CC BIN NGU NHIN RI RC THNG DNG

0.1 Bin ngu nhin Bernoulli

nh l 0.1. BNN X c gi l BNN Bernoulli nu X() = f0; 1gv (0) = 1 p; (1) = p vi 0 < p < 1. K hiu X B(p).

V d 0.1. Nu A l mt bin c th ta c th nh ngha hm ch boca A nh sau:

IA (!) =

1 nu ! 2 A0 nu ! =2 A

Ta thy IA B(p) vi p = P (A).V d 0.2. Tung mt ng xu (khng nht thit cn i v ng cht).Gi s ng xu xut hin mt sp vi xc sut p, suy ra xc sut ngxu xut hin mt nga l 1 p. Ta t cc bin c

N : xut hin mt nga. S: xut hin mt sp.Hin nhin = fN ;Sg. Ta xt nh x

X : ! RS 7! X(S) = 1N 7! X(N) = 0

Khi X B(p).

nh ngha 0.1. (Dy php th Bernoulli) Mt dy php th clp m mi php th c mt bin c A no c quan tm, sxy ra vi xc sut P (A) = p khng i v khng xy ra vi xc sutP (A) = 1 p, c gi l dy php th Bernoulli.

0.2 Bin ngu nhin c phn phi nh thc

nh ngha 0.2. Xt mt dy gm n php th Bernoulli, bin cc quan tm trong mi php th l bin c A. Gi X l s ln xuthin bin c A trong n php th. Khi , ta ni X l mt BNN cphn phi nh thc vi hai tham s p v n, k hiu X B(n; p).

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nh l 0.2. Nu X N(n; p) th X() = f0; 1; 2; : : : ;ng. P (X = k) = Cknpk(1 p)nk.

Chng minh. u tin ca nh l l hin nhin. Ta ch cn chngminh th hai, tc P (X = k) = Cknpk(1 p)nk.

Dy kt qu ca n php th c biu th bi mt dy n k t gm Av A: AAAA : : : A.

Bin c (X = k) xy ra khi trong dy c k k t A v n k k tA. Do , xc sut c c mt dy nh vy s bng pk(1p)nk. Mtkhc, c tt c Ckn dy khc nhau vi k k t A v n k k t A. Do ,P (X = k) = Cknp

k(1 p)nk. V d 0.3. Xc sut mt bnh nhn c cha bnh thnh cng vi kthut mi l p = 0; 8. Gi s c 10 bnh nhn.

1. Tnh xc sut c 6 bnh nhn c cha bnh thnh cng vi kthut mi.

2. Tnh xc sut c nhiu nht 8 bnh nhn c cha bnh thnhcng vi k thut mi.

3. Tnh xc sut c t nht mt bnh nhn c cha bnh thnhcng vi k thut mi.

Gii. 1. Gi X l s bnh nhn c cha khi bng k thut mi. Khi, X B (10; 0; 8). Ta cn tnh P (X = 6).

S dng nh l 0.2 ta tnh c

P (X = 6) = C610(0; 8)6(0; 2)4 = 0; 0881:

2. Ta cn tnh P (X 8). Ta c

P (X 8) = 1 P (X = 9) P (X = 10)= 1 C910(0; 8)9 (0; 2) (0; 8)10 = 0; 6242:

3. Ta cn tnh P (X 1). Ta cP (X 1) = 1 P (X = 0) = 1 (0; 2)10:

V d 0.4. Mt nh tuyn dng kim tra kin thc ln lt n ng vin,vi xc sut c chn ca mi ng vin 0,56. Nu xc sut nhtuyn dng chn ng 8 ng vin l 0,1794 th s ngi phi kim tral bao nhiu ?

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Gii. Gi X l s ng c vin c tuyn dng. Khi , X B(n; 0; 56).Ta cn tm n P (X = 8) = 0; 1794. Ta c

P (X = 8) = 0; 1794, C8n(0; 56)8(0; 44)n8 = 0; 1794, n = 12:Vy n = 12 l gi tr cn tm.

V d 0.5. Mt my sn xut ln lt tng sn phm vi xc sut xuthin ph phm l 4%. Cho my sn xut n sn phm th thy xc sutc t nht 1 ph phm ln hn 30%. Xc nh gi tr nh nht ca n.

Gii. Gi X l s ph phm c sn xut. Khi , X B(n; 0; 04). Tacn tm n nh nht P (X 1) > 0; 3.

P (X 1) > 0; 3, 1 P (X = 0) > 0; 3, P (X = 0) < 0; 7, 0; 96n < 0; 7, n > 8; 7:

Vy n = 9 l s nh nht tha yu cu bi.

V d 0.6. Bnh Tay-Sachs l mt bnh di truyn gy ra bi mt gen btthng (ta gi l gen Tay-Sachs). Nhng ngi c gen bt thng nykhng c mt enzym quan trng c gi l hexosaminidase A (Hexa)gip ph v mt cht bo c gi l ganglioside GM2. Cht ny tcht ln no v gy tn hi n cc t bo thn kinh. Tr s sinh bt uc du hiu bnh t 3 n 6 thng. Tr em b bnh Tay-Sachs thngb ic, b m, b bi lit v thng cht tui ln 5. Bnh Tay-Sachsph bin nht ngi Do Thi. Mt s nhm khng phi ngi Do Thicng c nguy c mc bnh cao nh: bao gm nhng ngi c t tin lngi Php-Canada, t Louisiana Bayou, hoc t cc qun th Amish Pennsylvania.

Nu mt cp v chng khng mc bnh Tay-Sachs nhng c manggen Tay-Sachs th xc sut a tr sinh ra mang bnh Tay-Sachs l 1

4.

Gi s cp v chng ny c 4 a con, hy tnh xc sut khng c atr no b bnh.

Gii. Gi X l s a tr trong gia nh mc bnh Tay-Sachs. Khi ,X B(4; 1

4). Ta cn tnh P (X = 0).

Ta c P (X = 0) = C04(34)4 = 0; 3164.

nh l 0.3. Cho Xi; i = 1; n l n BNN Bernoulli c lp v Xi B(p); i = 1; n. Khi , BNN X = X1 + X2 + + Xn c phn b nhthc vi hai tham s n v p.

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Vic chng minh nh l 0.3 kh n gin nn dnh cho bn c.nh l 0.3 gip cho vic tm hiu tnh cht ca cc BNN c phn phinh thc tr nn n gin hn.

Mt trong nhng kh khn ln nht, khi s dng BNN c phn phinh thc, l vic tnh xc sut P (X = k) = Cknpk(1 p)nk vi n ln v pnh rt kh khn. Chng ta th xt v d sau: Mt nh my sn xutbng n vi t l bng hng l 0; 4%. Bit nh my sn xut 10000bng n trong ngy, tnh xc sut trong ngy c 30 bng b hng.

Sau y l li gii ca bi ton:GiX l s bng n b hng trong ngy. Khi ,X B (10000; 0; 004).

Ta cn tnh xc sut P (X = 30).Ta c P (X = 30) = C3010000(0; 004)

30(0; 996)9970. Cng vic cn li cata l n gin gi tr C3010000(0; 004)

30(0; 996)9970. y thc s l iu rtkh nu chng ta khng dng my tnh.

Hn na, nu ta gp cc bi ton dng ny vi s lng php thcn ln hn gi tr trn (i khi ln ti hng triu) th d ta dng mytnh cng kh lng n gin gi tr tm c. khc phc nhc imtrn, ta s tm hiu kt qu quan trng sau:

nh l 0.4. Cho dy s dng fpngn=1;+1 (0; 1) tha iu kinlimn!1

npn = p > 0. Khi , vi mi k 2 Z>0 ta c

limn!+1

Cknpkn(1 pn)nk =

eppk

k!:

Chng minh. t Tn = Cknpkn(1 pn)nk vi mi n 2 Zk. Khi Tn = C

knp

kn(1 pn)nk

=n!

k! (n k)!pkn(1 pn)nk

=n (n 1) : : : (n k + 1)

k!pkn(1 pn)nk

=1

k!

1 1

n

: : :

1 k 1

n

(npn)

k(1 pn)nk:

Ta tnh c

limn!+1

1 1

n

: : :

1 k 1

n

= 1.

limn!+1

(npn)k = pk.

Hn na, v limn!+1

npn = p nn limn!+1

pn = 0. Do ,

limn!+1

(1 pn)nk = limn!+1

eln(1pn)(nk) = limn!+1

epn(nk) = ep:

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Vy limn!+1

Cknpkn(1 pn)nk =

eppk

k!.

H qu 0.1. Cho n; k 2 Z>0 sao cho n ln hn k rt nhiu. Khi ,vi p l s dng kh gn 0 (c th hn l p 0; 01 ) ta c

Cknpn(1 p)nk e

(np)(np)k

k!:

Vi H qu 0.1, gi tr C3010000(0; 004)30(0; 996)9970 c th c tnh n

gin nh sau:

C3010000(0; 004)30(0; 996)9970 e

404030

30!= 0; 0185:

0.3 Bin ngu nhin c phn phi hnh hc

nh ngha 0.3. Xt mt dy v hn cc php th Bernoulli, binc c quan tm trong mi php th l bin c A. Gi X l s phpth cn thit xut hin bin c A. Khi , ta ni BNN X c phnphi hnh hc vi tham s p, k hiu X G(p).

nh l 0.5. Nu X G(p) th X() = Z>0. P (X = k) = (1 p)k1p; k 2 Z>0.

Chng minh. u ca nh l l hin nhin. Ta s chng minh th hai, tc P (X = k) = (1 p)k1p; k 2 Z>0.

Bin c (X = k) xy ra khi v ch khi trong k 1 php th u tin,bin c A khng xut hin, trong ln th th k th mi xut hin binc A. Do , P (X = k) = (1 p)k1p. V d 0.7. Gieo mt con xc xc n khi no xut hin mt 1 chm thngng li. Tnh xc sut s ln gieo dng li con s 5.

Gii. Gi X l s ln gieo cn thit xut hin mt 1 chm. Khi X G(1

6). Ta cn tnh P (X = 5).

Ta c P (X = 5) = (1 16)4 1

6= 0; 0804.

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nh l 0.6. Nu X G(p) th P (X > k) = (1 p)k vi mi k 2 Z>0.

Gii. Ta c

P (X > k) =+1X

i=k+1

P (X = i) =+1X

i=k+1

(1 p)i1p:

S dng ng thc+1Pi=0

ai = 11a vi jaj < 1 ta c

+1Xi=k+1

(1 p)i1p = (1 p)kp+1Xi=0

(1 p)i = (1 p)kp1p= (1 p)k:

Do P (X > k) = (1 p)k. Trng hp k ln v p nh ta c th dng xp x

(1 p)k = ek ln(1p) ekp:V d 0.8. Gi s xc sut c ma trong mt ngy vng A l 1

100. Tnh

xc sut khng c ma trong mt nm vng A.

Gii. Gi X l s ngy cn thit xut hin ma vng A. Khi ,X G( 1

100). Ta cn tnh P (X > 365).

Ta c P (X > 365) = (1 1100

)365 e 1100365 = e3;65 = 0; 03.

0.4 Bin ngu nhin c phn phi nh thc m

nh ngha 0.4. Xt mt dy v hn cc php th Bernoulli, bin cc quan tm trong mi php th l bin c A. GiX l s php thcn thit bin c A xut hin r ln r 2 Z>0. Khi , ta ni BNN Xc phn b nh thc m vi hai tham s r v p, k hiuX NB(r; p).

nh l 0.7. Nu X NB(r; p) th X() = fr; r + 1; r + 2; : : :g. P (X = k) = Cr1k1pr(1 p)kr; k 2 Zr.

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Chng minh. u ca nh l l hin nhin. Ta s chng minh th hai, tc P (X = k) = Cr1k1p

r(1 p)kr; k 2 Zr.Bin c (X = k) xy ra khi v ch khi trong k 1 php th u tin,

bin c A xut hin r 1 ln, trong ln th th k cng xut hin binc A. Do , P (X = k) = Cr1k1p

r(1 p)kr; k 2 Zr. V d 0.9. Xc sut mt x th bn trng mc tiu l 0,8. X th chngng bn khi no mc tiu b trng 3 vin n. Tnh xc sut x thngng bn ln bn th 5.

Gii. Gi X l s ln bn cn thit mc tiu b trng 3 vin n.Khi , X NB(3; 0; 8). Ta cn tnh P (X = 5).

Ta c P (X = 5) = C24(0; 8)3(0; 2)2 = 0; 1229.

nh l 0.8. Xt mt dy v hn cc php th Bernoulli, bin c cquan tm trong mi php th l bin c A. Gi G1 l s php th cnthit ln u xut hin bin c A, G2 l s php th c cngthm ln th 2 xut hin bin c A, G3 l s php th c cngthm ln th 3 xut hin bin c A, . . . , Gr l s php th ccng thm ln th r xut hin bin c A. Khi , Gi; i = 1; r l ccBNN c lp v Gi G(p). Hn na, G1 +G2 + +Gr NB(r; p):

Chng minh nh l 0.8 kh n gin nn dnh cho bn c. Ktqu thu c trong nh l 0.8 rt tin li khi nghin cu tnh cht caBNN c phn phi nh thc m.

0.5 Bin ngu nhin c phn phi Poisson

nh ngha 0.5. BNN X c gi l BNN c phn phi Poisson vitham s > 0 nu hai iu kin sau y c tha mn:

X() = Z0.

P (X = k) = ek

k!; k 2 Z0.

Da vo H qu 0.1, ta thy phn phi Poisson l gii hn ca phnphi nh thc vi cc tham s p =

nv n khi n! +1. Do , phn phi

Poisson cng c nhiu ng dng trong thc t.M hnh phn phi Poisson l m hnh thng c dng cho cc

BNN dng s bin c xy ra trong mt khong thi gian no . Taxt v d sau y:

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V d 0.10. Quan st mt ca hng ngi ta thy c trung bnh 1 gi c30 khch n mua hng. Tnh xc sut c 4 khch vo ca hng trong10 pht.

Sau y l li gii ca bi ton: Gi X l s khch vo ca hngtrong 10 pht. Ta cn tnh P (X = 4).

Ta chia 10 pht thnh n (n ln) khong thi gian bng nhau[t0; t1]; [t1; t2]; : : : ; [tn1; tn] vi tn t0 = 10 sao cho mi khong c ti amt v khch vo. V trung bnh 1 gi c 30 khch vo siu th nn trong10 pht c trung bnh 5 khch vo siu th. Do , xc sut mt khongthi gian [ti; ti+1] c khch vo siu th l pn = 5n . Ta suy ra tng skhch n ca hng trong 10 pht tun theo lut phn phi nh thcvi tham s p = 5

nv n. Do , nu cho n ! +1 ta c ngay BNN X c

phn b Poisson vi tham s = 5. Vy P (X = 4) = e5544!

= 0; 1755.

V d 0.11. Mt bn xe khch trung bnh c 70 xe xut bn trong 1 gi.Tnh xc sut trong 5 pht c 3 xe xut bn.

Gii. GiX l s xe xut bn trong thi gian 5 pht. Khi ,X P (356).

Ta cn tnh P (X = 3).

Ta c P (X = 3) =e

356 35

6

33!

= 0; 0969.

V d 0.12. Quan st thy trung bnh 5 pht c 15 khch hng vo 1siu th nh. Tm xc sut c nhiu hn 2 khch vo siu th trong30 giy ?

Gii. Gi X l s khch vo siu th trong thi gian 30 giy. Khi ,X P (1; 5). Ta cn tnh P (X > 2).

P (X > 2) = 1 P (X = 0) P (X = 1) P (X = 2)= 1 e

1;5 (1; 5)00!

e1;5 (1; 5)1

1! e

1;5 (1; 5)22!

= 0; 1912

Vy P (X > 2) = 0; 1912.

0.6 Bin ngu nhin c phn phi siu bi

nh ngha 0.6. Mt tp hp gm c M phn t, trong c Nphn t (1 N < M ) mang tnh cht A. Ly ngu nhin mt lt nphn t thuc tp hp . Gi X l s phn t c tnh cht A trongn phn t ly ra. Khi , ta ni BNN X c phn b siu bi vi batham s N;M v n, k hiu X H(M ;N ;n).

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nh l 0.9. Nu X H(M ;N ;n) th X () = fmin fn (M N) ; 0g ; : : : ;min fn;Ngg.

P (X = k) = CkNC

nkMN

CnM; k 2 X().

Gii. chng minh u ca nh l 0.6 ta xt cc trng hp sau:+ Nu n M N ;n N th tp gi tr ca X l

X () = f0; 1; : : : ; ng :

+ Nu n M N ;n > N th tp gi tr ca X l

X () = f0; 1; : : : ; Ng :

+ Nu n > M N ;n N th tp gi tr ca X l

X () = fn (M N) ; n (M N) + 1; : : : ; ng :

+ Nu n > M N ;n > N th tp gi tr ca X l

X () = fn (M N) ; n (M N) + 1; : : : ; Ng :

Kt hp cc trng hp trn ta suy ra

X () = fmin fn (M N) ; 0g ; : : : ;min fn;Ngg :

chng minh hai ta c nhn xt: bin c (X = k) xy ra khitrong n phn t ly ra c k phn t c tnh cht A cn n k phn tcn li khng c tnh cht A. C CkN cch ly k phn t t N phn tc tnh cht A, CnkMN cch ly n k phn t t M N phn t khngc tnh cht A. Do , s bin c s cp thun li cho bin c (X = k)l CkN CnkMN . Hn na, s cch ly n phn t tp hp ban u l CnM .Chnh v th, ta c P (X = k) = C

kNC

nkMN

CnM; k 2 X().

V d 0.13. Mt thng bia c 24 chai trong ln 3 chai qu hn sdng. Chn ngu nhin t thng ra 4 chai bia (chn 1 ln). Tnh xcsut chn c c 4 chai bia khng qu hn s dng.

Gii. Gi X s chai bia khng qu hn dng. Khi , X H(24; 21; 4).Ta cn tnh P (X = 4).

Ta c P (X = 4) =C421C424

= 0; 5632.

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V d 0.14. Mt ca hng bn 50 con c chp, trong c 18 con cchp Nht. Mt khch hng chn ngu nhin 4 con c chp (chn 1ln). Tnh xc sut c t 2 ti 4 con c chp Nht.

Gii. GiX l s con c chp Nht c ly ra. Khi ,X H(50; 18; 4).Ta cn tnh P (2 X 4). Ta c

P (X = 2) =C232C

218

C450= 0; 3295

P (X = 3) =C132C

318

C450= 0; 1134

P (X = 4) =C418C450

= 0; 0133

Do , P (2 X 4) = 0; 4562. Nhn xt 0.1. Mt trong nhng kh khn khi s dng phn phi siubi l trng hp M;N nhn nhng gi tr kh ln. Khi , xc sutP (X = k) rt kh tnh ton ngay c khi s dng my tnh. Chng tath xt bi ton sau:

V d 0.15. Mt hp ng 100000 vin bi, trong c 40000 bi . Chnngu nhin 10 bi t hp, tnh xc sut c 7 bi .

Sau y l li gii ca bi ton:Gi X l s bi trong 10 bi ly ra, suy ra X H (100000; 40000; 10).

Ta cn tnh P (X = 7). Ta c

P (X = 7) =C740000C

360000

C10100000:

Cng vic cn li ca ta l n gin gi tr C740000C

360000

C10100000. y thc s l

iu rt kh nu chng ta khng c s h tr ca my tnh. Hn na,nu ta gp cc bi ton dng ny vi s lng c th c kho st cnln hn gi tr trn (i khi ln ti hng triu) th d chng ta c ccmy tnh cng kh lng n gin gi tr tm c. khc phc nhcim trn, ta s tm hiu kt qu quan trng sau:

nh l 0.10. Cho M;N; n 2 Z>0, nu n N; n M N (k hiua b c hiu l t s a

brt gn s 0) th vi mi s t nhin k

khng ln hn n ta c

CkNCnkMN

CnM Cknpk(1 p)k

vi p = NM

(k hiu a b c hiu l t s abrt gn 1).

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Chng minh. Ta bin i

CkNCnkMN

CnM=

n!N ! (M N)! (M n)!k! (n k)! (N k)!M ! (M N n+ k)! :

V n N; nM N nn N !

(Nk)! = Nk1 k1

N

: : :1 1

N

Nk. (Mn)!

M != 1

Mn(1n1M ):::(1 1M ) 1

Mn.

(MN)!(MNn+k)! (M N)nk.

Khi , CkNC

nkMN

CnM Ckn N

k(MN)nkMn

= Cknpk(1 p)nk vi p = N

M.

Tr li V d 0.15, v 10 40000; 10 60000 nn

P (X = 7) =C740000C

360000

C10100000 C710(0; 4)7(0; 6)3 = 0; 042467:

T nh l 0.10 ta c th rt ra h qu quan trng sau:

H qu 0.2. Cho X H (M ;N ;n), nu n N; n M N th ta cth xem X B n; N

M

.

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