cng vic i l chn mt

sinh vin nm th i( i = 1, 2, 3, 4 ). V th, t tnh cht c bn tng qut, chng ta c 3.4.2.5 = 120tiu ban khc nhau c th lp.

V d 1.1.3 S hiu ca bng li xe mt gm 7 k t, trong 3 k t u l cc ch ci v 4 k tsau l cc ch s. Hi c th c bao nhiu bng li xe mt khc nhau ?

Giip dng tnh cht c bn tng qut, chng ta c s bng li khc nhau c th c l:

26.26.26.10.10.10.10 = 175.760.000

Nu cc ch ci v ch s trong s hiu bng khc nhau th c bao nhiu bng li khc nhau?

V d 1.1.4 Mt hm s xc nh trn mt tp n phn t v ch nhn hai gi tr 0 v 1. Hi c th

lp c bao nhiu hm khc nhau.

Giit cc phn t l 1, 2, 3, . . . , n. V f(i) bng 1 hoc 0 cho mi i = 1, 2, . . . , n nn ta c 2n hm

khc nhau c th lp.

1.2. Hon v

C bao nhiu cch khc nhau khi sp xp c th t 3 k t a,b,c? Bng cch lit k trc tip

chng ta thy c 6 cch, c th l: abc, acb, bac, bca, cab v cba. Mi cch sp xp nh vy c gil mt hon v. V th c 6 hon v c th ca mt tp 3 phn t. Kt qu ny cng c th suy ra ttnh cht c bn, v phn t th nht trong hon v c th l mt trong 3 k t, phn t th 2 tronghon v c th chn mt trong 2 k t cn li v phn t th 3 c chn t mt phn t cn li. Vth, c 3.2.1 = 6 hon v c th.

Chng ta nh ngha khi nim hon v mt cch tng qut nh sau:

nh ngha 1.2.1 Cho n phn t khc nhau. Mt hon v ca n phn t l mt cch sp xp c tht n phn t cho.

Gi Pn l s hon v khc nhau c th lp t n phn t cho. Ta c

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

V d 1.2.5 Hi c bao nhiu cch sp xp v tr cc cu th(th mn, tin v phi, tri,. . . ) khcnhau trong mt i bng gm 9 cu th?

GiiC 9! = 362880 cch sp xp cc cu th.

V d 1.2.6 Mt lp hc l thuyt xc sut gm 6 nam v 4 n. Mt k thi c t chc, Cc sinhvin c xp hng theo kt qu lm bi ca h. Gii s khng c hai sinh vin no t cng mt

im.a) C th c bao nhiu cch xp hng khc nhau?b) Nu nam c xp hng trong nhm nam v n c xp hng trong nhm n th c th c

bao nhiu cch xp hng khc nhau?

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1.2. Hon v 3

Giia) Mi cch xp hng tng ng vi mt cch sp xp c th t 10 ngi, chng ta c cu tr li

trong phn ny l 10! = 3.628.800.b) V c 6! cch xp hng khc nhau trong 6 ngi nam v 4! cch xp khc nhau trong 4 ngi

n nn p dng tnh cht c bn, chng ta c 6!.4! = 17.280 cch sp xp khc nhau c th c.

V d 1.2.7 C Nga nh t 10 cun sch ln mt ci gi sch. Trong c 4 cun sch Ton, 3cun Ho hc, 2 cun Lch s v 1 cun Ngoi ng. C Nga mun sp xp nhng cun sch ca ccc cun sch ca minh sao cho cc cun cng mt mn thi k nhau. C th c bao nhiu cch spxp 10 cun sch khc nhau?

GiiC 4!.3!.2!.1! cch sp xp sao cho cc sch Ton u hng sau n cc sch Ho ri n

sch S v cui cng l sch Ngoi ng. Tng t, vi mi th t cc mn hc, chng ta c 4!.3!.2!.1!

cch sp xp khc nhau. y c 4! cch sp xp th t cc mn hc nn p n ca cu hi l c4!.4!.3!.2!.1! = 6912.By gi chng ta s xc nh s cc hon v ca mt tp n phn t khi m mt s phn t trong

hon v trng vi nhng phn t khc. i thng vo vn chng ta quan tm, hy xem xt v dsau:

V d 1.2.8 Hi c bao nhiu cch sp xp cc k t khc nhau t cc k t PE PPE R ?

GiiTrc ht chng ta ch rng c 6! hon v ca cc k t P1E1P2P3E2R khi 3 k t Pi v 2 k

t Ei c xem l khc nhau. Tuy nhin chng ta xem xt mt hon v bt k trong nhng hon vny, chng hn P1P2E1P3E2R. By gi nu chng ta hon v cc k t P vi nhau v hon v cc kt Evi nhau th kt qu vn s c dng PPEPER. l 3!.2! hon v

P1P2E1P3E2R P1P2E2P3E1RP1P3E1P2E2R P1P3E2P3E1RP2P1E1P3E2R P2P1E2P3E1RP2P3E1P1E2R P2P3E2P1E1RP3P2E1P1E2R P3P2E2P1E1RP3P1E1P2E2R P3P1E2P2E1R

c cng hnh thc nh PPEPER. V vy, c 6!/(3!.2!) = 60 cch sp xp cc k t khc nhau tcc k t PPEPER.

Cc hon v trong cc phn t c lp li nh trn c gi l hon v lp. Chng ta c nhngha chnh xc nh sau:

nh ngha 1.2.2 Mt hon v chp lp l mt cch xp xp c th t n phn t khng nht thitphn bit.

T v d (1.2.8), chng ta ch ra mt cch tng qut rng, c

n!

n1!.n2!. . . . nk!

hon v lp khc nhau ca n phn t, trong n1 phn t nh nhau, n2 phn t nh nhau,.. . , nkphn t nh nhau.

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V d 1.2.9 Mt vng thi u c vua c 10 u th. Trong c 4 ngi Nga, 3 ngi M, 2 ngiAnh v 1 ngi Brazil. Kt qu vng thi u ch ghi cc quc tch ca cc u th theo v tr m ht c. Hi c bao nhiu kt qu c th?

GiiC10!

4!.3!.2!.1!= 12600

kt qu c th.

V d 1.2.10 C bao nhiu tn hiu khc nhau, trong mi tnh hiu gm 9 c treo trn mt hng,c to ra t mt tp gm 4 c trng, 3 c v 2 c xanh nu tt c cc c cng mu l ging htnhau?

Gii

C 9!4!.3!.2!

= 1260

tn hiu khc nhau.

1.3. T hp

Chng ta thng quan tm n vic xc nh s cc nhm khc nhau gm k phn t c xydng t mt tng th gm n phn t. V d, c bao nhiu nhm gm 3 ch ci c chn t 5 chci A,B,C,D v E? tr li cu hi ny ta l gii nh sau: V c nm cch chn phn t u tin,

4 cch chn phn t tip theo v 3 cch chn phn t cui cng. V th c 5.4.3 cch chn nhmgm 3 phn t khi th t trong mi nhm c chn c lin quan. Tuy nhin, v mi nhm gm3 phn t, chng hn nhm gm ba ch ci A,B,Cs c m 6 ln(ngha l tt c cc hon vABC,ACB,BAC,CAB v CBA s c m khi th t la chn l quan trng). T suy rarng s cc nhm phn bit gm 3 ch ci c th to ra c l

5.4.3

3!= 10

Mi nhm con gm 3 phn t nh trn c gi l mt t hp chp 3 ca 5 phn t v s ccnhm con gm 3 phn t c gi l s cc t hp chp 3 ca 5. Ta c nh ngha tng qut nh sau

nh ngha 1.3.3 Cho mt tp n phn t. Mt t hp chp k ca n phn t(0 k n) l mt tpcon gm k phn t c ly ra t tp n phn t cho.

S cc t hp chp k ca n phn t, k hiu Ckn , c xc nh bi

Ckn =n(n 1) . . . (n k + 1)

k!

Cn nhn mnh rng trong mt tp con gm k phn t th khng phn bit th t ca cc phnt c chn.

V d 1.3.11 Mt hi ngh gm 3 ngi c thnh lp t mt nhm 20 ngi. Hi c th thnhlp c bao nhiu hi ngh khc nhau ?

GiiC C320 =

20.19.183.2.1

= 1140 hi ngh khc nhau c th thnh lp.

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1.3. T hp 5

V d 1.3.12 T mt nhm gm 5 n v 7 nam, hi c th thnh lp c bao nhiu hi ngh khcnhau gm 2 nv 3 nam? Trong trng hp c hai ngi nam hn th nhau v khng chu tham giacng mt hi ngh th c th thnh lp c bao nhiu hi ngh ?

GiiV c th thnh lp c C25 nhm gm 2 ph n v C37 nhm gm 3 nam nn t tnh cht c

bn ta suy ra c th lp c C25 .C37 = 350 hi ngh gm 2 n v 3 nam.

Mt khc, nu c hai ngi n ng t chi tham gia cng mt hi ngh th khi c C02C25 cch

chn nhm 3 ngi n ng khng c hai ngi hn th nhau v c C12 .C25 cch chn nhm 3 ngi

mi nhm cha ch mt trong hai ngi n ng hn th nhau. Nh vy c C02 .C35 + c

12.C

25 = 30

cch chn nhm ba ngi n ng khng c mt c hai ngi hn th nhau trong mt nhm. V cC25 cch chn 2 ngi n nn trong trng hp ny c 30.C

25 = 300 cch thnh lp hi ngh.

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Chng 2

PHP TNH XC SUT

2.1. PHP TH V S KIN

2.1.1. Php th v s kin

nh ngha 2.1.4 Php th l mt th nghim c th lp li trong cc iu kin bn ngoi ging htnhau v kt qu l mt phn t khng on trc c ca mt tp hp cc nh.

Vy d kin ca mt php th gm c: - Vic m t b my th nghim v vic ch dn cc iukin tin hnh.

- Vic xc nh tp hp cc kt qu ca th nghim.Ta xt cc v du sau:

V d 2.1.13 Ta gieo mt ng tin ng cht xung mt phng v quan st mt no xut hin lmt php th. Php th c hai kt qu l ng tin xut hin mt sp( S) hoc mt nga( N).

V d 2.1.14 Gieo mt con xc xc cn xng v ng cht trn mt mt phng v quan st mt noxuthinlmtphpth. Ccktqucaphpthlsxuthinmttrong6mtcaconxcxcm ta c th k hiu bng cc s trn mt: 1, 2, 3, 4, 5, 6.

V d 2.1.15 Trong mt hp kn c m bi , n bi xanh hon ton ging nhau v kch thc, trnglng. Ly ngu nhin mt bi v quan st xem bi c mu g l mt php th. Php th c hai kt qu:bi ly ra mu xanh v bi ly ra mu .

2.1.2. S kin lin kt vi php thS kin (hay cn gi bin c) l mt khi nim thng gp trong l thuyt xc sut. Ta khng c

mt nh ngha cht ch khi nim ny. S kin c hiu nh l mt s vic, mt hin tng no ca cuc sng t nhin v x hi.

nh ngha 2.1.5 Mt s kin ln kt vi mt php th l s kin c th xy ra hay khng xy raty thuc vo kt qu ca php th .

S kin thng c k hiu bng cc ch ci in hoa A , B , C , . . . .Mt s kin xy ra khi v ch khi c mt kt qu c th trong s nhng kt qu ca php th th

c gi l s kin c bn hay cn gi l s kin s cp. Tp hp tt c cc s kin s cp gi l

khng gian s cp, k hiu .S kin tt yu l s kin lun xy ra khi thc hin php th.S kin bt kh l s kin khng bao gi xy ra khi thc hin php th.S kin ngu nhin l s kin c th xy ra hoc khng xy ra khi thc hin php th.

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8 Chng 2. PHP TNH XC SUT

V d 2.1.16 Ta gieo mt ng tin ng cht xung mt phng v quan st mt no xut hin. GiN l s kin xut hin mt nga, S l s kin xut hin mt sp. Ta c S, N l cc s kin s cp vkhng gian s cp l = {S, N}.

Gi A l s kin khng xut hin mt no c th A l s kin bt kh. Gi B l s kin xut hinmt no ca ng tin, B l s kin tt yu.

V d 2.1.17 Gieo mt con xc xc cn xng v ng cht trn mt mt phng v quan st mt noxut hin. Gi Mi l s kin xut hin mt i chm ( i = 1, . . . , 6 ), Mi l cc s kin s cp. Khnggian s cp = {M1; M2; M3; M4; M5; M6}.

Gi A lskinxuthinmtcschmlschn. Khi A xyrakhivchkhi M2 hoc M4hoc M6 xy ra. Ta ng nht s kin A vi tp hp {M2; M4; M6}. Ta vit

A = {M2; M4; M6} cc s kin s cp M2; M4; M6 gi l cc s kin thun li cho s kin A v A xy ra khi v ch

khi mt trong cc s kin s cp thuc n xy ra.Tng t, nu gi B l s kin con xc xc xut hin mt c s chm l s l, C l s kin con

xc xc xut hin mt c s chm ln hn 4, D l s kin tt yu, E l s kin bt kh. Ta c:

B = {M1; M3; M5} C = {M5; M6} D = E =

Nh vy vi cch k hiu trn ta thy:- Mi s kin tng ng vi mt tp hp con ca khng gian s cp v ngc li, mt tp con ca

xc nh duy nht mt s kin no . Nh vy, mi s kin c xem nh mt tp con ca khnggian s cp.

- Nu s kin A th cc s kin s cp thuc A gi l cc s kin thun li cho s kin A .2.1.3. Cc php ton v quan h ca cc s kin

Tng: Tng ca hai s kin A v B, k hiu A + B (hoc A B), l mt s kin xy ra khi tnht mt trong hai s kin A, B xy ra.

Tch: Tch ca hai s kin A v B, k hiu A.B (hoc A B), l mt s kin xy ra khi c Av B ng thi xy ra.

Hiu: Hiu ca hai s kin A v B, k hiu A B (hayA \ B), l s kin xy ra khi A xy ra

v B khng xy ra, tc l A B = A.B. i lp: i lp ca A, k hiu A, l s kin khng xy ra s kin A. Ta suy ra A = A v

A + A = : s kin tt yu, A.A = : s kin bt kh, = . Xung khc: Hai s kin A v B gi l xung khc nu chng khng th xy ra, tc A.B = . Ko theo: S kin A gi l ko theo s kin B, k hiu A B, nu s kin A xy ra th s

kin B xy ra, tc l A B. Tng ng: Hai s kin A v B gi l tng ng, k hiu A = B, nu s kin A xy ra

th s kin B xy ra v ngc li, tc l A

B v B

A.

Khi ta xem mi s kin nh l mt tp con ca khng gian s cp th cc php ton trn ccs kin tng ng vi cc php ton v tp hp m chng ta quen bit v c th minh ha chngbng cc biu Ven.

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2.1. PHP TH V S KIN 9

V d 2.1.18 Gieo hai ng tin cn i v ng cht ln mt phng. Gi:A = S kin xut hin mt sp (S) trn ng tin th 1.

B = S kin xut hin mt nga (N) trn ng tin th 2.C= S kin xut hin mt nga (N) trn ng tin th 1.D = S kin xut hin t nht mt mt sp (S).E= S kin xut hin nhiu nht mt mt sp (S).a) Xc nh khng gian s cp v biu din cc s kin trn theo ngn ng tp hp.

b) Hy din t cc s kin sau bng ngn ng thng thng v ngn ng tp hp:A B, A C,BC,BD,CE,A,B,D,E,AB C.

c) Gi F l s kin khng xut hin mt nga. F tng ng vi s kin no.

Gii

a) Ta k hiu XY ngha l: X l mt xut hin ca ng tin th nht, Y l mt xut hin cang tin th 2. X, Y nhn hai gi tr l sp (S) v nga (N). Khi ta c khng gian s cp l:

= {SS,SN,NN,NS}

A = {SS,SN}, B = {SN,NN}, C = {N N , N S }, D = {SS,SN,NS}, E = {S N, NN, NS }b) Ta c:A B: l s kin ng tin th 1 xut hin mt sp hoc ng tin th hai xut hin mt nga.

A B = {SS,SN,NN}.A

C: l s kin ng tin th nht xut hin mt sp hoc nga. y l s kin tt yu,

A C = .BC: l s kin c hai ng tin xut hin mt nga, BC = {NN}.BD: l s kin ng tin th 1 xut hin mt sp v ng tin th hai xut hin mt nga.

BD = {SN}.CE: lskinngtinth1xuthinmtnga(chC E , C E = C). CE = {NS,NN}.Cc trng hp khc lm tng t, v dnh li nh mt bi tp.c) F tng ng vi D.

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2.2. CC NH NGHA V XC SUT

2.2.1. nh ngha xc sut theo li c in

nh ngha 2.2.6 Xt php th vi khng gian s cp bao gm n kt qu ng kh nng. Gi s skin A bao gm m kt qu thun li cho A xy ra. Khi , xc sut ca s kin( bin c) A, k hiuP(A), c nh ngha bng cng thc

P(A) =m

n=

S kt qu thun li cho A xy raTng s kt qu ca khng gian s cp

V d 2.2.19 Gieo ng thi hai ng tin cn xng v ng cht. Tnh xc sut hai ng xuthin khc nhau?

Gii

Ta c khng gian s cp = {(S, N); (S, S); (N, S); (N, N)}. Trong ,S, N ln lt k hiuchosxuthinmtspvsxuthinmtngavktqu (S, N) ngha l ng tin th nht xuthin mt Sv ng tin th hai xut hin mt N, cc k hiu khc tng t.

Gi A l s kin hai mt ng tin xy ra khc nhau, ta c:

A = {(S, N); (N, S)}

Vy xc xut ca s kin A l: P(A) = 24 = 0, 5.

V d 2.2.20 Mt ngi gi in thoi nhng qun mt hai s cui ca s in thoi cn gi m ch

nh l hai s khc nhau. Tm xc sut ngi quay ngu nhin mt ln trng s cn gi?GiiGi A l s kin ngi quay ngu nhin mt ln trng s cn gi.Ta c, mi kt qu l mt cch gi 2 s cui nn khng gian s cp c s kt qu: n = A210 = 90.

Trong s kt qu thun li cho A: m = 1.Vy xc sut ca s kin A : P(A) = 1

90.

V d 2.2.21 Mt hp c 7 chnh phm v 3 ph phm. Ly ngu nhin t hp 3 sn phm. Tmxc sut c 3 sn phm ly ra l chnh phm.

GiiMi kt kt qu l mt cch ly ra 3 sn phm khc nhau t 10 sn phm nn khng gian s cp

csktqul: n = C310 = 120. Sktquthunlicho A lscchlyra3chnhphmt7chnhphm: m = C37 = 35.

Vy xc sut ca s kin A l P(A) = 35120

= 724

.T nh ngha c in ca xc sut, ta d dng suy ra c cc tnh cht sau:

0 P(A) 1; P() = 1; P() = 0;

Nu A, B xung khc (AB = ) th P(A + B) = P(A) + P(B); P(A) = 1 P(A); Nu A B th P(A) P(B).

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2.3. CC NH L C BN V PHP TNH XC SUT 11

nh ngha c in v xc sut ch p dng cho cc php th c hu hn kt qu ng kh nng.Trong thc t, c nhng php th c v s kt qu ng kh nng. Khi , nh ngha c in v xcsut khng p dng c. khc phc hn ch , ngi ta a ra nh ngha hnh hc ca xcsut nh sau:

Xt mt php th c v hn cc kt qu ng kh nng. Mi kt qu ca php th c biudin mi mt im trong mt phng( hoc trong khng gian). Gi s tt c cc kt qu ca php thc biu din bi mt min hnh hc G(chng hn on thng, mt min mt cong hoc mt khikhng gian ...), Cn tp cc kt qu thun li cho s kin A bi min con no S G. Khi

P(A) = o S o G

y ty thuc vo Sv G m o c th l di, din tch hoc th tch v lun gi s rngSv G u l cc tp o c v o ca G khc khng.

V d 2.2.22 ng dy in thoi ngm ni mt tng i vi mt trm di 1 km. Tnh xc sut dy t ti ni cch tng i khng qu 100m.

GiiR rng nu dy in thoi l ng cht th kh nng n b t ti mt im bt k l nh nhau

nn khng gian s cp c th biu din bng mt on thng MN = 1km ni tng i vi trm.Gi A l s kin dy t ti ni cch tng i khng qu 100m. S kin A c biu din bng mton thng MKc di 100m. T P(A) = 100

1000= 0, 1.

2.2.2. nh ngha xc sut theo li thng k

iu kin ng kh nng ca cc kt qu ca mt php th khng phi lc no cng c mbo. C nhiu hin tng xy ra khng theo cc yu cu ca nh ngha c in, chng hn, tnh xcsut mt a tr sp sinh l con trai, ngy mai tri mua lc 5 gi,.... C mt cch khc xc nhxc sut ca mt s kin nh sau:

Xt mt php th v s kin A lin kt vi php th . Gi s, php th c thc hin n lnv c m ln xut hin s kin A. Khi m c gi l tn sxut hin ca s kin Av t s m

nc

gi l tn sutxut hin s kin A. Tng t, nu php th thc hin li ln th 2, th 3, ...th tnsut xut hin s kin A tng ng l m1

n1v m2

n2.

Trn c s quan st lu di cc thc nghim khc nhau, ngi ta nhn thy rng tn sut xuthin mt s kin c tnh n nh, thay i rt t trong cc lot php th khc nhau v thay i xung

quanh mt hng s xc nh. S khc bit cng t khi s php th cng ln. Ni cch khc, khi sphp th tng ln v hn, tn sut xut hin s kin A dn n mt s xc nh, s gi l xc sutca s kin A

P(A) = limn

m

nTrong thc t, xc sut ca s kin A c ly gn ng bng tn sut xut hin ca s khi s

ln thc hin php th ln.

2.3. CC NH L C BN V PHP TNH XC SUT

2.3.1. nh l cng xc sutnh l 2.3.1 Nu A, B l hai s kin xung khc (ngha l A B = ) th

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

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Proof: Tnh cht ny l mt h qu ca nh ngha xc sut theo phng php tin . y, ara chng minh cho trng hp php th c hu hn kt qu ng kh nng(nh ngha c in caxc sut).

Gi s khng gian s cp c n kt qu ng kh nng. Gi mA l s kt qu thun li cho s kinA xy ra, mB l s kt qu thun li cho s kin B xy ra. V A v B xung khc nn khng c ktqu no thun li cho c Av B nn s kt qu thun li cho A + B xy ra l mA + mB . V th ta c

P(A + B) =mA + mB

nP(A) =

mAn

P(B) =mB

n

T ta c P(A + B) = P(A) + P(B).

H qu 2.3.1 1. Nu A, A l hai s kin i lp th P(A) = 1 P(A).2. Nu A1, A2, . . . , An l n s kin i mt xung khc th

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

V d 2.3.23 Mt hp c 6 bi v 4 bi xanh. Ly ngu nhin ra 3 bi, tnh xc sut a) 3 bi ly ra cng mu.b) 3 bi ly ra c t nht mt bi .

Giia) Gi A l s kin 3 bi ly ra cng mu; B l s kin 3 bi ly ra mu xanh v Cl s kin 3 bi ly

ra mu . Ta c A = B + C, hai s kin B, Cxung khc nn ta c

P(A) = P(B) + P(C) =C34C310

+C36C310

=1

30+

1

6=

1

5

b) Gi Ai l s kin ly ra c i bi (i=1,2,3), gi D l s ly ra t nht mt bi . Ta cD = A1 + A2 + A3, trong A1, A2, A3 i mt xung khc nn

P(D) = P(A1) + P(A2) + P(A3) =C16 .C

24

C310+

C26 .C14

C310+

C36 .C04

C310=

29

30

Ch bi ny cng c th c gii nh sau:P(D) = 1 P(D) = 1 C34C310

= 2930

. yD l skin 3 vin bi ly ra mu xanh.

V d 2.3.24 Mt l hng gm 10 sn phm, trong c 2 ph phm. Ly ngu nhin khng honli t l hng ra 6 sn phm. Tm xc sut trong 6 sn phm ly ra c nhiu nht l 1 ph phm?

GiiGi A l s kin ly ra 6 sn phm v khng c ph phm; B l s kin ly ra 6 sn phm v cng 1 ph phm; Cl s kin ly ra 6 sn phm v c nhiu nht l 1 ph phm. Ta c C = A + B,trong hai s kin A, B xung khc. p dng cng thc cng, ta c

P(C) = P(A) + P(B) =C68C610

+C58C

12

C610=

2

15+

8

15=

2

3

nh l 2.3.2 (nh l cng m rng) Nu A, B l hai s kin bt k lin kt vi cng mt php thth

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

Mt cch tng qut: Nu A1, A2, . . . , An l cc s kin lin kt vi cng mt php th th

P(ni=1Ai) =ni=1

P(Ai) i


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Chng hn khi n = 3, ta c

P(A1 + A2 + A3) = P(A1) + P(A2) + P(A3) P(A1A2) P(A2A3) P(A1A3) + P(A1A2A3)

V d 2.3.25 Mt lp c 100 sinh vin, trong c 40 sinh vin gii ngoi ng, 30 sinh vin gii tinhoc, 20 sinh vin gii c ngoi ng v tin hc. Chn ngu nhin mt sinh vin. Tnh xc sut sinh

vin chn ra hc gii t nht mt mn l ngoi ng hoc tin hc?

GiiGi A l s kin sinh vin c chn ra gii ngoi ng hoc tin hc; B l s kin sinh vin chn

ra gii ngoi ng; Cl s kin sinh vin chn ra gii tin hc. Ta c A = B + C. V B, Ckhng xungkhc nn p dng cng thc cng m rng, ta c

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

P(BC) =

10

1000

+30

100

20

100

=1

2

2.3.2. Xc sut c iu kin. nh l nhn xc sut

Cho A, B l hai s kin lin kt vi cng mt php th. Khi , k hiu A/B l s kin A xy rakhi bit s kin B xy ra.

nh ngha 2.3.7 (Xc sut c iu kin) Cho A, B l hai s kin lin kt vi cng mt php thv P(B) > 0. Xc sut c iu kin ca s kin A khi bit s kin B xy ra, k hiu P(A/B),c xc nh nh sau

P(A/B) =P(AB)

P(B)

V d 2.3.26 Mt hp kn c 2 bi xanh v 1 bi . Ly ngu nhin ln lt 2 vin bi khng hon li.Tm xc sut bi ly ra ln th 2 l bi , bit rng bi ly ra ln th nht l bi xanh?

GiiTa k hiu hai vin bi xanh l 1, 2 v bi l 3. Khi mi kt qu ng kh nng (i, j) vi

i = j,i,j = 1, 2, 3 nn khng gian s cp l

=

{(1, 2); (1, 3); (2, 1); (2, 3); (3, 1); (3, 2)

}Gi A l s kin bi ly ln 1 l bi xanh; Gi B l s kin bi ly ln th 2 l bi . Ta c

A = {(1, 2); (1; 3); (2, 1); (2, 3)}; B = {(1, 3); (2, 3)}; A B = {(1, 3); (2; 3)}

Vy xc sut ln 2 ly c bi khi bit ln th nht ly c bi xanh l P(B/A) = P(AB)P(A)

=26

: 46

= 12.

V d 2.3.27 C 6 ngi( gm 2 nam v 4 n) np n xin vic vo mt cng ty. Gi s rng cngty ch tuyn 2 ngi v kh nng tuyn mi ngi l nh nhau.

a) Tnh xc sut c ng 2 n c chn?b) Gi s c t nht mt n c chn. Tnh xc sut 2 n c chn?c) Trong 4 n c mt ngi tn Hu. Tnh xc sut Hu c chn khi bit c t nht mt n

c chn?

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Giia) Gi l l s kin 2 n c chn. Ta c

P(A) =C24

C26=

2

5

b)Gi B l skin c t nhtmt n c chn. Khi , skin 2 nc chn khi bit c t nhtmt n c chn l A/B. V s kin A xy ra th s kin B xy ra, ngha l A B A B = A.

Ta c: P(B) = C22

C26= 1

15suy ra P(B) = 1 1

15= 14

15.

Vy xc sut 2 n c chn khi bit c t nht mt n c chn l

P(A/B) =P(AB)

P(B)=

P(A)

P(B)=

2

5:

14

15=

3

7

c) Gi C l s kin Hu c chn. Khi s kin Hu c chn khi bit t nht mt n cchn l C/B . V C B nn C B = C. Do

P(C/B) =P(CB)

P(B)=

P(C)

P(B)=

C14C26

:14

15=

2

7

Cc tnh cht:

P() = P(/B) = 0; P() = P(/B) = 1. P((A C)/B) = P(A/B) + P(C/B) P(AC/B)

c bit: Nu AC = th P((A C)/B) = P(A/B) + P(C/B) P(A/B) = 1 P(A/B)T nh ngha xc sut c iu kin, ta suy ra c nh l sau(V sao?):

nh l 2.3.3 (nh l nhn xc sut) Gi sA, B l hai s kin lin kt vi cng mt php th vP(A) > 0. Khi

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

Ta c mt cng thc tng t khi P(B) > 0 l: P(AB) = P(B).P(A/B)Mt cch tng qut, nh l nhn c pht biu nh sau:

Gi s n s kin A1, A2, . . . , An lin kt vi cng mt php th v P(A1A2 . . . An1) > 0. Khi, ta c

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

V d 2.3.28 Mt hp kn c 2 chnh phm v 3 ph phm. Ly ngu nhin ln lt 2 sn phm.Tnh xc sut

a) Hai sn phm ly ra u l chnh phm?b) Hai sn phm ly ra c t nht mt chnh phm?

Gii

a) Gi Ai l s kin ly c chnh phm ln th i(i = 1, 2); A l s kin ly c hai chnhphm. Ta c A = A1A2 do

P(A) = P(A1).P(A2/A1) =2

5.1

4=

2

20

14


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b) Gi B l s kin ly c t nht mt chnh phm. Ta c B = A1A2 do

P(B) = P(A1).P(A2/A1) =3

5

2

4=

6

10

VyP(B) = 1 610

= 710

.

V d 2.3.29 Mt th kho c mt chm cha kha gm 8 cha, trong ch c 3 cha m c kho.Th kho ly ngu nhin tng cha mt cho n khi m c kho th dng li. Tnh xc sut :

a) n ln th 2 th m c kho?b) M c kho khng qu 3 ln?

Giia) Gi Ai l s kin m c kho ln th i(i = 1, 8). ; A l s kin n ln th hai thi m c

kha. Ta c A = A1A2 v o

P(A) = P(A1)P(A2/A1) =C15C18

.C15 .C

13

C17=

5

8.3

7=

15

56

b) Gi B l s kin m c kho khng qu 3 ln. Khi B l s kin m c kho t nht 4ln. Ta c B = A1A2A3 nn

P(B) = P(A1).P(A2/A1).P(A3/A1A2) =5

8.4

7.3

6=

5

28

Ch rng ta cng tnh c P(B) t cng thc B = A1 A1A2 A1A2A3 v cc s kinA1, A

1A2,v A

1A2A3

i mt xung khc.

2.3.3. Tnh c lp ca cc s kin

nh ngha 2.3.8 Cho hai s kin A, B lin kt vi cng mt php th. Hai s kin A, B gi l clp nu P(AB) = P(A).P(B)

Nhn xt:

Nu P(A) = 0 th A, B c lp vi mi s kin B trong cng mt php th. V AB A P(AB) P(A) = 0 nn P(AB) = P(A).P(B) = 0

Nu A, B l hai s kin trong cng mt php th sao cho P(A) > 0, P(B) > 0 th A, B clp khi v ch khi A, B khng xung khc.

Nu A, B l hai s kin c lp v P(A) > 0 th P(B/A) = P(B). iu ny c ngha l skin A xy ra khng em li mt thng tin no cho bit s kin B c xy ra hay khng. Tngt nu P(B) > 0.

nh l 2.3.4 Cho A, B l hai s kin lin kt vi cng mt php th v A, B l hai s kin i lpca A, B. Khi cc mnh sau l tng ng

A, B c lp; A, B c lp; A, B c lp;

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A, B c lp.Mt cch tng qut, ta c nh ngha sau y:

nh ngha 2.3.9 Cho n s kin A1, A2, . . . , An lin kt vi cng mt php th.- H n s kin A1, A2, . . . , An gi l c lp vi nhau tng i mt nu P(Ai Aj) =P(Ai).P(Aj), i, j = 1, n , i = j.

- H n s kin A1, A2, . . . , An gi l c lp ton b nu vi bt kk s kin Ai1, Ai2, . . . , Aiktrong n s kin u tha mn

P(Ai1Ai2 . . . Aik) = P(Ai1)P(Ai2) . . . P (Aik)

vi {i1, i2. . . . , ik} {1, 2, . . . , n}, 2 k nc bit khi k = n ta c P(A1A2 . . . An) = P(A1)P(A2) . . . P (An).

Chrngtnhclptonbthsuyraclptnginhngiungclinichungkhngng. thy iu ny ta xt v d sau.

V d 2.3.30 Mt hp c 4 qu cu gm 1 cu xanh, 1 cu , 1 cu trng v 1 cu gm 3 mu trn.Ly ngu nhin mt qu cu. Gi A,B,Cl s kin ly ra c qu cu xanh, , trng. Xt tnhc lp ca h 3 s kin {A,B,C}.

Ta c P(A) = P(B) = P(C) = 12

v

P(AB) =1

4=

1

2.1

2= P(A).P(B)

P(AC) =1

4=

1

2.1

2= P(A).P(C)

P(BC) =1

4=

1

2.1

2= P(B).P(C)

P(ABC) =1

4= P(A).P(B).P(C) =

1

8

Vy 3 s kin A,B,Cc lp tng i nhng khng c lp ton b.

V d 2.3.31 Mt nh my c 3 phn xng hot ng c lp. Xc sut ngng hot ng ca phnxng th nht, th hai v th ba trong khong thi gian T tng ng l 0, 1; 0, 2; 0, 3. Tm xcsut trong khong thi gian T:

a) C 3 phn xng u ngng hot ng?b) C t nht mt phn xng ngng hot ng?c) C ng mt phn xng ngng hot ng?

Giia) Gi Ai l s kin phn xng i ngng hot ng trong khong thi gian T(i = 1, 2, 3). Theo

gi thit A1, A2, A3 c lp ton b v

P(A1) = 0, 1; P(A2) = 0, 2; P(A3) = 0, 3

Gi A l s kin c 3 phn xng ngng hot ng trong khong thi gian T. Ta c A = A1A2A3v P(A) = P(A1)P(A2)P(A3) = 0, 1.0, 2.0, 3 = 0, 006.

b) Gi B l s kin c t nht mt phn xng ngng hot ng trong khong thi gian T. Ta cB = A1A2A3. Do A1, A2, A3 c lp ton b nn

P(B) = P(> A1)P(A2)P(A3) = 0, 9.0, 8.0, 7 = 0, 504

16


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VyP(B) = 1 P(B) = 1 0, 504 = 0, 496.ChtacthgiicunytbiuthcB = A1A2A3 vpdngcngthccngtngqut.c) Gi C l s kin c ng mt phn xng ngng hot ng trong khong thi gian T. Ta c

C = A1A2A3

A1A2A3

A1A2A3 v 3 s kin A1A2A3, A1A2A3, A1A2A3 i mt xung khc nn

P(C) = P(A1A2A3) + P(A1A2A3) + P(A1A2A3)

Mt khc, ta cng c h cc s kin {A1, A2, A3}; {A1, A2, A3}; {A1, A2, A3} c lp ton b.Do :

P(C) = P(A1)P(A2)P(A3) + P(A1)P(A2)P(A3) + P(A1)P(A2)P(A3)

= 0, 1.0, 8.0, 7 + 0, 9.0, 2.0, 7 + 0, 9.0, 8.0, 3 = 0, 398

2.3.4. Cng thc xc sut ton phn v nh l Bayes

Cng thc xc sut ton phn

nh ngha 2.3.10 (H s kin y ) Cho n s kin A1, A2, . . . , An lin kt vi cng mt phpth. H n s kin A1, A2, . . . , An c gi l h s kin y nu

i) H n s kin cho i mt xung khc, tc l Ai Aj = , i, j(i = j);ii) Hp tt c n s kin l s kin tt yu, tc l ni=1Ai =

V d tp cc s kin s cp ca mt php th l mt h s kin y .

nh l 2.3.5 Gi s cc s kin A1, A2, . . . , An lin kt vi cng mt php th to thnh mt h y cc s kin sao cho p(Ai) > 0, i = 1, n. Khi vi mi s kin A ta c

P(A) =ni=1

P(Ai)P(A/Ai)

ng thc trn c gi l cng thc xc sut ton phn.

V d 2.3.32 C hai hp ging nhau, hp I c 6 bi v 4 bi xanh, hp II c 8 bi v 4 bi xanh.Ly ngu nhin mt hp ri t ly ra 2 vin bi. Tm xc sut hai bi ly ra u l bi ?

GiiGi Ai(i = 1, 2) l s kin hp th i c chn; Gi A l s kin hai bi ly ra l bi .Ta c, hai s kin A1, A2 to thnh mt h y cc s kin v P(A1) = P(A2) = 12 . p dng

cng thc xc sut ton phn, ta c

P(A) = P(A1)P(A/A1) + P(A2)P(A/A2) =1

2

C26C210

+1

2

C28C212

=25

66

V d 2.3.33 Mt ca hng bn bng n, trong c 20% do nh my th nht sn xut, 46% donh my th 2 sn xut, 34% do nh my th 3 sn xut. Bit rng t l bng n b hng ca nhmy th nht, th hai, th ba ln lt l: 3%; 1%; 2%. Mt ngi mua ngu nhin mt bng n.Tnh xc sut bng n ngi mua b hng?

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Gi Ai l s kin bng n c sn xut nh my th i(i = 1, 2, 3). Gi A l s kin ngimua c bng n hng.

Ta c cc s kin A1, A2, A3 to thnh h y cc s kin v

P(A1) = 0, 2; P(A2) = 0, 46; P(A3) = 0, 34p dng 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)

= 0, 2.0, 03 + 0, 46.0, 01 + 0, 34.0, 02 = 0, 174

V d 2.3.34 Mt hp c 10 qu bng tennis, trong c 7 qu mi v 3 qu c. Ln mt ly ra 2qu thi u, sau b tr li. Sau , ln 2 ly ra 2 qu thi u. Tnh xc sut hai qu ly raln th 2 l qu bng mi?

GiiGi Ai l s kin 2 bi ly ra ln 1 c i bi mi. Ta c A0, A1, A2 to thnh h s kin y v

P(A0) =C23C210

=1

15; P(A1) =

C17C13

C210=

7

15; P(A2) =

C27C210

=7

15

Gi A l s kin 2 qu cu ly ra ln hai l qu cu mi. p dng cng thc xc sut ton phn,ta c

P(A) = P(A0).P(A/A0) + P(A1).P(A/A1) + P(A2).P(A/A2)

=

1

15

C27C210 +

7

15

C26C210 +

7

15

C25C210 =

1

15 .

7

15 +

7

15

1

3 +

7

15

2

9 =

196

675

nh l Bayes

nh l 2.3.6 Xt mt php th. Gi sA1, A2, . . . , An l h y cc s kin v P(Ai) > 0.i =1, n v A l mt s kin bt k, P(A) > 0. Khi

P(Ai/A) =P(Ai)P(A/Ai)ni=1 P(Ai)P(A/Ai)

V d 2.3.35 Mt nh my sn xut thp tm gm 2 phn xng sn xut. Phn xng 1 v phn

xng 2 sn xut vi lng sn phm l 60% v 40%. Bit t l ph phm ca phn xng 1 v 2tng ng l 3% v 4%. Ly ngu nhin mt tm thp ca nh my th thy tm thp l mt phphm. Tm xc sut tm thp do phn xng th nht sn xut?

GiiGi Ai l s kin tm thp ly ra do phn xng th i xn xut(i = 1, 2). A 1, A2 to thnh h

y cc s kin vP(A1) = 0, 6; P(A2) = 0, 4

Gi A l s kin tm thp ly ra l ph phm. p dng cng thc xc sut ton phn, ta c

P(A) = P(A1)P(A/A1) + P(A2)P(A/A2) = 0, 6.0, 03 + 0, 4.0, 04 = 0, 034

p dng cng thc Bayes, ta c xc sut ph phm ly ra do phn xng 1 sn xut l

P(A/A1) =P(A1)P(A/A1)

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

0, 6.0, 03

0, 034=

9

17

18


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V d 2.3.36 Chailhng: lIc50snphm,trongc20snphmxu; lIIc40snphm,trong c 15 sn phm xu. Ly ngu nhin mt l v t ly ngu nhin mt sn phm.

a) Tnh xc sut sn phm ly ra l sn phm tt?b) Bit sn phm ly ra l tt. Tnh xc sut sn phm thuc l II?

Giia) Gi A1, A2 l s kin sn phm ly ra o l I, II. Ta c A1, A2 to thnh h s kin y v

P(A1) = P(A2) =1

2

Gi A l bin c sn phm ly ra l sn phm tt. p dng cng thc xc sut ton phn, ta c

P(A) = P(A1)P(A/A1) + P(A2)P(A/A2) =1

2

30

50

+1

2

25

40

=49

80

b) p dng cng thc Bayes, ta c xc sut sn phn tt ly l II l

P(A2/A) =P(A2)P(A/A2)

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

1225404980

=25

49

2.3.5. Dy php th c lp v cng thc Bernoulli

Dyphpthclp-DyphpthBernoulli Xt mt php th . Thc hin php th n ln v

gi i l php th thc hin ln th i.

nh ngha 2.3.11 Cc php th 1, 2, . . . , n c gi l c lp nu xc sut xy ra ca cc skin lin kt vi php th i no khng ph thuc vo kt qu ca cc php th khc.

Nh vy, nu Ai lskinlinktviphpth i(i = 1, n) thccskin A1, A2, . . . , An lclp ton b.

nh ngha 2.3.12 Cho dy n php th c lp. Trong mi php th, ta xt s kin A v A. Gi sxc sut s kin A xy ra trong mi php th l khng i v bng p(0 < p < 1) v xc sut

xy ra bin c A = 1 p. Khi n php th c lp trn c gi l n php th Bernoulli. K hiuB(n; p).

nh l 2.3.7 (nh l Bernoulli) Thc hin n php th c lp. Trong mi php th s kin A xyra vi xc sut khng i P(A) = p(0 < p < 1). Khi , xc sut s kin A xy ra ngk ln trongn php th l

Pn(k) = Cknp

k(1 p)nk (k = 0, n)

H qu 2.3.2 Vi nhng gi thit nh trong nh l Bernoulli, xc sut trongn php th s kin Axy ra t nhtk1 ln v nhiu nhtk2 ln l

Pn(k1 k k2) =k2i=k1

Cknpk(1 p)nk

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S ln c kh nng xy ra nhiu nht

nh ngha 2.3.13 Cho n php th Bernoulli. Trong mi php th, xc xut s kin A xy raP(A) = p v P(A = 1

p. S m gi l s ln xy ra s kin A nhiu nht nu

Pn(m) Pn(k), k = 0, n

hayPn(m) = max{Pn(0), Pn(1), . . . , P n(n)}

nh l 2.3.8 Cho nphp th Bernoulli. Trong mi php th, xc xut s kin A xy ra P(A) = pv P(A = 1 p = q. Gi m l s ln s kin A xy ra nhiu nht, ta c

np q m np + q

V d 2.3.37 C 10 sinh vin thi mn xc sut. Kh nng thi t ca cc sinh vin u nh nhau v

bng 70%.a) Tm xc sut c 8 sinh vin thi t?b) Tm xc sut c t nht 1 sinh vin thi trt?c) Tm xc sut c t nht 8 sinh vin thi khng t?d) Tm s sinh vin c kh nng thi t nhiu nht trong 10 sinh vin?

GiiBi ton tng ng vi mt dy php th Bernoulli vi n = 8, p = 0, 7. p dng cc nh l trn

gii bi ton.

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Chng 3

BIN NGU NHIN

3.1. BIN NGU NHIN

3.1.1. Bin ngu nhin

Khi nim

Bin ngu nhin l mt i lng c gi tr thc bin i ph thuc vo kt qu ca php thngu nhin. K hiu bin ngu nhin l X , Y , Z , . . .. Ta c nh ngha chnh xc ca bin ngu nhinnh sau:

nh ngha 3.1.14 Binngunhinlmtnhxttp ccktqucamtphpthvotpccs thc

R.

Bin ngu nhin c min gi tr hu hn hoc m c gi l bin ngu nhin ri rc.Bin ngu nhin c min gi tr l mt khong(hoc on) gi l bin ngu nhin lin tc.

bin ngu nhin cn c gi l i lng ngu nhin.

nh ngha 3.1.15 (Bin ngu nhin c lp) Cho X, Y l hai bin ngu nhin lin kt vi mtphp th. X, Y gi l c lp nhau nu x1, x2, y1, y2 R :

P((x1 X < x2).(y1 Y < y2)) = P(x1 X < x2).P(y1 Y < y2)

V d 3.1.38 Gieo mt con xc xc. Gi X l s chm xut hin ca con xc xc th X l mt bin

ngu nhin nhn cc gi tr c th l 1, 2, 3, 4, 5, 6.X l bin ngu nhin ri rc.

V d 3.1.39 Xt php th l vic o thi gian sng(tnh bng gi) ca mt con transitor. Gi Y lthi gian sng ca mt con transitor th Y l mt bin ngu nhin c min gi tr l [0; +).

Y l bin ngu nhin lin tc.

3.1.2. Hm phn phi xc sut ca bin ngu nhin

nh ngha

Cho X l bin ngu nhin lin kt vi php th T c khng gian s cp l . Hm s k hiu vxc nh nh sau

F(x) = P(X < x) Vi (X < x) = {in : X() < x}21


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22 Chng 3. BIN NGU NHIN

gi l hm phn phi xc sut ca bin ngu nhin X.Hm phn phi xc sut phn nh mc tp trung xc sut v bn tri ca im x. T cc tnh

cht ca xc sut ta suy ra cc tnh cht sau ca hm phn phi xc sut.

Tnh cht 3Gi s F(x) l hm phn phi xc sut ca bin ngu nhinX, ta c:

x R : 0 F(x) 1. lim

x+F(x) = 1; lim

xF(x) = 0.

x1, x2 R : x1 < x2 F(x1) F(x2). F(x) lin tc bn tri ti mi im, tc l lim

xx0F(x) = F(x0).

H qu 3.1.3 , R : P( X < ) = F() F() Nu Xlbinngunhinlintcth F(x) lintctimiimtrn R v P(X = ) = 0,

R

Bng phn phi xc sut v hm mt

Bng phn phi xc sut: Bng phn phi xc sut dng thit lp lut phn phi xc sut cabin ngu nhin ri rc, n gm 2 hng: hng th nht lit k cc gi tr c th x1, x2, . . . , xn cabin ngu nhin ri rc X v hng th 2 lit k cc xc sut tng ng p1, p2, . . . , pn ca cc gi trc th .

X x1 x2 . . . xnP p1 p2 . . . pn

NuccgitrcabinngunhinXgmhuhns x1, x2, . . . , xn thccskin X = x1, X =x2, . . . , X = xn lp thnh mt h y cc s kin. Do , ta c

ni=1pi = 1 F(x) = xi


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3.1. BIN NGU NHIN 23

nu F(x) khviti x0 th F(x0) = f(x0). NuF(x) khvitrongkhong(a, b) th trong khong(a, b) ta c F(x) = f(x).

, R, < : P(

X < ) =

f(x)dx.

y cn nhn mnh rng: hm phn phi xc sut F(x) c xc nh da vo bng phnphi xc sut nu X l bin ngu nhin ri rc v c xc nh thng qua hm mt nu X lbin ngu nhin lin tc.

V d 3.1.41 Cho bin ngu nhin ri rc Xc bng phn phi xc sut

X 0 1 2P 1

412

14

Tm hm phn phi xc sut F(x) ca Xv v th ca n.GiiNu x 0 th F(x) = 0Nu 0 < x 1 th F(x) = p0 = 14Nu 1 < x 2 th F(x) = p0 + p1 = 14 +

12

= 34

Nu x > 2 th F(x) = p0 + p1 + p2 = 14 +12

+ 14

= 1Vy hm phn phi xc sut ca X l

F(x) =

0 nu x 014

nu 0 < x 134 nu 1 < x 21 nu x > 2

V d 3.1.42 Bin ngu nhin Xc hm phn phi xc sut nh sau

F(x) =

0 nu x 134

x + 34

nu 1 < x 13

1 nu x > 13

Tm xc sut Xnhn gi tr trong khong [0, 13

)

GiiTheo tnh cht ca hm phn phi xc sut, ta c

P(0 X 1

a) Tm h s a?b) Tm hm mt xc sut f(x)?c) Tm xc sut X (0, 25; 0, 75)?

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Giia) V hm phn phi F(x) lin tc ti x = 1 tc l lim

x1+0F(x) = F(1) a = 1.

b) Theo nh ngha ca hm mt xc sut, ta c

f(x) = F(x) =

0 nu x 02x nu 0 < x 10 nu x > 1

c) P(0, 25 X < 0, 75) = F(0, 75) F(0, 25) = (0, 75)2 (0, 25)2 = 0, 5

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3.2. CC THAM S CA BIN NGU NHIN 25

3.2. CC THAM S CA BIN NGU NHIN

3.2.1. K vng ton

nh ngha 3.2.17 Gi s X l mt bin ngu nhin. Ta gi k vng ton ca bin ngu nhin X lmt s, k hiu E(X), v c xc nh nh sau:

Nu X l bin ngu nhin ri rc c bng phn phi xc sut:X x1 x2 . . . xnP p1 p2 . . . pn

hocX x1 x2 . . . xn . . .P p1 p2 . . . pn . . .

vi pi = P(X = xi)(i = 1, n)

th

E(X) =

ni=1

xipi hoc E(X) =

i=1

xipi

Nu X l bin ngu nhin lin tc vi hm mt f(x) th

E(X) =

+

x.f(x)dx

(vi iu kin tch phn suy rng v phi hi t tuyt i)

nghacakvng: K vng ca E(X) c trng cho gi tr trung bnh ca bin ngu nhin X.

V d 3.2.44 Gi s X l bin ngu nhin c bng phn phi xc sut:

X 1 0 1 2P 0, 1 0, 2 0, 2 0, 5

Ta c E(X) = 1.0, 1 + 0.0, 2 + 1.0, 2 + 2.0, 5 = 1, 1V d 3.2.45 Cho X l bin ngu nhin lin tc c hm mt

f(x) = {ax2 vi 0 x 1

0 vi x / (0; 1)Tm hm mt v tnh E(X).

Gii

Theo tnh cht ca hm mt , ta c+

f(x)dx = 1 a = 3Vy hm mt

f(x) =

{3x2 vi 0 x 10 vi x / (0; 1)

T , ta c:

E(X) =

+

xf(x)dx =

0

xf(x)dx +

0

1xf(x)dx +

+1

xf(x)dx

25


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=

10

x.3x2dx =3

4

V d 3.2.46 Theo thng k c bit t l cht ca mt ngi tui trn 30 l 11000

. Mt cng tybo him bn bo him nhn mng cho ngi tui trn 30 vi s tin l 100000 ng. Cngty mun li trung bnh khi bn mt bo him l nh th l 70000. Hi s tin m cng ty bo himphi tr nu ngi mua bo him cht l bao nhiu?

GiiGi a l s tin m cng ty bo him phi tr nu ngi mua bo him cht trong nm . Gi X

lstinlikhibnmtbohimthXlbinngunhincthnhnhaigitr 100000a, 100000v c bng phn phi l

X 100000

a 100000

P 11000 9991000

Ta c E(X) = (100000 a) 11000

+ 100000. 9991000

= 100000 a1000

tin li trung bnh l 70000 khi bn mt bo him th

E(X) = 100000 a1000

= 70000 a = 30.000.000Tnh cht 5

E(c) = 0(c = const) E(cX) = cE(X)

E(X

Y) = E(X)

E(Y)

Nu X, Y l hai bin ngu nhin c lp thE(XY) = E(X)E(Y)

3.2.2. Phng sai:

omcphntncabinngunhin Xquanh gia tr k vng E(X),ngitaarakhinim phng sai nh sau.

nh ngha 3.2.18 Gi s X l mt bin ngu nhin c k vng E(X) = a. Nu bin ngu nhin(X a)2 c k vng th gi tr k vng E[(X a)2] c gi l phng sai ca bin ngu nhin X.K hiu D(X).

Ta c: D(X) = E[(X a)2]T nh ngha, ta suy ra cng thc tnh phng sai nh sau:a) Nu X l bin ngu nhin c bng phn phi xc sut

X x1 x2 . . . xnP p1 p2 . . . pn

hocX x1 x2 . . . xn . . .P p1 p2 . . . pn . . .

D(X) =ni=1

(xi a)2.pi hoc D(X) =i=1

(xi a)2.pi

b) Nu X l bin ngu nhin lin tc th

D(X) =

+

(x a)2f(x)dx

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3.3. MT S BIN NGU NHIN LIN TC QUAN TRNG 27

Tnh cht 6T nh ngha ca phng sai, ta suy ra c cc tnh cht sau:

D(X) = E(X2)

E2(X)

D(c) = 0(c = const) D(cX) = c2D(X) Nu X, Y l hai bin ngu nhin c lp thD(X Y) = D(X) D(Y)

V d 3.2.47 Cho X l bin ngu nhin ri rc c bng phn phi xc sut

X 1 0 1 2P 0, 1 0, 2 0, 3 0, 4

Tnh E(X), D(X)

GiiTa c

E(X) = 1.0, 1 + 0.0, 2 + 1.0, 3 + 2.0, 4 = 1

D(X) = (1 1)2.0, 1 + (0 1)2.0, 2 + (1 1).0, 3 + (2 1)2.0, 4 = 1Ta cng c th tnh c D(X) t cng thc D(X) = E(X2) E2(X) trong

E(X2) = (1)2.0, 1 + 02.0, 2 + 12.0, 3 + 22.0, 4 = 2VyE(X) = 1 v D(X) = 1.

3.2.3. lch chun

nh ngha 3.2.19 Gi s bin ngu nhin c phng sai D(X). Kh , bin (X) =

D(X) gil lch chun ca X.

lch chun l mt bin c trng cho tnh n nh ca bin ngu nhin X. Trong lnh vcu t, lch chun c trng cho mc ri ro.

3.3. MT S BIN NGU NHIN LIN TC QUAN TRNG

3.3.1. Bin chun N(a, )

nh ngha 3.3.20 Bin ngu nhin lin tc Xgi l chun N(a, ) nuhmmtcancdng

f(x) =1

2e

(xa)2

22 , x R; a, : tham s ; > 0

Tnh cht 7 (Cc tham s c trng)Cho X l chun N(a, ), ta c

E(X) = Med(X) = M od(X) = a; D(X) = 2; (X) =

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Cch xc nh xc sut ca mt s kin lin kt vi bin chun

nh ngha 3.3.21 (Hm Laplace) Hm laplace l hm s

(x) =

1

2 x

0 e

t2

2

dt

nh l 3.3.9 Nu F(x) l hm phn phi xc sut ca bin chun N(a, ) th

F(x) = (x a

)

nh l 3.3.10 Nu X l bin chun N(a, ), , R, < thP( X < ) = (

a

) ( a

)

H qu 3.3.4 Nu X l bin chun N(a, ), R, > 0 thP(

|X

a

|< ) = 2(

)

Cc nh l v bin chun

nh l 3.3.11 Nu X l bin chun N(a, ) th bin ngu nhin cX,X c(vi c = const) l binchun vi tham N(ca, | c | ), N(a c, ).nh l 3.3.12 Nu Xi l cc bin chun N(ai, i), i = 1, n v cc Xi, i = 1, n c lp ton b th

bin ngu nhin X =n

i=1 Xi l bin chun N(a, ) vi

a = a1 + a2 + . . . + an; 2 = 21 +

22 + . . . +

2n

.H qu 3.3.5 Nu Xi l cc bin chun N(a, ), i = 1, n v cc Xi, i = 1, n c lp ton b th binngu nhin X = 1

n

ni=1 Xi l bin chun N(a,

n

)

nh l 3.3.13 (inh l Lindebeg- Levi) Nu Xi, i = 1, n l n bin ngu nhin c lp ton b, cngphn phi vi k vngE(Xi) = a, phng sai D(Xi) = 2, i = 1, n th bin ngu nhin X =

ni=1 Xi

v bin X = 1n

ni=1 Xi xp x bin chun N(na,

n) v N(a,

n).

Tc l vi n kh ln, ta c

P(X < x) 1n

2

x

e(tna)2

2n2 dt

P(X < x) n

2

x

e(ta)2n

22 dt

H qu 3.3.6 (nh l Moivre - Laplace) Nu Xi, i = 1, n l n bin ngu nhin n gin A(a) clp ton b c k vngE(Xi) = a, D(Xi) = a(1 a), i = 1, n th bin ngu nhin X =

ni=1 Xi l

bin nh thc B(n, a) xp x bin chun N(na,

na(1 a))Tc l vi n kh ln, ta c

P(B(n, a) < x) 1

na(1 a).2x

e(tna)2

2na(1a)dt

hayP( B(n, a) < ) ( na

na(1 a) ) ( na

na(1 a) )vi (x) l hm Laplace.

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3.3. MT S BIN NGU NHIN LIN TC QUAN TRNG 29

3.3.2. Bin khi bnh phng 2n

nh ngha 3.3.22 Cho X1, X2, . . . , X n l n bin ngu nhin chun N(0, 1) c lp vi nhau. Khi

bin ngu nhin X =n

i=1

X2i c gi l bin khi bnh phng vi n bc t do. K hiu Kn hoc

2n.

nh l 3.3.14 Hm mt f(x) ca phn phi khi bnh phng2n vi n bc t do l

f(x) =

0 vi x 01

2n2 (n2 )

ex2

2 xn

21 vi x > 0

Tnh cht 8 (Cc tham s c trng)Cho X l bin khi bnh phng2n vi n bc t do, ta c

E(X) = n; D(X) = 2n

Cc nh l v bin 2n

nh l 3.3.15 Nu X, Y l cc bin 2n, 2m th X+ Y l bin

2n+m.

nh l 3.3.16 Nu X l cc bin 2n th bin ngu nhin Z =2nn

2nxp x bin chun N(0, 1) khi n

kh ln (n > 30).

3.3.3. Bin Student Tnnh ngha 3.3.23 Cho X l bin ngu nhin chun N(0, 1), Y l bin 2n vi n bc t do v X, Yc lp. Khi , bin ngu nhin X

Y

n

c gi l bin Student vi n bc t do. K hiu Tn = XYn

.

nh l 3.3.17 Hm mt f(x) ca phn phi StudentTn vi n bc t do l

f(x) =(n+1

2)

(n2

)

n(1 +

x2

n)

n+12

Tnh cht 9 (Cc tham s c trng)

Cho X l bin Student Tn vi n bc t do, ta c

E(Tn) = 0; D(Tn) =n

n 2nh l 3.3.18 Bin StudentTn s xp x bin chun N(0, 1) khi n kh ln (n > 30)

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30


31/50

Chng 4

MU V CC THAM S MU

4.1. MU V CC PHNG PHP XY DNG MU

4.1.1. Tng th v mu

nh ngha 4.1.24 Tp hp ton b cc i tng cn nghin cu, kho st "c tnh" no cachng gi l tng th(hay tp hp tng qut hay tp sinh). K hiu tp tng th l .

S phn t(lc lng) ca gi l kch thc ca tng th.

nh ngha 4.1.25 T tng th, ta chn ngu nhin(theo mt cch chn quy nh trc) n phnt(i tng), tp n phn t c chn gi l mt mu. Khi , n gi l kch thc mu

4.1.2. Cc phng php xy dng muMu lp

Ly mu c lp l ly mu m phn t ly ra, sau khi ghi gi tr c trng, c tr tr li tngth trn u ri ly tip phn t khc.(phn phi nh thc).

Mu khng lp

Ly mu khng lp l ly mu m phn t ly ra, sau khi ghi gi tr c trng, khng tr trli tng th m ly tip phn t khc.(phn phi siu bi).

Ta bit rng, phn phi siu bi hi t v phn phi nh thc nn khi s phn t ca tng th lN rt ln so vi kch thc mu n(N > 100n) th vic ly mu khng lp li xem nh mu c lp.Do , trong l thuyt, ta thng nghin cu mu lp.

Xy dng mu theo li in hnh

V d 4.1.48 c lng chiu cao trung bnh ca hc sinh lp 4 ti a phng A c 20000 hcsinh lp 4. Trong , thnh ph 7000, nng thn 8000 v min ni 5000 hc sinh. Ly mu2000 hc sinh nh sau: ly700 hc sinh thnh ph, 800 hc sinh nng thn v 500 hc sinh min ni. Khi mu c chn nh trn c xy dng theo li in hnh.

Xy dng mu theo li my mcV d 4.1.49 kim tra mt on ng AB di 3000m. Bt u t A c cch 30m ta ly mtmu. Khi , ta c mt mu c kch thc n = 100 xy dng theo li my mc.

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32 Chng 4. MU V CC THAM S MU

4.2. Cc phng php trnh by s liu

4.2.1. Mu ngu nhin v mu thc nghim

Ta chn ngu nhin mt phn t t tp . Khi c xem nh l khng gian cc s kin scp. Gi X l bin ngu nhin biu th c trng nghin cu trn tp (X lin kt vi php th lyra mt phn t). K hiu l php th ly ra mt phn t.

Lp li php th n ln. Gi Xi l gi tr c trng ca phn t c ly ra ln th i(i =1, n. Khi cc bin X1, X2, . . . , X n c lp c cng quy lut phn phi vi X, n bin ngu nhin(X1, X2, . . . , X n) gi l mu ngu nhin ca X.

Sau khi ly mu, ta c X1 = x1, X2 = x2, . . . , X n = xn. B n s (x1, x2, . . . , xn) c gi lmu c th(mu thc nghim) ca X.

nh ngha 4.2.26 Ta gi mu ngu nhin kch thc n ca bin ngu nhin X l mt b n th t(X

1, X

2, . . . , X n), trong X

1, X

2, . . . , X n l n bin ngu nhin c lp c cng phn phi xc sut

vi X.Sau khi ly mu, ta c X1 = x1, X2 = x2, . . . , X n = xn. B n s (x1, x2, . . . , xn) c gi l

mu c th(mu thc nghim) ca X.

4.2.2. Cc phng php trnh by mu

Trnh by mt mu c t gi tr khc nhau

Gi s khi ly mu kch thc n ca bin ngu nhin X c mu c th vi s liu ban u(x1, x2, . . . , xn) nhng trong ch c k gi tr khc nhau: a1 < a2 < .. . < ak

Gi ni l s ln ai(i = 1, n c trong mu thc nghim. ni gi l tn s.Gi fi = nin l tn sut ca gi tr ai trong mu thc nghim.Khi , ta c bng thng k(Bng phn phi tn s khng chia lp) sau:

ai a1 a2 . . . akni n1 n2 . . . nk

V d 4.2.50 Ta ly mu kch thc n = 20, ta c 1, 3, 2, 1, 5, 3, 4, 1, 4, 3, 2, 5, 4, 3, 12, 1, 4, 3, 3Ta c bng thng k

ai 1 2 3 4 5ni 5 3 6 4 2

Trnh by mt mu c nhiu gi tr khc nhau

Trong trng hp ly mu kch thc n c nhiu gi tr khc nhau hoc do ngha thc t m tachia mu thnh nhiu lp.

Khng c quy tc chia lp. Tuy nhin, theo mt s nh thng k ngh chia lp nh sau:1) Xc nh s lng lp k {

1 + log2n k 5lgn

6 k 20

2) B rng ca lp

b = amax amink

3) Tn s ni ca lp ai1 ai l s ln gi tr ca mu m ai1 x < aifi =

nin

l tn sut ca lp ai1 ai32


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4.2. Cc phng php trnh by s liu 33

4) Gi tr chnh gia(trung tm) ca lp ai1 ai l: ai = ai1+ai2Ta c bng thng k(Bng phn phi tn s chia lp) nh sau:

Lp[ai, ai) a0 a1 a1 a2 . . . ak1 akni n1 n2 . . . nk

Ch , nu trong cc bng phn phi tn s thc nghim trn ta thay tn s ni bi tn sut tngng fi ta c bng gi l bng phn phi tn sut(chia lp hoc khng chia lp) thc nghim.

Hm phn phi thc nghim

nh ngha 4.2.27 Cho X l mt bin ngu nhin v ly mu kch thc n ca X. Hm phn phithc nghim ng vi mu c chn, k hiu Fn(x), v c xc nh nh sau:

+ Nu mu thc nghim cho theo bng khng chia lp (4.2.2.) th

Fn(x) =ai ak

+ Nu mu thc nghim cho theo bng chia lp (??) th

Fn(x) =

ai1 ak1

nh l 4.2.19 GisF(x) lhmphnphixcsutcaXv Fn(x) l hm phn phi thc nghimca X. Khi , vi n kh ln Fn(x) F(x).V d 4.2.51 Tm hm phn phi thc nghim ca Xbit

a)ai 1 3 7

ni 2 5 3; b)

Lp[ai, ai) 0

4 4

8 8

12

ni 1 5 3Giia) Ta c

F10(x) =1

10

ni 5

b) Ta c

F9(x) =

0 Nu x 019 Nu 0 < x 423

Nu 4 < x 81 Nu x > 8

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4.3. CC THAM S C TRNG CA MU

Gi s (X1, X2, . . . , X n) l mu ngu nhin ca X v sau khi ly mu ta c mu thc nghim(x1, x2, . . . , xn)

4.3.1. Cc tham s ca mu ngu nhin

1. Bin ngu nhin X = 1n

ni=1 Xi c gi l trung bnh ca mu ngu nhin

2. Bin ngu nhin 2n =1n

ni=1(Xi X)2 c gi l phng sai ca mu ngu nhin

2n1 =n

n12n =

1n1

ni=1(XiX)2 cgil phngsaiiuchnhcamungunhin

n =

2n : lch chun ca mu ngu nhin.

n1 = 2n1 : lch chun iu chnh ca mu ngu nhin.

4.3.2. Cc tham s ca mu thc nghim

1. S trung bnh ca mu thc nghim:

x =1

n

ni=1

xi

2. S phng sai ca mu thc nghim:

2n = 1n

ni=1

(xi x)2

3. S phng sai iu chnh mu thc nghim:

2n1 =n

n 12n =

1

n 1ni=1

(xi x)2

n =

2n : lch chun ca mu thc nghim.

n1

= 2n1

: lch chun iu chnh ca mu thc nghim.

T cc cng thc trn, ta suy ra cng thc tnh i vi mu thc nghim c bng phn phikhng chia lp v chia lp nh sau:

+ Nu mu thc nghim c bng phn phi tn s khng chia lp dng

ai a1 a2 . . . akni n1 n2 . . . nk

(ki=1

ni = n)

th

x =1

n

k

i=1

niai

2n =1

n

ki=1

ni(ai x)2 = 1n

ki=1

nia2i x2

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4.3. CC THAM S C TRNG CA MU 35

+ Nu mu thc nghim c bng phn phi tn s chia lp

Lp[ai, ai) a0 a1 a1 a2 . . . ak1 akni

n1

n2

. . . nk

(k

i=1

ni = n)

t ai =ai1+ai

2, ta c bng

ai a1 a

2 . . . a

k

ni n1 n2 . . . nk

Khi

x =1

n

ki=1

niai

2n = 1n

ki=1

ni(ai x)2 = 1nki=1

nia2i x2

V d 4.3.52 Tnh x, 2n ca mu trong cc trng hp sau:

a)ai 1 3 5ni 3 5 2

; b)[ai, ai) 0 2 2 4 4 6 6 8 8 10 10 12ni 5 10 10 5 10 20

Giia) Lp bng tnh

ai ni ai.ni nia2i

1 3 3 33 5 15 455 2 10 50

n = 10 28 98

S trung bnh mu l x = 1n

ki=1 niai =

110

.28 = 2, 8

S phng sai mu 2n =1n

ki=1 nix

2i x2 = 11098 (2, 8)2 = 1, 96

b) t xi =xi1+xi

2, ta c

xi 1 3 5 7 9 11ni 5 10 10 5 10 20

Lp bng tnhxi ni x

i .ni nix

i2

1 5 5 53 10 30 905 10 50 2507 5 35 2459 10 90 810

11 20 220 2420n = 60 430 3820

S trung bnh mu l x = 1n

ki=1 nix

i =

160

.430 = 436

S phng sai mu 2n =1n

ki=1 nix

i2 x2 = 1

603820 (43

6)2 = 12, 31

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Cng thc tnh ton Khi tnh ton cc tham s c trng ca mu thc nghim trnh victnh ton cc s c gi tr ln phc tp, ngi ta thng s dng cc tnh cht sau

x0 R, d = 0,ki=1

ni = n ta c

x =1

n

ki=1

niai =d

n

ki=1

niai x0

d+ x0

2n =1

n

ki=1

ni(ai x)2 = d2

n

ki=1

ni(ai x0

d) (x x0)2

Thng thng ta chn x0 l gia tr ti tn s ln nht, d l khong cch u(nu c).

V d 4.3.53 Tm x, 2n, n1 vi

xi 3, 94 3, 97 4, 00 4, 03 4, 06ni 1 7 10 5 2

GiiTa chn x0 = 4, 00 = 4, d = 0, 03. Ta c bng tnh:

xi nixi40,03

nixi40,03

ni(xi40,03

)2

3, 94 1 2 2 43, 97 7 1 7 74, 00 10 0 0 04, 03 5 1 5 54, 06 2 2 4 8

n = 25 0 0 24

S trung bnh mu l

x =d

n

k

i=1ni

ai x0d

+ x0 =0, 03

25.0 + 4 = 4

S phng sai mu v phng sai iu chnh

2n =d2

n

ki=1

ni(ai x0

d) (x x0)2 = 0, 03

2

25.24 (4 4)2 = 0, 000864

n1 =

n

n 12n =

25

24.0, 000864 = 0, 03

V d 4.3.54 Tm x, 2n, n1 vi

[ai1, ai) 10500 10550 10550 10600 1060 10650 10650 10700ni 15 55 20 10

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4.3. CC THAM S C TRNG CA MU 37

Giit xi =

xi1+xi2

, ta c

xi 10525 10575 10625 10675

ni 15 55 20 10

Ta chn x0 = 10575, d = 50. Ta c bng tnh:

xi nixi1057550

ni.xi1057550

ni(xi1057550

)2

10525 15 1 15 1510575 55 0 0 010625 20 1 20 2010675 10 2 20 40

n = 100 2 25 75S trung bnh mu l

x =50

100.25 + 10575 = 10587, 5

S phng sai mu v phng sai iu chnh

2n =502

100.75 (10587, 5 10575)2 = 1618, 75

n1 = n

n 12n =

100

99

.1618, 75

40, 44

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Chng 5

C LNG THAM S

5.1. PHNG PHP C LNG IM - HM C LNG

5.1.1. Phng php c lng im

Gi s X l bin ngu nhin biu th c trng nghin cu trn tp , c phn phi xc sut bit nhng cn ph thuc vo tham s cha bit.

c lng , ta ly mu kch thc n. Khi , ta c mu ngu nhin (X1, X2, . . . , X n). Saukhi ly mu, ta c mu thc nghim (x1, x2, . . . , xn) vi Xi = xi(i = 1, n).

ngvimimuthcnghim (x1, x2, . . . , xn) tacmts n(x1, x2, . . . , xn) dngclngcho . Phng php c lng gi l phng php c lng im. V s n(x1, x2, . . . , xn) ng

vi mt im trn ng thng s.

Ta gi U l tp hp cc mu thc nghim (x1, x2, . . . , xn) hay ni cch khc U lmingitrcamu ngu nhin (X1, X2, . . . , X n). Khi ta c mt hm

n : U R, (x1, x2, . . . , xn) n(x1, x2, . . . , xn)

Bin ngu nhin n(x1, x2, . . . , xn) gi l hm c lng ca tham s .Vn l phi tm hm n(x1, x2, . . . , xn) sao cho c lng c tt, tc l c lng khng

mc sai s h thng v hiu qu.

5.1.2. Cc tiu chun c lng tham s c trng ca X

c lng khng chch

nh ngha 5.1.28 Hm c lng n(x1, x2, . . . , xn) ca tham s c gi l c lng khngchch ca tham s nu E(n(x1, x2, . . . , xn)) = .

Ngc li nu E(n(x1, x2, . . . , xn)) = th ta ni hm c lngn(x1, x2, . . . , xn) l hm clng chch ca .

c lng bn vng

nh ngha 5.1.29 Hm c lng n(x1, x2, . . . , xn) cathams cgilclngbnvng

ca tham s nu > 0 ta c limn+

P(| n |< ) = 1

nh l 5.1.20 Nu hm c lngn(x1, x2, . . . , xn) tha mn

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40 Chng 5. C LNG THAM S

i) E(n(x1, x2, . . . , xn)) = hay limn E(n(x1, x2, . . . , xn)) = ii) limn D(n(x1, x2, . . . , xn)) = 0 th n(x1, x2, . . . , xn) l c lng bn vng ca .

c lng hiu qunh ngha 5.1.30 c lng khng chch n(x1, x2, . . . , xn) ca c gi l c lng hiuqu nu vi mi c lng khng chch

n(x1, x2, . . . , xn) ca th D(n(x1, x2, . . . , xn)) D(

n(x1, x2, . . . , xn)).

nh ngha 5.1.31 Hm c lng n(x1, x2, . . . , xn) ca tham s c gi l c lng tt catham s nu n l c lng khng chch, bn vng v hiu qu ca

> 0 ta c limn+

P(| n |< ) = 1

5.1.3. c lng ca k vng v phng sai ca bin ngu nhin Xnh l 5.1.21 Cho X l mt bin ngu nhin v (x1, x2, . . . , xn) l mu ngu nhin ca X. Khi

a) X = 1n

ni=1 Xi l c lng khng chch, bn vng v hiu qu ca k vngE(X) ca bin

ngu nhin X.b) Phng sai iu chnh 2n1 =

1n1

ni=1(Xi X)2 ca mu ngu nhin l c lng khng

chch, bn vng ca phng sai D(X) i vi bin ngu nhin X.

ngha: - Mun c lng k vng E(X) ta ly trung bnh mu c lng cho n.- Mun c lng phng sai D(X) ta ly phng sai mu iu chnh c lng cho n.

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5.2. C LNG KHONG 41

5.2. C LNG KHONG

5.2.1. Nguyn l xc sut nh v ln

Nguyn l xc sut nhVi mt s > 0 kh b v P(A) = (thng thng 0 < 0, 05) th trong thc t ta tha

nhn s kin A khng xy ra trong mt ln thc hin php th.

Nguyn l xc sut ln

Vi mt s > 0 kh b v P(B) = 1 (thng thng 0 < 0, 05) th trong thc t tatha nhn s kin B lun xy ra trong mt ln thc hin php th.

5.2.2. Khong tin cy v tin cynh ngha

Gi s X l bin ngu nhin biu th c trng nghin cu trn tp , c phn phi xc sut bit nhng cn ph thuc vo tham s cha bit. c lng ta ly mu kch thc n. Gi s(x1, x2, . . . , xn) l mu thc nghim ng vi mu ngu nhin (X1, X2, . . . , X n) ca X. Nu ta tmc hai bin ngu nhin f1(X1, X2, . . . , X n), f2(X1, X2, . . . , X n) sao cho

P[f1(X1, X2, . . . , X n) < < f2(X1, X2, . . . , X n)] =

,trong = 1

cho trc v gn bng 1(thng thng 0 < 0, 05), th khong s thc

(f1(X1, X2, . . . , X n), f2(X1, X2, . . . , X n)) hayf1(X1, X2, . . . , X n) < < f2(X1, X2, . . . , X n) gil khong tin cy ca tham s vi tin cy.

Trong trng hp khong tin cy i xng c dng

(g(X1, X2, . . . , X n) , g(X1, X2, . . . , X n) + )

Khi c gi l chnh xc hay sai s c lng.Cn ch rng, cng mt tin cy ta c th tm c nhiu khong tin cy khc nhau ca

tham s . Khong tin cy no c di ngn nht th xem khong tin cy l tt nht. Trong thchnh, ta ch tm khong tin cy tt nht.

Nhn xt

T nh ngha trn, nu ta chn = 1% = 99% v

P[f1(X1, X2, . . . , X n) < < f2(X1, X2, . . . , X n)] =

iu ny ni ln rng xc sut ly ra mt mu thc nghim (x1, x2, . . . , xn) mf1(x1, x2, . . . , xn) < < f2(x1, x2, . . . , xn) l 99%

Nh vy nu ta ly ra mt mu thc nghim (x1, x2, . . . , xn). Ta c khong tin cy c th

f1(x1, x2, . . . , xn) < < f2(x1, x2, . . . , xn)

v theo nguyn l xc sut ln, ta lun tha nhn tham s tha mn

f1(x1, x2, . . . , xn) < < f2(x1, x2, . . . , xn)

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5.2.3. Khong tin cy cho k vng E(X) = vi X N(, )

X l i lng ngu nhin biu th c trng nghin cu trn tp c phn phi chun vi kvng E(X) = cha bit. c lng , ta ly mu ngu nhin kch thc n. Ta c mu ngu

nhin (X1, X2, . . . , X n) v mu thc nghim tng ng (x1, x2, . . . , xn). Ta tnh s trung bnh mux. Ty theo lch chun bit hay cha m tnh lch chun mu iu chnh 2n1. Vi tincy = 1 , ta c

Khong tin cy i xng E(X) = l (x , x + ) hayx < < x + . c lng E(X) = ta ly s trung bnh mu x c lng cho n. Khi

P(x < < x + ) = P(| X |< ) =

Trng hp lch chun bit Khi =n

1

2Trng hp lch chun cha bit Tnh s 2n1. Ta c

P(| X |< ) = 1 P

| X n1

n |<

n

n1

=

= n1n

.t(n 1; 1 )

VX

n1

n c phn phi Student vi n

1 bc t do.

Ch rng, khi n > 30:X

n1

n c phn phi xp x phn phi chun N(0, 1) nn t(n

1; 1 ) 1

2

.

V d 5.2.55 c lng cng trung bnh ca mt loi bi bng thp, ngi ta ly ra 25 bi kim tra. Tnh c cng trung bnh x = 10, vi tin cy99%. Hy tm khong tin cy ixng ca cng trung bnh ca vin bi thp . Bit rng cng ca vin bi c phn phi chunN(, 2) trong hai trng hp:

a) lch chun = 1b) lch chun cha bit v t mu thc nghim tnh c n1 = 1, 2

GiiGi X l cng ca vin bi bng thp th X l i lng ngu nhin c phn phi chun. Khi

k vng E(X) = l cng trung bnh ca mt vin bi.a) Ta c n = 25, x = 10, = 1, tin cy = 99%

Tra bng phn v chun, ta c 1

2

= 2, 567.

T , ta c =n

1

2

=

125

2, 567 0, 515Vy khong tin cy i xng ca cng trung bnh l

(x , x + ) = (9, 485; 10, 515) hay9, 485 < < 10, 515

b) Ta c n = 25, x = 10, n1 = 1, 2, tin cy = 99%

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5.2. C LNG KHONG 43

TrabngphnvStudentvin1 = 24 bctdo,mcphnv1 = 0, 05, tac t(24; 0, 05) =2, 797 nn

=n1

nt(n 1; 1 ) = 1, 2

52, 797 0, 67

Vy khong tin cy i xng ca cng trung bnh l

(x , x + ) = (9, 33; 10, 67) hay9, 33 < < 10, 67V d 5.2.56 c lng hao ph xng ca mt loi t chy trn on ng AB, ngi ta theodi 100 chuyn xe v tnh c x = 10 lt, n1 = 0, 5 lt, vi tin cy95%. Gi s xng bin ngunhin X ch lng xng hao ph ca mt loi t trn trn on AB c phn phi chun. Hy clng khong tin cy i xng ca ca mc hao ph xng trung bnh ?

Giia) Ta c n = 100, x = 10, tin cy = 95%. Tra bng phn v chun, ta c 1 (2) =

1(95%2)

= 1, 96.T , ta c =

n

1

2

=

0, 05100

1, 96 0, 1Vy khong tin cy i xng ca cng trung bnh l

(x , x + ) = (9, 9; 10, 1) hay9, 9 < < 10, 1

5.2.4. c lng t l phn trm hay xc sut

Gi s A l mt s kin lin kt vi mt php th. Mun bit xc sut xy ra bin c A hayp = P(A), ta lp li php th n ln(ly mu c lp kch thc n).

Gi Xi l s ln xy ra s kin A php th th i(i = 1, n) th X1, X2, . . . , X n l cc i lngngu nhin c lp c cng phn phi

Xi 0 1P 1 p p (i = 1, n)

Khi X = 1n

ni=1 Xi = Fn(A)

Ta c E(X) = pBi ton c lng xc sut chnh l bi ton c lng k vng E(X) = p.Khi n kh ln:X = 1

nni=1 Xi = Fn(A) c phn phi xp x phn phi chun N(p,p(1 p)).

c lng t l

Gi s tng th c N phn t, trong c M phn t mang c tnh A. Do khng iu traton b nn ta khng bit t l p = M

N. Php th ly ra mt phn t. Gi C l s kin phn t ly ra

mang c tnh A. Khi P(C) = MN

= p.Suy ra bi ton c lng t l chnh l bi ton c lng xc sut. c lng t l p, ta ly mu kch thc n, ta tnh c t l fn v vi tin cy = 1 , ta

c:

Khong tin cy cho t l p l (fn

, fn + ) c xc nh nh sau:

Vi n > 50 : nfn > 5 v n(1 fn) > 5. Ta c

=

fn(1 fn)

n.1

2

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V d 5.2.57 c lng t l gch loi 2 ca mt nh my, ngi ta ly ra 10000 vin v thy c300 vin gch loi 2. Vi tin cy99%

a) Hy c lng t l gch loi 2?b) Cn ly thm t nht bao nhiu gch na t l gch loi 2 ton b so vi t l mu c sai s

khng vt qu 0, 001. Gi thit ly t l mu fn thay cho t l mu ca mu cn thm.

GiiGi p l t l gch loi 2 ca nh my.Ta c n = 10000, t l mu fn = 30010000 = 0, 03 tin cy = 1 = 99%a)

=

fn(1 fn)

n.1

2

=

0, 03.0, 07

10000.2, 576 0, 0044

Vy khong c lng i xng ca t l l

fn < p < fn + 0, 03 0, 0044 < p < 0, 03 + 0, 0044

hay0, 0256 < p < 0, 0344b) Gi N l s gch cn ly thm n = N + 10000T l mu fn = 0, 03 v tin cy = 0, 99, ta c 1

(2

)= 2, 576

Theo gi thit 0, 001

=

fn(1 fn)

n1

2

=

0, 03.0, 07

n.2, 576 0, 001 n 193101, 18

suy ra nmin = 193102. S gch cn ly thm t nht l N = 193102 10000 = 193102V d 5.2.58 c lng c trong h, ngi ta bt 1000 con lm du ri th li. Sau li bt 900con thy c 90 con lm du. Hy c lng s lng c trong h vi tin cy95%

GiiGi N l s c trong h.Gi p l t l c lm du trong h, ta c p = 1000

N

Ta c lng p. Ta c n = 900, t l mu fn = 90900 = 0, 1 tin cy = 1 = 95%Tra bng phn v chun, ta c 1 (

2) = 1, 96

=

fn(1 fn)

n1

2

=

0, 1.0, 9

900.1, 96 0, 0196

Suy ra fn < p < fn + 0, 0804 < 1000N < 0, 1196Vy s c trong h l 8362 N 12437

5.2.5. c lng phng sai D(X) vi X N(, )

Gis Xlbinngunhinbiuthctrngnghincutrntp cphnphichunD(X) =

2 cha bit. c lng D(X) talymukchthc n,tacmungunhin (X1, X2, . . . , X n)v mu thc nghim (x1, x2, . . . , xn).Ta c Xi N(, 2) Xi N(0, 1). Do :n

i=1 (Xi

)2 =ns202

c phn phi khi bnh phng vi (n 1) bc t do.44


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5.2. C LNG KHONG 45

Trong

2n =1

n

ni=1

(xi x)2; x = 1n

ni=1

xi

Ta xt cc trng hp sau:

Vi k vng E(X) = bit

Tnh sn = 1nn

i=1(xi )2Tra bng phn v khi bnh phng vi n bc t do ng vi cc mc phn v

2, 1

2ta c

22

(n), 212

(n)

Khi , khong tin cy ca phng sai

ns2n21

2

(n)< 2 0, ta ly ngu nhin hai im B v C viOB = x,OC= y. Tnh xc sut ly -c hai im B, C sao cho BC 4

1. Xc inh a v hm mt xc sut f(x)?

2. Tnh cc tham s c tr-ng ca X?

3. Tnh xc sut trong ba ln thc hin php th v X c 2 ln X nhn gi tr trong

khong (0,8

)?

Cu 16. Cho bin ngu nhin X c hm mt xc sut

f(x) =

ax2, x (0, 2]

0, x / (0, 2]

a) Xc nh a v hm phn phi F(x) ?

b) Tnh E(X), D(X) ?

c) Tnh P(1 < X < 9) ?

Cu 17. di X ca chi tit my l N(5cm, 0.81cm). Tnh xc sut ly -c mtchi tit my c chiu di trong khong (4cm, 7cm)?

Cu 18. Chiu cao Y ca nam gii tr-ng thnh l N(160cm, cm). Bit rng P(|X160| < 2) = 0.2569, tm v tnh xc sut ly ngu nhin 4 nam th c t nht mt

ng-i c chiu cao trong khong 158 cm n 162 cm?

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