TRNG CAO NG CNG NGH THNG TIN TPHCMKHOA I CNGB MN TON

BI TP

XC SUT V THNG K TON

Ngi bin son: THS. DNG TH XUN ANTHS. NGUYN TH THU THY

LU HNH NI BNM 2013

LI NI U

Nhm gip sinh vin Trng Cao ng Cng Ngh Thng Tin TP. HCMhc tp mn XC SUT THNG K t kt qu cao, chng ti tin hnhbin son quyn Bi Tp Xc Sut v Thng K Ton. Quyn bi tp ny cbin son ph hp vi sinh vin bc Cao ng v lu hnh ni b trong phmvi Nh trng. Ti liu c bin son da trn cng mn hc v c sdng km theo quyn Gio Trnh Xc Sut v Thng K Ton. Ni dung gmhai phn:

Phn Xc sut gm 2 chng:Chng 1. BIN C NGU NHIN V XC SUTChng 2. I LNG NGU NHIN V LUT PHN PHI XC SUT

Phn Thng k gm 3 chng:Chng 3. MU NGU NHIN V CC C TRNG MUChng 4. C LNG THAM SChng 5. KIM NH GI THUYT THNG K

Ti liu gii quyt hu ht cc bi tp t c bn n phc tp vi li gii rrng, d hiu. Cc bi tp c phn loi theo tng dng c th t n ginn tng hp. M u mi dng bi tp, tc gi tm tt ni dung l thuyttrng tm ngi c vn dng thc hnh. Phn cui mi chng l bi tpt gii sinh vin c c hi t rn luyn.

Vi mc ch nh trn, chng ti hy vng y s l ti liu b ch cho ccem sinh vin trong qu trnh hc tp bc Cao ng cng nh qu trnh hclin thng sau ny. Ti liu s c cp nht thng xuyn trong qu trnhging dy v hc tp. Chng ti rt mong nhn c kin gp , xy dng tqu Thy C v cc em sinh vin ti liu ngy cng hon thin hn. Trntrng cm n!

TP. HCM, thng 6 nm 2013

Cc tc gi

2

Mc lc

LI NI U 2

1 BIN C NGU NHIN V XC SUT 5

1.1 B tc v Gii tch T hp . . . . . . . . . . . . . . . . . . . . 5

1.1.1 Php m . . . . . . . . . . . . . . . . . . . . . . . . . . 5

1.1.2 Hon v . . . . . . . . . . . . . . . . . . . . . . . . . . . 6

1.1.3 Chnh hp . . . . . . . . . . . . . . . . . . . . . . . . . . 6

1.1.4 T hp . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6

1.2 Biu din mi lin h gia cc bin c . . . . . . . . . . . . . . 8

1.3 Tnh xc sut theo nh ngha c in . . . . . . . . . . . . . . 11

1.4 p dng cc cng thc tnh xc sut . . . . . . . . . . . . . . . 13

1.4.1 Cng thc cng . . . . . . . . . . . . . . . . . . . . . . . 13

1.4.2 Cng thc nhn . . . . . . . . . . . . . . . . . . . . . . 15

1.4.3 Cng thc xc sut ton phn v cng thc Bayes . . . 19

1.4.4 Cng thc Bernoulli . . . . . . . . . . . . . . . . . . . . 22

1.5 Bi tp tng hp . . . . . . . . . . . . . . . . . . . . . . . . . . 24

1.6 Bi tp t gii . . . . . . . . . . . . . . . . . . . . . . . . . . . 37

2 I LNG NGU NHIN V CC QUY LUT PHNPHI XC SUT C BIT 39

2.1 Lut phn phi xc sut ca LNN . . . . . . . . . . . . . . . 39

3

2.1.1 Bng phn phi xc sut (PPXS) . . . . . . . . . . . . . 40

2.1.2 Hm mt ca LNN lin tc . . . . . . . . . . . . . 43

2.2 Hm phn phi xc sut . . . . . . . . . . . . . . . . . . . . . . 46

2.3 Cc c trng bng s ca LNN . . . . . . . . . . . . . . . . 51

2.4 Cc quy lut phn phi xc sut c bit . . . . . . . . . . . . 57

2.4.1 Quy lut phn phi siu bi . . . . . . . . . . . . . . . . 57

2.4.2 Quy lut phn phi nh thc . . . . . . . . . . . . . . . 59

2.4.3 Quy lut phn phi Poisson . . . . . . . . . . . . . . . . 62

2.4.4 Quy lut phn phi chun . . . . . . . . . . . . . . . . . 65

2.4.5 Xp x cc quy lut phn phi xc sut . . . . . . . . . 68

2.5 Bi tp tng hp . . . . . . . . . . . . . . . . . . . . . . . . . . 72

2.6 Bi tp t gii . . . . . . . . . . . . . . . . . . . . . . . . . . . 98

3 MU NGU NHIN V CC C TRNG MU 101

4 C LNG THAM S 107

4.1 c lng khong cho trung bnh tng th . . . . . . . . . . . . 110

4.2 c lng khong cho t l tng th . . . . . . . . . . . . . . . 116

4.3 Bi tp tng hp . . . . . . . . . . . . . . . . . . . . . . . . . . 118

5 KIM NH GI THIT THNG K 124

5.1 So snh trung bnh tng th vi mt s . . . . . . . . . . . . . 127

5.2 So snh t l tng th vi mt s . . . . . . . . . . . . . . . . . 131

5.3 Bi tp tng hp . . . . . . . . . . . . . . . . . . . . . . . . . . 132

5.4 Bi tp t gii . . . . . . . . . . . . . . . . . . . . . . . . . . . 147

MT S THI THAM KHO 149

PH LC: BNG TRA THNG K 155

TI LIU THAM KHO 161

4

Chng 1

BIN C NGU NHINV XC SUT

1.1 B tc v Gii tch T hp

1.1.1 Php m

Giai tha: n! = 1.2.3 . . .n. Quy c: 0! = 1.

Quy tc cng: Gi s c mt cng vic c thc hin theo 1 trong ktrng hp khc nhau.

Trng hp th 1 c n1 cch thc hinTrng hp th 2 c n2 cch thc hin

. . . . . . . . .Trng hp th k c nk cch thc hin

=Cng vic ny c

n = n1 + n2 + . . .+ nkcch thc hin

Quy tc nhn: Gi s c mt cng vic c thc hin thng qua k giaion lin tip.

Giai on 1 c n1 cch thc hinGiai on 2 c n2 cch thc hin

. . . . . . . . .Giai on k c nk cch thc hin

=C n = n1.n2 . . .nk

cch hon thnh cng vic

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1.1.2 Hon v

Hon v trn ng thng: S cch sp xp n phn t khc nhau vo nv tr cho l Pn = n! Hon v trn ng trn: S cch sp xp n phn t khc nhau vo n vtr cho trn mt ng trn l Pn1 = (n 1)!

1.1.3 Chnh hp

Cho tp A c n phn t. Mi b k phn t khc nhau (k n) rt t tpA, c phn bit th t c gi l mt chnh hp (khng lp) chp k ca n

phn t. S chnh hp chp k ca n phn t l Akn =n!

(n k)! Cho tp A c n phn t. Mi b k phn t rt t tp A, c phn bit tht trong mi phn t c th c mt n k ln, c gi l mt chnh hplp chp k ca n phn t. S chnh hp lp chp k ca n phn t l Bkn = n

k.

Ch . S chnh hp lp chp k ca n phn t c th tnh c bng cch pdng quy tc nhn, trong c k giai on, mi giai on c n cch.

1.1.4 T hp

Cho tp A c n phn t. Mi b k phn t khc nhau (k n) rt t tp A,khng phn bit th t c gi l mt t hp chp k ca n phn t. S t

hp chp k ca n phn t l Ckn =n!

k!.(n k)!

Bi 1. Vui i d m ci ca hai bn Bnh v Yn. Trc khi ra v, Vui v4 ngi bn na cng chp hnh lu nim vi c du ch r. Hy tnh s cchxp cc bn thnh 1 hng chp hnh sao cho:

a) C du ng cnh ch r.b) C du khng ng cnh ch r.c) Vui ng cnh bn phi c du.

BI GIIa) Ta xem c du v ch r l mt "b" th c P6 = 6! = 720 cch xp 5 ngibn cng c du ch r thnh mt hng.ng vi mi cch xp ny, c P2 = 2! = 2 cch hon v trong ni b "b" cdu v ch r.Vy c 6!.2! = 1440 cch xp theo yu cu.

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b) Xp ngu nhin 7 ngi bn thnh mt hng th c P7 = 7! = 5040 cch.Theo cu a), c 1440 cch xp 7 ngi bn ny thnh mt hng sao cho c duv ch r ng cnh nhau.Vy c 5040 1440 = 3600 cch xp sao cho c du khng ng cnh ch r.c) Tng t cu a) nhng trong ni b gia Vui v c du ch c 1 cch xp.Do , c 1.6! = 720 cch xp sao cho Vui ng cnh bn phi c du.

Bi 2. Lp hc c 45 sinh vin, trong c 43 bn l on vin. C bao nhiucch chn ngu nhin 5 on vin bu vo Ban cn s lp gm 1 lp trng,1 lp ph hc tp, 1 lp ph i sng, 1 b th v 1 ph b th.

BI GIIV vic chn 5 thnh vin c phn bit v tr nn s cch chn 5 on vin bu vo Ban cn s lp l s chnh hp (khng lp) 43 chp 5:

A543 =43!

38!= 115511760 (cch)

Bi 3. Xp ngu nhin 12 hnh khch ln 5 toa tu. C bao nhiu cch:

a) Xp ngu nhin 12 hnh khch ln 5 toa tu mt cch ty .b) Xp ngu nhin 12 hnh khch ln 5 toa tu sao cho toa th nht c 3

hnh khch.

BI GIIa) S cch xp ngu nhin 12 hnh khch ln 5 toa tu mt cch ty l schnh hp lp 12 chp 5: B125 = 512 = 244140625 (cch).b) Ta xem bi ton gm 2 giai on:- Giai on 1: Xp ngu nhin 3 hnh khch (t 12 hnh khch) vo toa thnht, c C312 = 220 (cch).- Giai on 2: Xp ngu nhin 9 hnh khch vo 4 toa tu cn li mt cchty , c B94 = 49 = 262144 (cch).Vy c C312.B94 = 57671680 cch xp ngu nhin 12 khch ln 5 toa tu sao chotoa th nht c 3 hnh khch.

Bi 4. Mt thng c 50 quyn sch, trong c 20 quyn sch Ting Vit v30 quyn sch Ton. C bao nhiu cch:

a) Ly ngu nhin ra 10 quyn sch.b) Ly ngu nhin ra 9 quyn sch trong c 5 quyn sch Ton.c) Ly ra 8 quyn sch Ton trao cho 8 em hc sinh.

7

BI GIIa) S cch ly ngu nhin ra 10 quyn sch t 50 quyn sch l s t hp 50chp 10: C1050 = 10272278170 (cch).b) C C420 cch ly ra 4 quyn sch Ting Vit.

C C550 cch ly ra 5 quyn sch Ton.Vy c C420.C530 = 690441570 cch ly ngu nhin ra 9 quyn sch trong c5 quyn sch Ton.c) C C830 cch ly ra 8 quyn sch Ton t 30 quyn sch Ton.ng vi mi cch ny, ta c 8! cch trao 8 quyn sch cho 8 em hc sinh.Vy c C830.8! = 235989936000 cch tha yu cu.

Bi 5. Mt l hng c 10 sn phm, trong c 8 sn phm tt v 2 phphm. C bao nhiu cch:

a) Ly ngu nhin ra 4 sn phm.b) Ly ra ngu nhin 4 sn phm, trong c 3 sn phm tt.c) Ly ra ngu nhin 4 sn phm, trong c t nht 1 ph phm.

BI GIIa) S cch ly ngu nhin 4 sn phm t 10 sn phm l s t hp 10 chp 4:C410 = 210 (cch).b) Ta c th xem bi ton gm 2 giai on:- Giai on 1: Ly 3 sn phm tt t 8 sn phm tt, c C38 = 56 (cch).- Giai on 2: Ly 1 ph phm t 2 ph phm, c C12 = 2 (cch).Vy c C38 .C12 = 112 cch ly ra ngu nhin 4 sn phm, trong c 3 snphm tt.c) V l hng ch c 2 ph phm nn bi ton gm 2 trng hp:- Trng hp 1: Ly c 1 ph phm (v 3 sn phm tt)

C: C12 .C38 = 112 (cch).- Trng hp 2: Ly c 2 ph phm (v 2 sn phm tt)

C: C22 .C28 = 28 (cch).Vy c C12 .C38 + C22 .C28 = 140 cch ly ra ngu nhin 4 sn phm, trong ct nht 1 ph phm.

1.2 Biu din mi lin h gia cc bin c

Cho hai bin c A v B. Tng hai bin c: Tng ca hai bin c A v B l mt bin c C nu Cxy ra A xy ra hoc B xy ra (t nht mt trong hai bin c xy ra).K hiu: C = A+B.

8

Tch hai bin c: Tch ca hai bin c A v B l mt bin c C nuC xy ra A xy ra v B xy ra (ng thi (c hai) bin c xy ra).K hiu: C = A.B.

Bin c i lp: Bin c B c gi l bin c i lp ca bin c Anu B xy ra A khng xy ra. K hiu: B = A.

Bi 1. Ba sinh vin d thi mn Ton cao cp. t cc bin c: Ai l "Sinhvin th i thi t"; Bi l "C i sinh vin thi t", i = 0, 3. Nu ngha cacc bin c sau:

a) A1.A2.A3 b) A1.A3 c) A1 + A2 +A3d) B0 e) A2.B1 f) A3B2

BI GII

a) A1.A2.A3: bin c c ba sinh vin thi t.b) A1.A3: bin c sinh vin th nht thi t v sinh vin th ba thi hng.c) A1 + A2 + A3: bin c c t nht mt sinh vin thi t.d) B0: bin c khng c sinh vin no khng t c ba sinh vin u t.e) A2.B1: bin c sinh vin th hai thi t v c mt sinh vin thi t

ch c sinh vin th hai thi t.f) A3B2: bin c sinh vin th ba thi hng v c hai sinh vin thi t ch

c sinh vin th ba thi hng.

Bi 2. Hai x th cng bn vo mt bia, mi ngi bn mt vin. Gi Ai lbin c "x th th i bn trng bia", i = 1, 2. Hy biu din cc bin c sauqua bin c A1, A2:

a) A: bin c x th th nht bn trng bia v x th th hai bn trt.b) B: bin c bia b trng n.c) C: bin c bia khng b trng n.

BI GIIa) A = A1.A2 b) B = A1 + A2 c) C = A1.A2 A1 + A2Bi 3. Kim tra 3 sn phm. Gi Ak l bin c sn phm th k tt. Hy trnhby cch biu din qua Ak cc bin c sau:

a) A: bin c tt c u xu.b) B: bin c c t nht mt sn phm tt.

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c) C: bin c c t nht mt sn phm xu.d) D: bin c khng phi tt c sn phm u tt.e) E: bin c c ng mt sn phm xu.f) F : bin c c t nht hai sn phm tt.

BI GII

a) A = A1.A2.A3b) B = A1 + A2 + A3c) C = A1 +A2 + A3d) D = A1.A2.A3 A1 + A2 + A3e) E = A1.A2.A3 + A1.A2.A3 + A1.A2.A3f) F = A1.A2.A3 + A1.A2.A3 + A1.A2.A3 + A1.A2.A3

Bi 4. Ba sinh vin thi mn xc sut thng k. Gi Ai l bin c sinh vin thi thi t, i = 1, 3. Hy biu din qua Ai cc bin c sau:

a) A: bin c c ba sinh vin u thi t.b) B: bin c c khng qu hai sinh vin thi t.c) C: bin c c t nht mt sinh vin thi t.d) D: bin c c mt sinh vin thi t.e) E: bin c sinh vin th nht v sinh vin th hai thi t.f) F : bin c ch c sinh vin th hai thi t.g) G: bin c c t nht mt sinh vin thi hng.

BI GII

a) A = A1.A2.A3b) B = A1.A2.A3+A1.A2.A3+A1.A2.A3+A1.A2.A3+A1.A2.A3+A1.A2.A3+

A1.A2.A3

c) C = A1 + A2 + A3 A1.A2.A3 + A1.A2.A3 + A1.A2.A3 + A1.A2.A3 +A1.A2.A3 + A1.A2.A3 + A1.A2.A3 \A1.A2.A3

d) D = A1.A2.A3 + A1.A2.A3 + A1.A2.A3e) E = A1.A2f) F = A1.A2.A3g) G = A1 + A2 + A3 A1.A2.A3 + A1.A2.A3 + A1.A2A3 + A1.A2.A3 +

A1.A2.A3 + A1.A2.A3 + A1.A2.A3 A1.A2.A3

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Bi 5. (Sinh vin t gii) C 3 ngi cng bn vo mt mc tiu, mi ngibn mt pht. Gi Ai l bin c ngi th i bn trng (i = 1, 3). Hy biudin qua Ai cc bin c sau:

a) A: bin c ch c ngi th nht bn trng.b) B: bin c ngi th nht bn trng cn ngi th hai bn trt.c) C: bin c c t nht mt ngi bn trng.d) D: bin c c 3 ngi u bn trng.e) E: bin c c t nht hai ngi bn trng.f) F : bin c ch c 2 ngi bn trng.g) G: bin c khng ai bn trng.h) H: bin c khng c hn hai ngi bn trng.i) I : bin c ngi th nht bn trng hoc ngi th hai bn trng.

1.3 Tnh xc sut theo nh ngha c in

Xt mt php th vi khng gian mu gm n bin c s cp c ngkh nng xy ra. Gi s A l mt bin c bt k. Ta c xc sut binc A xy ra l:

P (A) =S trng hp thun li cho AS trng hp ng kh nng

=m

n

Tnh cht:

P () = 1 P () = 0 Vi A l bin c bt k th 0 P (A) 1

Bi 1. Gieo mt con xc xc cn i v ng cht. Tnh xc sut xut hinmt ln hn 4 chm.

BI GIIGi A: bin c xut hin mt ln hn 4 chm.S trng hp ng kh nng l n = 6.S trng hp thun li cho A l m = 2.Do , P (A) =

m

n=

1

3

11

Bi 2. Mt l hng gm 50 my vi tnh, trong c 20 my do cng ty A snxut v 30 my do cng ty B sn xut. Mt khch hng chn ngu nhin 5my t l hng ny. Tnh xc sut :

a) C 5 my do cng ty A sn xut.b) C 3 my do cng ty B sn xut.

BI GIIL hng c 50 my vi tnh, chn ngu nhin 5 my th c n = C550 cch.a) Gi A: bin c chn c 5 my do cng ty A sn xut. Ta tnh P (A).S trng hp thun li cho A: m = C520.

Vy P (A) =m

n=

C520C550

0, 0073.b) Gi B: bin c chn c 3 my do cng ty B sn xut (v 2 my do cngty A sn xut). Ta tnh P (B).S trng hp thun li cho B: m = C330.C220.

Vy P (B) = mn

=C330.C

220

C550 0, 3641.

Bi 3. Mt nh phn tch th trng chng khon a ra mt danh sch cth 5 loi c phiu. Gi s xp c bng th t tng trng ca 5 loi cphiu ny vo nm ti v kh nng xp hng u nh nhau. Tnh xc sut d on ng 3 loi c phiu xp u bng ny:

a) Khng k th t.b) Xp theo th t nht, nh, ba.

BI GIIGi A: bin c d on ng 3 loi c phiu xp u bng.S trng hp thun li cho A l m = 1.a) Trong trng hp khng k th t th tng s cch d on l n = C35 = 10.Do , P (A) =

m

n=

1

10= 0, 1.

b) Trong trng hp c k th t nht, nh, ba th tng s cch d on ln = A35 = 60.Vy P (A) = m

n=

1

60 0, 0167.

Bi 4. T mt nhm bn gm 5 ngi: An, Bnh, Yn, Lnh, Phc. Chn ngunhin 3 bn t nhm ny, tnh xc sut trong c "Bnh".

BI GIIT nhm bn c 5 ngi, chn ngu nhin 3 bn th c n = C35 = 10 cch.

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Gi B: bin c trong 3 ngi c chn c bn Bnh.Sau khi chn Bnh th ta cn chn thm 2 bn t 4 ngi cn li, nn s trnghp thun li cho B l m = C24 = 6.

Vy P (B) =C24C35

=6

10= 0, 6.

Bi 5. C 10 khch bc ngu nhin vo mt ca hng c 3 quy. Tm xcsut c 3 ngi n quy s 1.

BI GIITa xem vic mi khch hng n 1 quy tnh tin l mt giai on, mi giaion c 10 cch nn s trng hp ng kh nng l n = 310.Gi A: bin c c 3 ngi n quy s 1. Ta cn tnh P (A).S trng hp thun li cho A l s cch chn ngu nhin 3 ngi t 10 ngi xp vo quy s 1 v 7 ngi cn li s n ngu nhin hai quy 2 v 3, tcl m = C310.27.

Vy P (A) =C310.2

7

310=

5120

19683 0, 2601.

Ch . Ta c th tnh s trng hp ng kh nng bng cch p dng chnhhp lp: B103 = 310.

1.4 p dng cc cng thc tnh xc sut

1.4.1 Cng thc cng

Hai bin c xung khc: Hai bin c A v B c gi l xung khc vinhau trong mt php th nu A v B khng cng xy ra.

H bin c xung khc tng i: H bin c {A1, A2, . . . , An} c gi lxung khc tng i nu hai bin c bt k trong h xung khc vi nhau.

Cng thc cng: Cho A v B l hai bin c bt k, ta c: P (A+B) = P (A)+P (B)P (AB). Nu A v B l hai bin c xung khc th: P (A+B) = P (A) + P (B).

Cng thc trn c th m rng cho n bin c, chng hn: Cho ba bin c bt k A,B,C, ta c:

P (A+B+C) = P (A)+P (B)+P (C)P (AB)P (BC)P (CA)+P (ABC)

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Cho {A, B, C} l h bin c xung khc tng i th:P (A+B + C) = P (A) + P (B) + P (C)

Cho A l mt bin c bt k, ta c: P (A) = 1 P (A).

Bi 1. C 5 con nga nh s t 1 n 5 tham gia mt cuc ua. Gi s khnng chy nhanh thng trn ca 5 con nga l nh nhau. Tnh xc sut con nga s 3 t nht l v nh.

BI GIIGi A: bin c con nga s 3 t nht l v nh.

Ai: bin c con nga s 3 v th i, i = 1, 5.Ta c: A = A1 + A2 v {A1, A2} xung khc, nn

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

5+

1

5= 0, 4

Bi 2. Rt ngu nhin ng thi 3 l bi t b bi 52 l. Tnh xc sut rtc t nht 1 l bi t.

BI GIIGi A: bin c rt c t nht 1 l bi t.= A: bin c khng rt c l bi t no.Ta c: P (A) =

C348C352

=4324

5525 0, 7826.

Suy ra, P (A) = 1 P (A) = 1 43245525

=1201

5525 0, 2174.

Bi 3. Hai ngi cng bn vo mt tm bia mt cch c lp, mi ngi bnmt pht. Xc sut ngi th nht v th hai bn trng bia ln lt l 0,6v 0,8. Tnh xc sut bia b trng n.

BI GIIGi A: bin c bia b trng n ( c t nht mt ngi bn trng bia).

Ai: bin c ngi th i bn trng bia, i = 1, 2.

Cch 1: Ta c A = A1 +A2 = P (A) = P (A1) + P (A2) P (A1.A2)V {A1, A2} c lp nn P (A) = P (A1) + P (A2) P (A1).P (A2)= P (A) = 0, 6 + 0, 8 0, 6.0, 8 = 0, 92.

Cch 2: Ta c A: bin c bia khng trng n.= A = A1.A2 = P (A) = P (A1).P (A2) (v {A1, A2} c lp)

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= P (A) = 0, 4.0, 2 = 0, 08Vy P (A) = 1 P (A) = 1 0, 08 = 0, 92.

Bi 4. Mt l hng c 50 sn phm, trong c 4 ph phm. Mt khch hngkim tra bng cch ly ngu nhin 10 sn phm t l hng ny. Nu trong 10sn phm ly ra c khng qu 1 ph phm th ngi ny s mua l hng. Tnhxc sut l hng c mua.

BI GIIGi A: bin c l hng c mua.

Ai: bin c trong 10 sn phm ly ra c i ph phm, i = 0, 10.Ta c: A = A0 + A1 v {A0, A1} xung khc= P (A) = P (A0) + P (A1) = C

04 .C

1046

C1050+C14 .C

946

C1050=

2717

3290 0, 82584.

Bi 5. Kho st v mc quan tm ca ngi dn trong mt khu ph i vi3 t bo A, B, C, ngi ta c s liu sau: c 20% ngi dn xem bo A; 15%ngi dn xem bo B; 10% ngi dn xem bo C; 5% xem A v B; 4% xem Av C; 3% xem B v C; 2% xem c A, B v C. Chn ngu nhin mt ngi dntrong khu ph, tnh xc sut ngi ny khng xem bo no trong s 3 tbo A, B, C ni trn.

BI GIIGi D: bin c ngi ny khng xem bo no trong 3 t bo A, B v C.= D: bin c ngi ny xem t nht mt trong 3 t bo A, B, C.Gi A,B,C ln lt l cc bin c ngi ny xem bo A, B, C.Ta c: D = A+B + C nn

P (D) = P (A) + P (B) + P (C) P (AB) P (AC) P (BC) + P (ABC)= 0, 2 + 0, 15 + 0, 1 0, 05 0, 04 0, 03 + 0, 02 = 0, 35

Vy P (D) = 1 P (D) = 1 0, 35 = 0, 65.

1.4.2 Cng thc nhn

Hai bin c c lp: Hai bin c A v B c gi l c lp vi nhautrong mt php th nu A c xy ra hay khng cng khng nh hng n khnng xy ra ca B v ngc li.

H bin c c lp tng i: H bin c {A1, A2, . . . , An} c gi lc lp tng i nu hai bin c bt k trong h c lp vi nhau.

15

H bin c c lp ton phn: H bin c {A1, A2, . . . , An} c gil c lp ton phn nu mi bin c trong h c lp vi mt tch bt kcc bin c cn li trong h.

Cng thc xc sut c iu kin: Cho hai bin c A v B. Xc sut bin c A xy ra vi iu kin bin c B xy ra c gi l cng thc xcsut c iu kin: P (A|B)Ngoi ra, xc sut c iu kin c th tnh theo cng thc: P (A|B) = P (AB)

P (B)

Cng thc nhn: Nu A v B l hai bin c bt k th: P (AB) = P (A).P (B|A) = P (B).P (A|B) Nu A v B l hai bin c c lp th: P (AB) = P (A).P (B)

Cng thc trn c th m rng cho n bin c, chng hn: Cho A, B, C l ba bin c bt k, ta c: P (ABC) = P (A).P (B|A).P (C|AB) Nu {A, B, C} l h c lp ton phn th: P (ABC) = P (A).P (B).P (C)

Bi 1. Cho A, B l hai bin c c lp v P (A) = 0, 3 v P (B) = 0, 7. Tnhcc xc sut: a) P (AB) b) P (A|B) c) P (B|A)

BI GIIV A,B l hai bin c c lp nn:a) P (A.B) = P (A).P (B) = 0, 3.0, 7 = 0, 21b) P (A|B) = P (A) = 0, 3c) P (B|A) = P (B) = 0, 7

Bi 2. Hai x th cng bn vo mt tm bia mt cch c lp, mi ngi bnmt pht. Xc sut x th th nht v th hai bn trng bia ln lt l 0,7v 0,8. Tnh xc sut :

a) C hai x th bn trng bia.b) Ch c x th th nht bn trng bia.

BI GIIGi Ai: bin c x th th i bn trng bia, i = 1, 2.Ta c: {A1, A2} c lp.a) Gi A: bin c c hai x th bn trng bia.V A = A1.A2 nn P (A) = P (A1).P (A2) = 0, 7.0, 8 = 0, 56.b) Gi B: bin c ch c x th th nht bn trng bia.V A = A1.A2 nn P (A) = P (A1).P (A2) = 0, 7.0, 2 = 0, 14.

16

Bi 3. C 3 hp: mi hp ng 5 vin bi, trong hp th i c i vin bi trng(i = 1, 3). Ly ngu nhin t mi hp ra mt vin bi.

a) Tm xc sut ly c 3 vin bi trng.b) Nu trong 3 bi ly ra c mt bi trng, tm xc sut vin bi trng

l ca hp th nht?

BI GIIGi Ai: bin c ly c vin bi trng t hp th i, i = 1, 2, 3Ta c: {A1, A2, A3} l h c lp ton phn.a) Gi A: bin c ly c 3 vin bi trng= A = A1.A2.A3= P (A) = P (A1).P (A2).P (A3) = 1

5 25 35=

6

125= 0, 048.

b) Gi B: bin c trong 3 bi ly ra c 1 bi trng

Ta cn tnh: P (A1|B) = P (A1.B)P (B)

Ta c: B = A1.A2.A3 +A1.A2.A3 + A1.A2.A3 nn

P (B) = P (A1.A2.A3) + P (A1.A2.A3) + P (A1.A2.A3)

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

=1

5 35 25+

4

5 25 25+

4

5 35 35=

58

125

Vy P (A1|B) = P (A1.A2.A3)P (B)

=P (A1).P (A2).P (A3)

P (B)=

6

125

58

125

=3

29

Bi 4. Mt tn hiu S c truyn t im A n im B. Tn hiu s cnhn ti B nu c hai cng tc I v II u ng. Gi s rng kh nng cngtc th nht v th hai ng, tng ng l 0,8 v 0,6. Cho bit hai cng tchot ng c lp nhau. Tnh xc sut:

a) Tn hiu c nhn ti B.b) Cng tc th I m, bit rng ti B khng nhn c tn hiu S.c) Cng tc th II m, bit rng ti B khng nhn c tn hiu S.

BI GIIGi Ai: bin c cng tc th i ng, i = 1, 2.Theo gi thit, {A1, A2} c lp.a) Gi B: bin c tn hiu c nhn ti B.= B = A1.A2 = P (B) = P (A1).P (A2) = 0, 8.0, 6 = 0, 48.b) Ta c: P (B) = 1 P (B) = 1 0, 48 = 0, 52.Ta cn tnh: P (A1|B) = P (A1.B)

P (B)=

P (A1)

P (B)=

0, 2

0, 52=

5

13 0, 3846

17

c) Ta cn tnh: P (A2|B) = P (A2.B)P (B)

=P (A2)

P (B)=

0, 4

0, 52=

10

13 0, 7692

Bi 5. Mt kin hng gm 20 chic tivi, trong c 16 chic t cht lngtt v 4 chic b li. Ln lt chn ngu nhin ra 3 chic tivi t kin hng ny.Tnh xc sut :

a) Chic tivi chn ln th 3 l tt, bit rng chic tivi chn ln th nht vth hai l tt.

b) Chic tivi chn ln th 3 b li, bit rng chic tivi chn ln th nht ltt v ln th hai b li.

c) Chic tivi chn ln th 3 b li, bit rng chic tivi chn ln th nht bli v ln th hai l tt.

d) Chic tivi chn ln th 3 b li, bit rng trong hai chic tivi chn ra lnth nht v th hai c mt chic tt v mt chic b li.

BI GIIGi Ai: bin c chic ti vi chn ln th i l tt, i = 1, 2, 3.Theo nh ngha xc sut c iu kin, ta c:a) P (A3|A1.A2) = 14

18=

7

9 0, 7778.

b) P (A3|A1.A2) = 318

0, 1667.c) P (A3|A1.A2) = 3

18 0, 1667.

d) Gi A: bin c trong hai chin ti vi chn ra ln th nht v th hai c mt

chic tt v mt chic b li. Ta cn tnh: P (A3|A) = P (A3.A)P (A)

V A = A1.A2 + A1.A2 nn

P (A) = P (A1).P (A2|A1) + P (A1).P (A2|A1) = 1620

419

+4

20 1619

=16

95

Ngoi ra,

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

= P (A1).P (A2|A1).P (A3|A1.A2) + P (A1).P (A2|A1).P (A3|A1.A2)=

16

20 419

318

+4

20 1619

318

=16

285

Vy P (A3|A) =16

285

16

95

=1

3 0, 3333.

18

1.4.3 Cng thc xc sut ton phn v cng thc Bayes

H bin c y v xung khc: H bin c {A1, A2, . . . , An} c gil h y v xung khc khi v ch khi ch c duy nht mt bin c trong hxy ra, ngha l: A1 + A2 + . . .+ An = v Ai.Aj = ,i 6= j.

Gi s {A1, A2, ..., An} l h y v xung khc. Khi , vi A l mtbin c bt k, ta c cng thc xc sut ton phn (y ):

P (A) = P (A1)P (A|A1) + P (A2)P (A|A2) + ...+ P (An)P (A|An) Cho bin c A bt k v {A1, A2, ..., An} l mt h y v xung khc. Khi, ta c cng thc Bayes:

P (Ai) =P (Ai)P (A|Ai)

P (A)=

P (Ai)P (A|Ai)n

k=1

P (Ak)P (A|Ak)

Bi 1. Mt m ng c s n ng bng na s n b. Xc sut n ngb bnh tim l 0,06 v xc sut n b b bnh tim l 0,036. Chn ngunhin 1 ngi t m ng, tnh xc sut ngi ny b bnh tim.

BI GIIGi A: bin c ngi c chn b bnh tim.

B: bin c ngi c chn l n b.C: bin c ngi c chn l n ng.

Ta c {B,C} l h y v xung khc v P (B) = 23;P (C) =

1

3p dng cng thc xc sut ton phn (CTXSTP):

P (A) = P (B).P (A|B) + P (C).P (A|C) = 23 0, 036 + 1

3 0, 06 = 0, 044

Bi 2. C hai hp ng bi: hp I c 5 bi trng v 7 bi en; hp II c 6 bi trngv 4 bi en. Ly ngu nhin 1 bi t hp I b sang hp II, ri t hp II ly ngunhin ra 1 bi. Tnh xc sut:

a) Bi ly t hp II l bi trng.b) Bi ly t hp I sang hp II l bi trng, bit rng bi ly ra t hp II l bi

trng.

BI GIIGi Ti: bin c bi ly t hp th i l bi trng, i = 1, 2.

19

a) Ta cn tnh P (T2). Ta c {T1, T 1} l h y v xung khc.p dng CTXSTP:

P (T2) = P (T1).P (T2|T1) + P (T 1).P (T2|T 1)=

5

12 711

+7

12 611

=7

12 0, 5833

b) Ta cn tnh:

P (T1|T2) = P (T1.T2)P (T2)

=P (T1).P (T2|T1)

P (T2)=

35

132

7

32

=5

11 0, 4545

Bi 3. C 3 hp: mi hp ng 5 vin bi, trong hp th i c i vin bi trng(i = 1, 3). Chn ngu nhin mt hp, ri t hp chn ly ngu nhin ngthi 3 vin bi.

a) Tm xc sut ly c 3 vin bi trng.b) Nu trong 3 bi ly ra c mt bi trng, tm xc sut vin bi trng

l ca hp th nht.

BI GIIa) Gi A l bin c ly c 3 bi trng. Ta cn tnh: P (A)

Gi Hi: bin c chn c hp th i, i = 1, 2, 3.Ta c P (T1) = P (T2) = P (T3) =

1

3v {T1, T2, T3} l h y v xung khc.

p dng CTXSTP:

P (A) = P (H1).P (A|H1) + P (H2).P (A|H2) + P (H3)P (A|H3)=

1

3 0 + 1

3 0 + 1

3 C

33

C35=

1

30

b) Gi B l bin c trong 3 bi ly ra c 1 bi trng.

Ta cn tnh: P (H1|B) = P (H1.B)P (B)

tnh P (B), ta p dng CTXSTP:

P (B) = P (H1).P (B|H1) + P (H2).P (B|H2) + P (H3)P (B|H3)=

1

3 C

11 .C

24

C35+

1

3 C

12 .C

23

C35+

1

3 C

13 .C

22

C35=

1

2= 0, 5

Do , P (H1|B) = P (H1.B)P (B)

=P (H1).P (B|H1)

P (B)=

1

3 C11 .C24

C35

0, 5=

2

5= 0, 4.

20

Bi 4. iu tra v gii tnh ca sinh vin mt trng hc, ngi ta thy rngc 65% nam v 35% n. Trong , t l hc sinh n v nam thch chi gametng ng l 20% v 25%. Chn ngu nhin mt sinh vin ca trng ny, tnhxc sut:

a) Sinh vin c chn thch chi game.b) Sinh vin c chn l nam, bit rng sinh vin ny thch chi game.

BI GIIa) Gi A l bin c chn c sinh vin thch chi game. Ta tnh P (A).

A1: bin c chn c sinh vin n.A2: bin c chn c sinh vin nam.

Ta c P (A1) = 35%; P (A2) = 65% v {A1, A2} l h y v xung khc.p dng CTXSTP:

P (A) = P (A1).P (A|A1) + P (A2).P (A|A2) = 35%.20% + 65%.25% = 0, 2325

b) Ta cn tnh: P (A2|A) = P (A2.A)P (A)

=P (A2).P (A|A2)

P (A)=

65%.25%

0, 2325= 0, 6989.

Bi 5. Mt nh my c hai phn xng. Sn lng ca phn xng I gp 3ln xn lng ca phn xng II. T l ph phm ca phn xng I v II lnlt l 3% v 5%. Chn ngu nhin 1 sn phm ca nh my, tnh xc sut:

a) Chn c sn phm tt do phn xng I sn xut.b) Chn c 1 ph phm.c) Gi s chn c sn phm tt, tnh xc sut sn phm ny do phn

xng I sn xut.

BI GIIGi A l bin c chn c sn phm tt.

Ai: bin c chn c sn phm do phn xng th i sn xut, i = 1, 2.= P (A1) = 3

4;P (A2) =

1

4

a) Ta cn tnh: P (A.A1) = P (A1).P (A|A1) = 34 (1 3%) = 0, 7275

b) Ta cn tnh: P (A). Ta c: {A1, A2} l h y v xung khc.p dng CTXSTP:

P (A) = P (A1).P (A|A1) + P (A2).P (A|A2) = 34 3% + 1

4 5% = 0, 035

c) Ta cn tnh: P (A1|A) = P (A1.A)P (A)

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

P (A)

21

Ta tnh P (A) bng cch p dng CTXSTP:

P (A) = P (A1).P (A|A1) + P (A2).P (A|A2) = 34 97% + 1

4 95% = 0, 965

Khi , P (A1|A) =3

4 97%0, 965

=291

396 0, 7539

Ch . Ta c th tnh P (A) bng cch p dng cng thc

P (A) = 1 P (A) = 1 0, 035 = 0, 965

1.4.4 Cng thc Bernoulli

Php th Bernoulli l php th m trong ta ch quan tm n hai bin cA v A vi P (A) = p. Xt lc Bernoulli gm:

n php th c lp (trong mi php th ch xy ra mt trong hai trnghp: A xy ra hoc A khng xy ra).

Xc sut bin c A xy ra trong mi php th l P (A) = p v P (A) =1 p = q.Khi , xc sut trong n php th, bin c A xy ra k ln c tnh theocng thc Bernoulli:

P(A) = Pn(k) = Ckn.p

k.qnk, k = 0, n

Bi 1. Gieo 100 ht u tng. Xc sut ny mm ca mi ht l 0,9. Tnhxc sut trong 100 ht:

a) C 85 ht ny mm.b) C t nht 99 ht ny mm.

BI GIITa c lc Bernoulli vi:- S php th: n = 100- Xc sut 1 ht u ny mm: p = 0, 9.Ta c cng thc Bernoulli :

P100(k) = Ck100(0, 9)

k(0, 1)100k , (k = 0, 100)

a) Gi A l bin c c 85 ht ny mm = k = 85P (A) = P100(85) = C

85100(0, 9)

85(0, 1)15 = 0, 0327

b) Gi B l bin c c t nht 99 ht ny mm. Ta cn tnh P (B).Gi C l bin c c 99 ht ny mm v D l bin c c 100 ht ny mm

22

Ta c B = C +D (C,D xung khc)= P (B) = P (C) + P (D) = P100(99) + P100(100)

= C99100(0, 9)99(0, 1)1 + C100100 (0, 9)

100(0, 1)0 = 0, 0003

Bi 2. Mt l hng cha rt nhiu sn phm vi t l ph phm l 0,02. Lyngu nhin 150 sn phm kim tra. Tnh xc sut ly c t nht 1 phphm.

BI GIITheo gi thit, ta c lc Bernoulli vi:- S php th: n = 150- Xc sut ly c 1 ph phm: p = 0, 02.Cng thc Bernoulli :

P150(k) = Ck150(0, 02)

k(0, 98)150k , (k = 0, 150)

Gi A l bin c ly c t nht mt ph phmB: bin c khng ly c ph phm no

Ta c: A = B = P (A) = 1 P (B)Theo cng thc Bernoulli : P (B) = P150(0) = C0150(0, 02)0(0, 98)150 = 0, 0483Vy xc sut ly c t nht mt ph phm:

P (A) = 1 P (B) = 1 C0150(0, 02)0(0, 98)150 = 0, 9517

Bi 3. Mt x th bn 10 vin n vo 1 tm bia mt cch c lp. Xc sutbn trng bia ca mi vin n bng nhau v bng 0,6. Tm xc sut c t 5n 7 vin n trng ch.

BI GIITheo gi thit, ta c lc Bernoulli vi:- S php th: n = 10- Xc sut 1 vin n bn trng ch: p = 0, 6.Khi , cng thc Bernoulli :

P10(k) = Ck10(0, 6)

k(0, 4)10k , (k = 0, 10)

Gi A: bin c c t 5 n 7 vin n trng ch. Ta cn tnh P (A)Gi Ai: bin c c i vin n bn trng ch, i = 0, 10Ta c A = A5 + A6 +A7 ({A5, A6, A7} xung khc)= P (A) = P (A5) + P (A6) + P (A7) = P10(5) + P10(6) + P10(7)

= C510(0, 6)5(0, 4)5 + C610(0, 6)

6(0, 4)4 +C710(0, 6)7(0, 4)3 = 0, 6665

Bi 4. Ngn hng thi mn L c 500 cu hi. Thy Bnh chn ngu nhin20 cu hi lm thi cui k. Mi cu hi c 4 phng n tr li trong

23

ch c 1 phng n ng. Mi cu tr li ng t 0,5 im. Bn n lm bithi bng cch chn h ha mt phng n cho mi cu hi. Tnh xc sut bn n t 8 im.

BI GIIGi A: bin c bn n t 8 im bn n tr li ng 16 cu hi.Trc ht ta nhn xt rng: 500 cu hi trong ngn hng thi khng nhhng n kt qu bi ton.Theo bi, ta c lc Bernoulli vi:- S php th: n = 20- Xc sut bn n tr li ng 1 cu hi: p = 0, 25.Cng thc Bernoulli : P20(k) = Ck20(0, 25)k(0, 75)20k vi k = 0, 20Cho ta P (A) = P20(16) = C1620 (0, 25)16(0, 75)4 = 0, 357.106

Bi 5. Trong mt hp c 9 vin bi, trong c 3 bi cn li l bi xanh. Lnlt ly ngu nhin c hon li 3 bi. Tm xc sut :

a) Ly c 2 bi xanh.b) Ly c t nht 1 bi .

BI GIITa c lc Bernoulli vi:- S php th: n = 3- Xc sut ly c 1 bi xanh trong mi php th: p =

6

9=

2

3

Ta c cng thc Bernoulli : P3(k) = Ck3(23

)k(13

)3k, vi k = 0, 3.

a) Gi A: bin c ly c 2 bi xanh = k = 2.Vy P (A) = P3(2) = C23

(2

3

)2(1

3

)1= 4

9 0, 4444

b) Gi B: bin c ly c t nht 1 bi = B: bin c khng ly c bi no ly c 3 bi xanh.Tng t, ta c: P (B) = P3(3) = C33

(2

3

)3(1

3

)0= 8

27 0, 2963

Vy P (B) = 1 P (B) = 1 C33(23

)3(13

)0= 0, 7037

1.5 Bi tp tng hp

Bi 1. Cho hai bin c A, B v P (A) = 0, 2;P (B) = 0, 5;P (A + B) = 0, 6.Tnh P (AB).

24

BI GIITa c cng thc cng xc sut: P (A+B) = P (A) + P (B) P (AB)= P (AB) = P (A) + P (B) P (AB) = 0, 2 + 0, 5 0, 6 = 0, 1.

Bi 2. Cho ba bin c A, B, C v P (A) = 0, 4;P (B) = 0, 2;P (C) = 0, 3. Tnhxc sut P (A+B + C), bit rng A,B,C c lp nhau.

BI GIIp dng cng thc cng xc sut:

P (A+B+C) = P (A)+P (B)+P (C)P (AB)P (AC)P (CA)+P (ABC)

Theo gi thit, A,B,C c lp nhau nn:P (AB) = P (A).P (B) = 0, 4.0, 2 = 0, 08P (BC) = P (B).P (C) = 0, 2.0, 3 = 0, 06P (CA) = P (C).P (A) = 0, 3.0, 4 = 0, 28P (ABC) = P (A).P (B).P (C) = 0, 4.0, 2.0, 3 = 0, 024

Do , P (A+B+C) = 0, 4+0, 2+ 0, 3 0, 08 0, 06 0, 12+0, 024 = 0, 664.

Bi 3. Rt ngu nhin ng thi 2 l bi t b bi 52 l. Tnh xc sut:

a) Rt c 2 l bi C.b) Rt c 2 l bi R mu en.c) Rt c 2 l bi C, bit rng hai l ny mu .

BI GIIRt ngu nhin ng thi 2 l bi t b bi 52 l, c n = C252 = 1326 cch.a) Gi A: bin c rt c 2 l bi C = C m = C213 = 78 cch.Suy ra, P (A) = m

n=

C213C252

=1

17 0, 05882.

b) Gi B: bin c rt c 2 l bi R mu en th P (B) = 0 v khng c lbi R no mu en.c) Gi C: bin c rt c 2 l bi mu th P (C) = C

226

C252=

25

102 0, 2451.

Ta cn tnh: P (A|C) = P (AC)P (C)

=P (A)

P (C)=

C213C226

=6

25= 0, 24.

Bi 4. Trong mt xng c 3 my lm vic. Trong mt ca, my th nht cth cn sa cha vi xc sut 0,12; my th hai vi xc sut 0,15 v my thba vi xc sut 0,18. Tm xc sut sao cho trong mt ca lm vic c t nhtmt my khng cn sa cha.

25

BI GIIGi A: bin c c t nht mt my khng cn sa cha. Ta cn tnh: P (A).Gi B: bin c c 3 my cn sa cha th A = B = P (A) = 1 P (B)Gi Ai: bin c my th i cn sa cha, i = 1, 3, th {A1, A2, A3} l h c lpTa c: B = A1.A2.A3= P (B) = P (A1).P (A2).P (A3) = 0, 12.0, 15.0, 18 = 0, 00324Vy P (A) = 1 P (B) = 1 0, 00324 = 0, 99676.

Bi 5. Cho hai hp: hp th nht cha 30 bi trng, 7 bi ; hp th hai cha10 bi trng, 6 bi . Ly ngu nhin t mi hp ra 1 bi. Tnh xc sut haibi ly ra cng mu.

BI GIIGi A: bin c hai bi ly ra cng mu. Ta cn tnh: P (A).Gi Ai: bin c bi ly ra t hp th i l bi , i = 1, 2.Ta c: A = A1.A2 +A1.A2 (V {A1, A2} l h c lp)= P (A) = P (A1.A2) + P (A1.A2) = P (A1).P (A2) + P (A1).P (A2)Do vy, P (A) = 7

37 616

+30

37 1016

=171

296 0, 5777.

Bi 6. Ba ngi chi bng r, mi ngi nm mt qu. Xc sut nm trng rca mi ngi ln lt l 0,5; 0,6; 0,7. Tnh xc sut :

a) C 3 ngi u nm trng r.b) C t nht mt ngi nm trng r.c) C t nht mt ngi nm khng trng r.d) C ng 2 ngi nm trng r.e) Ngi th 3 nm khng trng r, bit rng c hai ngi nm trng r.

BI GIIGi Ai: bin c ngi th i nm trng r, i = 1, 2, 3.Ta c: {A1, A2, A3} c lp.a) Gi A: bin c c 3 ngi u nm trng rTa c: A = A1.A2.A3= P (A) = P (A1).P (A2).P (A3) = 0, 5.0, 6.0, 7 = 0, 21.b) Gi B: bin c c t nht mt ngi nm trng r

C: bin c c 3 ngi u nm khng trng rTa c: B = C = P (B) = 1 P (C).Li c: C = A1.A2.A3 = P (C) = P (A1).P (A2).P (A3) = 0, 5.0, 4.0, 3 = 0, 06Vy P (B) = 1 P (C) = 1 0, 06 = 0, 94.

26

c) Gi D: bin c c t nht mt ngi nm khng trng r th D = A= P (D) = 1 P (A) = 1 0, 21 = 0, 79.d) Gi E: bin c c ng 2 ngi nm trng r.Ta c: E = A1.A2.A3 + A1.A2.A3 +A1.A2.A3 nn

P (E) = P (A1.A2.A3) + P (A1.A2.A3) + P (A1.A2.A3)

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

= 0, 5.0, 6.0, 3 + 0, 5.0, 4.0, 7 + 0, 5.0, 6.0, 7 = 0, 44

e) Ta cn tnh: P (A3|E) = P (A3.E)P (E)

=P (A1.A2.A3)

P (E)=

P (A1).P (A2).P (A3)

P (E)

Cho ta, P (A3|E) = 0, 5.0, 6.0, 30, 44

=9

44 0, 2045.

Bi 7. Hai bn Bnh v Yn cng d thi mn xc sut thng k mt cch clp. Kh nng Yn thi t mn ny l 0,6 v xc sut c t nht mttrong hai bn thi t l 0,9. Tnh xc sut :

a) Bn Bnh thi t.b) C hai bn u thi t.c) C t nht mt bn thi hng.d) Bn Yn thi t, bit rng ch c mt trong hai bn thi t.

BI GIIGi B: bin c Bnh thi t mn xc sut thng k

C: bin c Yn thi t mn xc sut thng kTheo gi thit: P (C) = 0, 6 v P (B + C) = 0, 9 v {B,C} l h c lp.a) Ta cn tnh: P (B). p dng cng thc cng v cng thc nhn xc sut:

P (B + C) = P (B) + P (C) P (BC) = P (B) + P (C) P (B).P (C)

Do , P (B) = P (B + C) P (C)1 P (C) =

0, 9 0, 61 0, 6 = 0, 75.

b) Gi A: bin c c hai bn u thi t mn xc sut thng kTa c: A = B.C = P (A) = P (B).P (C) = 0, 6.0, 75 = 0, 45.c) Gi D: bin c c t nht mt bn thi hng mn xc sut thng kV D = A nn P (D) = 1 P (A) = 1 0, 45 = 0, 55d) Gi E: bin c ch c mt trong hai bn thi t.

Ta cn tnh: P (C|E) = P (C.E)P (E)

Ta i tnh P (E). Do E = B.C +B.C nn ta c:

P (E) = P (B.C) + P (B.C) = P (B).P (C) + P (B).P (C)

= 0, 75.0, 4 + 0, 25.0, 6 = 0, 45

27

Vy P (C|E) = P (B).P (C)P (E)

=0, 25.0, 6

0, 45=

1

3 0, 3333.

Bi 8. Nam np h s d thi vo trng i hc A v trng cao ng B. Khnng Nam thi u vo trng i hc l 0,6 v trng cao ng l 0,8; khnng Nam khng u vo t nht mt trong hai trng l 0,3. Tnh xc sut Nam thi u vo t nht mt trong hai trng A v B.

BI GIIGi A,B: ln lt l bin c Nam thi u vo trng i hc v thi u votrng cao ng. Ta cn tnh: P (A+B) = P (A) + P (B) P (AB)Theo gi thit, P (A) = 0, 6;P (B) = 0, 8V P (A+B) = 0, 3 P (AB) = 0, 3 P (AB) = 0, 7Vy P (A+B) = 0.6 + 0, 8 0, 7 = 0, 7.

Bi 9. Mt l hng c 3 kin sn phm: kin th nht c 8 sn phm tt v2 ph phm; kin th hai c 10 sn phm tt v 3 ph phm; kin th ba c15 sn phm tt v 5 ph phm. Mt khch hng kim tra bng cch ly ngunhin t mi kin ra 2 sn phm. Ngi ny s mua kin hng nu c 2 snphm c ly ra u tt. Tnh xc sut c t nht mt kin hng c mua.

BI GIIGi A: bin c c t nht mt kin hng c mua. Ta cn tnh: P (A).Gi Ai: bin c kin hng th i c mua khch hng chn c 2 sn phmtt t kin th i, i = 1, 2, 3

B: bin c khng c kin hng no c mua.Ta c A = B = P (A) = 1 P (B).V B = A1.A2.A3 v {A1, A2, A3} l h c lp ton phn nn

P (B) = P (A1).P (A2).P (A3) = (1 P (A1)).(1 P (A2)).(1 P (A3))= (1 C

28

C210) (1 C

210

C213) (1 C

215

C220)

= (1 2845

) (1 1526

) (1 2138

) =3179

44460 0, 0715

Vy P (A) = 1 P (B) = 1 317944460

=41281

44460 0, 9285.

Bi 10. Mt ngi bun bn bt ng sn ang mun bn mt mnh t ln.ng ta d on rng: nu nn kinh t tip tc pht trin th kh nng mnht c mua l 80%; ngc li nu nn kinh t ngng pht trin th ng tach bn c mnh t vi xc sut 40%. Theo d bo ca mt chuyn giakinh t: xc sut nn kinh t tip tc tng trng l 65%. Tnh xc sut

28

ngi ny bn c mnh t.

BI GIIGi A: bin c ngi ny bn c mnh t. Ta cn tnh: P (A).Gi B: bin c nn kinh tt tip tc pht trin.Theo gi thit, P (B) = 65% v h {B,B} y , xung khc.p dng CTXSTP: P (A) = P (B).P (A|B) + P (B).P (A|B)= P (A) = 65%.80% + (1 65%).40% = 33

50= 0, 66.

Bi 11. Mt lp c s hc sinh nam bng 3 ln s hc sinh n. T l hc sinhn gii ton l 30% v t l hc sinh nam gii ton l 40%. Chn ngu nhin1 hc sinh trong lp ny. Tnh xc sut:

a) Hc sinh ny gii ton.b) Hc sinh ny l nam, bit rng hc sinh ny gii ton.

BI GIIGi A: bin c chn c hc sinh n = P (A) = 1

4

B: bin c chn c hc sinh nam = P (B) = 34

Ta c {A,B} l h y v xung khc.a) Gi G: bin c chn c hc sinh gii. p dng CTXSTP:

P (G) = P (A).P (G|A) + P (B).P (G|B) = 14 30% + 3

4 40% = 3

8= 0, 375

b) Ta cn tnh: P (B|G) = P (BG)P (G)

=P (B).P (G|B)

P (G)

Cho ta P (B|G) =3

4 40%0, 375

=4

5= 0, 8.

Bi 12. Mt phn xng c s lng nam cng nhn gp 4 ln s lng ncng nhn. T l cng nhn tt nghip THPT i vi n l 15%, nam l 25%.Chn ngu nhin 1 cng nhn ca phn xng ny. Tnh xc sut :

a) Chn c nam cng nhn tt nghip THPT.b) Chn c mt cng nhn tt nghip THPT.c) Chn c mt cng nhn n, bit rng ngi ny tt nghip THPT.

BI GIIGi A: bin c chn c cng nhn tt nghip THPT

B: bin c chn c cng nhn n = P (B) = 15

29

C: bin c chn c cng nhn nam = P (C) = 45

Ta c {B,C} l h y v xung khc.a) Gi D: bin c chn c cng nhn nam tt nghip THPTDo D = C.A nn P (D) = P (C).P (A|C) = 4

5 25% = 1

5= 0, 2

b) Ta cn tnh: P(A). p dng CTXSTP:

P (A) = P (B).P (A|B) + P (C).P (A|C) = 15 15% + 4

5 25% = 23

100= 0, 23

c) Ta cn tnh: P (B|A) = P (BA)P (A)

=P (B).P (A|B)

P (A)

Cho ta P (B|A) =1

5 15%0, 23

=3

23 0, 1304.

Bi 13. Mt nh my c 3 phn xng cng sn xut mt loi sn phm. Ssn phm ca phn xng th nht, th hai, th ba ln lt chim 25%, 25%v 50% tng sn lng ca nh my. T l ph phm ca tng phn xngtng ng l 1%, 2,5% v 4,5%. Ly ngu nhin 1 sn phm t nh my ny.

a) Tm xc sut ly c sn phm tt. Nu ngha thc t ca kt quny.

b) Nu ly c sn phm tt th theo bn, sn phm c kh nng l dophn xng no sn xut nht?

BI GIIGi Ai: bin c ly c sn phm do phn xng th i sn xut, i = 1, 2, 3= P (A1) = 25%;P (A2) = 25%;P (A3) = 50% v h {A1, A2, A3} y ,xung khc.a) Gi A: bin c ly c sn phm tt. p dng CTXSTP:

P (A) = P (A1).P (A|A1) + P (A2).P (A|A2) + P (A3).P (A|A3)= 25%.(1 1%) + 25%.(1 2, 5%) + 50%.(1 4, 5%) = 31

32= 0, 96875

T l sn phm tt ca nh my l 96,875%, chng t nh my hot ng tt.b) Ta tnh:

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

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

P (A)=

25%.99%

0, 96875=

198

775 25, 55%

P (A2|A) = P (A2.A)P (A)

=P (A2).P (A|A2)

P (A)=

25%.97, 5%

0, 96875=

39

155 25, 16%

P (A3|A) = P (A3.A)P (A)

=P (A3).P (A|A3)

P (A)=

50%.95, 5%

0, 96875=

198

775 49, 29%

30

So snh cc xc sut trn, ta c th kt lun nu ly c mt sn phm ttth sn phm c kh nng do phn xng 3 sn xut nht.

Bi 14. Trong mt k thi, mi sinh vin phi thi 2 mn. Bn Nga c lngrng: xc sut t mn th nht l 0,8. Nu t mn th nht th xc sutt mn th hai l 0,6; nu khng t mn th nht th xc sut t mn thhai l 0,3. Tnh xc sut Nga:

a) t mn th hai.b) t i mn, i = 0, 1, 2.c) t t nht 1 mn.d) t mn th 2, bit rng Nga t 1 mn.e) t mn th 2, bit rng Nga t t nht 1 mn.

BI GIIGi Ai l bin c Nga thi t mn th i, i = 1, 2Theo gi thit, P (A1) = 0, 8;P (A2|A1) = 0, 6;P (A2|A1) = 0, 3a) Ta cn tnh: P (A2).Ta c {A1, A1} l h y v xung khc. p dng CTXSTP:P (A2) = P (A1).P (A2|A1)+P (A1).P (A2|A1) = 0, 8.0, 6+ (1 0, 8).0, 3 = 0, 54b) Gi Bi l bin c Nga thi t i mn, i = 0, 1, 2 B0 = A1.A2 = P (B0) = P (A1).P (A2|A1) = 0, 2.(1 0, 3) = 0, 14.

B1 = A1.A2 + A1.A2 cho ta

P (B1) = P (A1.A2) + P (A1.A2) = P (A1).P (A2|A1) + P (A1).P (A2|A1)= 0, 8.(1 0, 6) + (1 0, 8).0, 3 = 0, 38

B2 = A1.A2 = P (B2) = P (A1).P (A2|A1) = 0, 8.0, 6 = 0, 48.c) Gi C l bin c Nga t t nht mt mn = C = B0= P (C) = 1 P (B0) = 1 0, 14 = 0, 86.d) Ta cn tnh: P (A2|B1) = P (A2.B1)

P (B1)=

P (A2).P (B1|A2)P (B1)

= P (A2|B1) = P (A1.A2)P (B1)

=P (A1).P (A2|A1)

P (B1)=

0, 2.0, 3

0, 38=

3

19 0, 1579.

e) Ta cn tnh: P (A2|C) = P (A2.C)P (C)

=P (A2)

P (C)=

0, 54

0, 86=

27

43 0, 6279.

Bi 15. Mt ngi bn ln lt 3 pht n vo mt mc tiu mt cch clp. Xc sut trng mc tiu mi ln bn ln lt l 0,55; 0,6; 0,7. Xc sutmc tiu b h khi trng 1, 2, 3 pht n ln lt l 0,2; 0,4 ; 0,8.

31

a) Tnh xc sut c i pht n trng mc tiu, i = 0, 1, 2, 3.b) Tnh xc sut c nhiu nht 2 pht n trng mc tiu.c) Tnh xc sut mc tiu b h.d) Gi s mc tiu b h, tnh xc sut c 1 pht n trng mc tiu.

BI GIIGi Ai: bin c pht n th i trng mc tiu, i = 1, 2, 3Theo gi thit, P (A1) = 0, 55;P (A2) = 0, 6;P (A3) = 0, 7 v h {A1, A2, A3}y , xung khc.a) Gi Bi: bin c c i pht n trng mc tiu, i = 0, 1, 2, 3. B0 = A1.A2A3= P (B0) = P (A1).P (A2).P (A3) = (1 0, 55).(1 0, 6).(1 0, 7) = 0, 054. Li c B1 = A1.A2.A3 + A1.A2.A3 +A1.A2.A3 nn

P (B1) = P (A1).P (A2).P (A3) + P (A1).P (A2).P (A3) + P (A1).P (A2).P (A3)

= 0, 55.0, 4.0, 3 + 0, 45.0, 6.0, 3 + 0, 45.0, 4.0, 7 = 0, 273

Tng t, B2 = A1.A2.A3 + A1.A2.A3 + A1.A2.A3 nn

P (B2) = P (A1).P (A2).P (A3) + P (A1).P (A2).P (A3) + P (A1).P (A2).P (A3)

= 0, 55.0, 6.0, 3 + 0, 55.0, 4.0, 7 + 0, 45.0, 6.0, 7 = 0, 442

B3 = A1.A2.A3 = P (B3) = P (A1).P (A2).P (A3) = 0, 55.0, 6.0, 7 = 0, 231.

Ch . H {B0, B1, B2, B3} l h y v xung khc nnP (B0) + P (B1) + P (B2) + P (B3) = 1

Do , ta c th tnh P (B2) = 1 P (B0) P (B1) P (B3).

b) Gi A: bin c c nhiu nht 2 pht n trng mc tiu. Ta tnh P (A).Cch 1: A = B3 = P (A) = 1 P (B3) = 1 0, 231 = 0, 769.Cch 2: A = B0 +B1 +B2 v h {B0, B1, B2} l h xung khc nn:

P (A) = P (B0) + P (B1) + P (B2) = 0, 054 + 0, 273 + 0, 442 = 0, 769

c) Gi B: bin c mc tiu b h. p dng CTXSTP:

P (B) = P (B0).P (B|B0) + P (B1).P (B|B1) + P (B2).P (B|B2) + P (B3).P (B|B3)= 0, 054.0 + 0, 273.0, 2 + 0, 442.0, 4 + 0, 231.0, 8 = 0, 4162

d) Ta cn tnh:

P (B1|B) = P (B1.B)P (B)

=P (B1).P (B|B1)

P (B)=

0, 273.0, 2

0, 4162=

273

2081 0, 1312

32

Bi 16. C hai chung th: chung I c 5 th en v 10 th trng; chung IIc 8 th en v 15 th trng. Quan st thy t chung I c 1 con th chy sangchung II; sau , t chung II c 1 con th chy ra ngoi. Tnh xc sut :

a) Con th chy t chung I sang chung II l con th trng.b) Con th chy ra t chung II l con th trng.c) Con th chy t chung I sang chung II l con th trng, bit rng con

th chy t chung II ra ngoi l con th trng.d) Con th chy t chung I sang chung II l th trng v con th chy t

chung II ra ngoi cng l th trng.

BI GIIa) Gi A: bin c con th chy t chung I sang chung II l con th trng= P (A) = 10

15=

2

3b) Gi B: bin c con th chy t chung II ra ngoi l con th trng.Ta c {A,A} l h y v xung khc. p dng CTXSTP:

P (B) = P (A).P (B|A) + P (A).P (B|A) = 23 1624

+1

3 1524

=47

72 0, 6528

c) Ta cn tnh: P (A|B) = P (AB)P (B)

=P (A).P (B|A)

P (B)=

2

3 1624

47

72

=32

47 0, 6808.

d) Ta cn tnh: P (A.B) = P (A).P (B|A) = 23 1624

=4

9 0, 4444.

Bi 17. C hai chung th: chung I c 8 th en v 12 th trng; chung IIc 6 th en v 15 th trng. Quan st thy t chung I c 1 con th chy sangchung II; sau , t chung II c 2 con th chy ra ngoi. Tnh xc sut :

a) Hai con th chy t chung II ra ngoi l hai con th trng.b) Trong 2 con th chy ra t chung 2 c 1 con th trng.c) Hai con th chy t chung II ra ngoi l hai con th en.

BI GIIGi A: bin c con th chy t chung I sang chung II l con th trng= P (A) = 12

20=

3

5;P (A) =

8

20=

2

5v {A,A} l h y v xung khc.

a) Gi B: bin c hai con th chy ra t chung II l 2 con th trng.p dng CTXSTP:

P (B) = P (A).P (B|A) +P (A).P (B|A) = 35 C

216

C222+

2

5 C

215

C222=

38

77 0, 4935.

33

b) Gi C: bin c hai con th chy ra t chung II l 2 con th trng.p dng CTXSTP:

P (C) = P (A).P (C|A) + P (A).P (C|A)=

3

5 C

116.C

12

C222+

2

5 C

115.C

17

C222=

166

385 0, 4312

c) Gi D: bin c hai con th chy ra t chung II l 2 con th trng.p dng CTXSTP:

P (D) = P (A).P (D|A) + P (A).P (D|A) = 35 C

26

C222+

2

5 C

27

C222=

29

385 0, 0753

Ch . Gi Bi l bin c trong 2 con th chy ra t chung II c i con thtrng, i = 0, 1, 2 v p dng CTXSTP th ta c cng thc tng qut:

P (Bi) = P (A).P (Bi|A) + P (A).P (Bi|A)

=3

5 C

i16.C

2i2

C222+

2

5 C

i15.C

2i7

C222, vi i = 0, 1, 2.

Bi 18. C hai chung g: chung I c 12 con g mi v 8 con g trng; chungII c 15 con g mi v 10 con g trng. Quan st thy c 2 con g chy tchung I sang chung II; sau , c 1 con g chy t chung II ra ngoi. Tnhxc sut :

a) Hai con g chy t chung I sang chung II l 2 con g mi.b) Trong hai con g chy t chung I sang chung II c 1 con g trng.c) Hai con g chy t chung I sang chung II l 2 con g trng.d) Con g chy t chung II ra ngoi l con g trng.

BI GIIa) Gi A: bin c hai con g chy t chung I sang chung II l 2 con g mi.

= P (A) = C212

C220=

33

95 0, 3474.

b) Gi B: bin c trong hai con g chy t chung I sang chung II c 1 cong trng.

= P (B) = C18 .C

112

C220=

48

95 0, 5052.

c) Gi C: bin c hai con g chy t chung I sang chung II l 2 con g trng.

= P (C) = C28

C220=

14

95 0, 1474.

34

Ch . Gi Ai l bin c trong 2 con g chy t chung I sang chung II c i

con g trng, i = 0, 1, 2 th P (Ai) =Ci8.C

2i12

C220, vi i = 0, 1, 2.

d) Gi D: bin c con g chy ra t chung II l con g trng.Ta c {A,B,C} l h y v xung khc. p dng CTXSTP:

P (D) = P (A).P (D|A) + P (B).P (D|B) + P (C).P (D|C)=

C212C220

1027

+C18 .C

112

C220 1127

+C28C220

1227

=2

5= 0, 4

Bi 19. C hai chung g: chung I c 12 con g mi v 8 con g trng;chung II c 15 con g mi v 9 con g trng. Quan st thy c 2 con g chyt chung I sang chung II; sau , c 2 con g chy t chung II ra ngoi. Tnhxc sut trong 2 con g chy t chung II ra ngoi c i con g mi, i = 0, 1, 2.

BI GIIGi Ai l bin c trong 2 con g chy t chung I sang chung II c i con g

mi, i = 0, 1, 2 th P (Ai) =Ci12.C

2i8

C220, vi i = 0, 1, 2.

Ta c {A0, A1, A2} l h y v xung khc. Gi A: bin c hai con g chy ra t chung II l 2 con g trng.p dng CTXSTP:

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

C012.C28

C220 C

211

C226+C112.C

18

C220 C

210

C226+C212.C

08

C220 C

29

C226=

4118

30875 0, 1334

Gi B: bin c trong hai con g chy ra t chung II c 1 con g mi.p dng CTXSTP:

P (B) = P (A0).P (B|A0) + P (A1).P (B|A1) + P (A2).P (B|A2)=

C012.C28

C220 C

115C

111

C226+C112.C

18

C220 C

116C

110

C226+C212.C

08

C220 C

117.C

19

C226=

15039

30875

Gi C: bin c trong hai con g chy ra t chung II c 1 con g mi.p dng CTXSTP:

P (C) = P (A0).P (C|A0) + P (A1).P (C|A1) + P (A2).P (C|A2)=

C012.C28

C220 C

215

C226+C112.C

18

C220 C

216

C226+C212.C

08

C220 C

217

C226=

11718

30875 0, 3795

35

Ch . Gi Bi l bin c trong 2 con g chy ra t chung II c i con g mi,i = 0, 1, 2. Ta c cng thc tng qut:

P (Bi) = P (A0).P (Bi|A0) + P (A1).P (Bi|A1) + P (A2).P (Bi|A2)

=C012.C

28

C220 C

i15C

2i11

C226+C112.C

18

C220 C

i16C

2i10

C226+C212.C

08

C220 C

i17.C

2i9

C226

Bi 20. Trong mt hp c 12 bng n, trong c 3 bng hng. Ln lt lyngu nhin c hon li 3 bng dng. Tm xc sut :

a) C 3 bng ly ra u hng.b) C 3 bng ly ra u khng hng.c) C t nht mt bng khng hng.d) Ch c bng th 2 hng.

BI GIITa c lc Bernoulli vi:- S php th: n = 3- Xc sut ly c 1 bng hng trong mi php th: p =

3

12= 0, 25

Khi , Pn(k) = Ck3 .(0, 25)k .(0, 75)3k , vi k = 0, 3a) Gi A: bin c ly c 3 bng hng. p dng cng thc Bernoulli :

P (A) = P3(3) = C33 .(0, 25)

3.(0, 75)0 =1

64= 0, 015625

b) Gi B: bin c ly c 3 bng khng hng. Tng t nh trn, ta c:P (B) = P3(0) = C

03 .(0, 25)

0.(0, 75)3 =27

64= 0, 421875

c) Gi C: bin c ly c t nht mt bng khng hng = C = AVy P (C) = 1 P (A) = 1 1

64=

63

64= 0, 984375.

d) Gi D: bin c ch c bng ly ra ln th 2 l hng. Ta cn tnh: P (D).Gi Ai: bin c bng ly ra ln th i l bng khng hng, i = 1, 2, 3.

V ly theo phng thc c hon li nn {A1, A2, A3} l h c lp ton phnv P (A1) = P (A2) = P (A3) =

9

12= 0, 75

Ta c: D = A1.A2.A3= P (D) = P (A1).P (A2).P (A3) = 0, 75.(1 0, 75).0, 75 = 0, 140625.

Bi 21. Mt l hng c 200 sn phm vi t l ph phm l 6%. Mt khchhng (khng bit t l ph phm trong l hng l bao nhiu) kim tra l hngv giao c: ngi ny rt ngu nhin 7 sn phm kim tra, nu c t nht1 ph phm th s l hng s b loi. Tnh xc sut l hng ny c mua.

36

BI GIIGi A: bin c l hng ny c mua trong 7 sn phm ly ra kim trakhng c ph phm no. Cn tnh P (A).Theo gi thit, ta c lc Bernoulli vi:- S php th: n = 7- Xc sut ly c 1 ph phm: p = 0, 06Ta c cng thc Bernoulli : Pn(k) = Ck7 .(0, 06)k .(0, 94)7k , vi k = 0, 7Cho ta P (A) = P7(0) = C07 .(0, 06)0.(0, 94)7 = 0, 6485

1.6 Bi tp t gii

Bi 1. Gieo ng thi hai con xc xc cn i v ng cht. Tm xc sut :

a) Tng s chm mt trn hai con xc xc bng 8.b) Hiu s chm mt trn hai con xc xc c gi tr tuyt i bng 2.c) S chm mt trn hai con xc xc bng nhau.

Bi 2. Cho hai bin c A, B v P (A) = 0, 4;P (B) = 0, 6;P (A + B) = 0, 7.Tnh P (AB).

Bi 3. Mt cng ty s dng hai hnh thc qung co: qung co trn bo vqung cao trn tivi. Qua kho st thng tin nhng ngi n mua sn phmca cng ty th thy rng: c 35% khch hng bit c thng tin qung cotrn tivi; 20% khch hng bit c thng tin qung co trn bo v 10%khch hng bit c thng tin qung co qua c hai hnh thc trn. Chnngu nhin mt khch hng n mua sn phm. Tnh xc sut ngi nybit c thng tin qung co ca cng ty.

Bi 4. Cho hai hp ng bi: hp th nht c 6 bi v 4 bi xanh; hp th haic 8 bi v 2 bi xanh. Chn ngu nhin t mi hp ra 1 bi. Tnh xc sut :

a) Trong 2 bi ly ra c 1 bi xanh.b) Trong 2 bi ly ra c t nht mt bi xanh.c) Hai bi ly ra cng mu.d) Vin bi l ca hp th hai, bit rng trong 2 bi ly ra c 1 bi xanh.

Bi 5. Bn ba vin n vo mt tm bia mt cch c lp. Xc sut trngch ca vin n th nht, th hai v th ba ln lt l 0,4; 0,5; 0,7.

a) Tnh xc sut trong 3 vin n c mt vin trng ch.b) Tnh xc sut c t nht mt vin trng ch.

37

c) Tnh xc sut c khng qu mt vin trng ch.

Bi 6. Hai sinh vin Vui v Phc cng d thi mn xc sut thng k mt cchc lp. Xc sut c 1 sinh vin thi t l 0,46. Kh nng Phc thi tl 0,6. Tnh xc sut :

a) Vui thi t.b) C hai sinh vin cng thi t.c) C t nht mt sinh vin thi t.d) Vui thi t, bit rng c mt sinh vin thi t.

Bi 7. Mt nh my sn xut bng n. T l bng hng ca my I, II, IIIln lt l 2%, 3%, 5%. Mt l hng ca nh my ny c 48% sn phm camy I, 22% sn phm ca my II, 30% sn phm ca my III.

a) Tnh t l bng n tt ca l hng.b) Chn ngu nhin mt bng n t l hng ny, nu bng n l bng

hng th n c kh nng do my no sn xut nht?

Bi 8. Nhm nghin cu th trng ca mt cng ty cho bit: trong s nhngngi tm hiu thng tin v sn phm ca h c 60% ngi trc tip nshowroom xem sn phm v 40% ngi hi thng tin qua in thoi. Ngoira, trong s nhng ngi n trc tip showroom c 2% ngi mua sn phm,trong nhng ngi hi qua in thoi th c 1% ngi mua sn phm. Chnngu nhin mt ngi tm hiu thng tin v sn phm ca cng ty ny. Tnhxc sut ngi ny mua sn phm ca cng ty.

Bi 9. C hai hp sn phm: hp th nht c 20 sn phm trong c 4 phphm; hp th hai c 18 sn phm trong c 3 ph phm.

a) Ly ngu nhin ng thi 2 sn phm ca hp th nht kim tra.Tnh xc sut ly c t nht mt ph phm.

b) Ln lt ly ngu nhin 3 sn phm (c hon li) ca hp th hai kim tra. Tnh xc sut ly c 2 ph phm.

c) Ly ngu nhin t mi hp 1 sn phm. Tnh xc sut ly c phphm.

d) Ly ngu nhin mt hp ri t hp ly ngu nhin ra 2 sn phm.Tnhxc sut ly c ph phm.

e) Ly ngu nhin 1 sn phm t hp th nht b sang hp th hai. Sau ly ngu nhin ng thi 2 sn phm ca hp th hai. Tnh xc sut ly c 1 ph phm t hp th hai.

38

Chng 2

I LNG NGUNHIN V CC QUYLUT PHN PHI XCSUT C BIT

2.1 Lut phn phi xc sut ca LNN

Lut phn phi xc sut ca i lng ngu nhin (LNN) l mt biu ,trong ch ra:- Cc gi tr c th nhn c ca LNN.- Xc sut tng ng LNN nhn cc gi tr . Lut phn phi xc sut ca LNN ri rc c th hin di dng bng, gil bng phn phi xc sut, nh sau:

X x1 x2 . . . xnp p1 p2 . . . pn

vi xi l cc gi tr ca X; pi = P (X = xi):ni=1

pi = 1 v pi > 0, i = 1, n

Ch . Cho X l LNN ri rc c bng phn phi xc sut nh trn thP (X = c) = 0 nu c / {x1, x2, . . . , xn}

39

Lut phn phi xc sut ca LNN lin tc c biu th bi th ca hms y = f(x) xc nh trn (,+), gi l hm mt xc sut, thamn:

i) f(x) 0, x

ii)

f(x)dx = 1

Tnh cht: Vi X l LNN lin tc v a, b l hai s thc bt k th

P (a < X < b) = P (a X < b) = P (a < X b) = P (a X b) = ba

f(x)dx

2.1.1 Bng phn phi xc sut (PPXS)

Bi 1. Mt kin hng c 10 sn phm, trong c 6 sn phm loi I v 4 snphm loi II. Chn ngu nhin (ng thi) t kin hng ra 2 sn phm. GiX l s sn phm loi II c ly ra. Lp bng phn phi xc sut ca X.

BI GII Ta c: X = 0, 1, 2. Tnh xc sut: P (X = k) = C

k4 .C

2k6

C210, k = 0, 1, 2

Vy bng PPXS ca X l:

X 0 1 2p 1

3

8

15

2

15

Bi 2. L hng I c 10 sn phm tt v 2 ph phm, l hng II c 14 sn phmtt v 5 ph phm. Chn ngu nhin t l hng I ra 1 sn phm v b vo lhng II. Sau , t l hng II chn ngu nhin ra 3 sn phm. Gi X l s snphm tt trong 3 sn phm ly ra t l hng II.

a) Lp bng phn phi xc sut ca X.b) Tnh P (1 < X 4).

BI GIIa) Lp bng PPXS ca X: Ta c: X = 0, 1, 2, 3. Gi A: bin c sn phm ly t l hng I b vo l hng II l sn phm tt.= {A,A} l h y v xung khc

40

p dng CTXSTP:

P (X = k) = P (A).P (X = k|A) + P (A).P (X = k|A)

=10

12 C

k15.C

3k5

C320+

2

12 C

k14.C

3k6

C320, k = 0, 3

Vy bng PPXS ca X l:

X 0 1 2 3p 7

684

8

57

1057

2280

2639

6840

b) Da vo bng PPXS, ta c:

P (1 < X 4) = P (X = 2) + P (X = 3) = 581684

0, 8494

Bi 3. Kin hng I c 12 sn phm trong c 3 ph phm, kin hng II c 15sn phm trong c 5 ph phm. Chn ngu nhin t mi kin hng ra 1 snphm. GiX l s sn phm tt chn c. Lp bng phn phi xc sut caX.

BI GII Ta c: X = 0, 1, 2. Gi Ai: bin c sn phm ly t kin hng I l sn phm tt, i = 1, 2.= {A1, A2} l h c lpp dng cng thc nhn:

P (X = 0) = P (A1.A2) = P (A1).P (A2) =3

12 515

=1

12

P (X = 2) = P (A1.A2) = P (A1).P (A2) =9

12 1015

=1

2

P (X = 1) = 1 P (X = 0) P (X = 2) = 1 112

12=

5

12

Vy bng PPXS ca X l:

X 0 1 2p 1

12

5

12

1

2

Bi 4. Mt x th c 4 vin n, bn ln lt tng vin vo mt mc tiumt cch c lp. Xc sut trng mc tiu mi ln bn l 0,7. Nu c 1 vintrng mc tiu hoc ht n th dng. Gi X l s vin n bn.

41

a) Lp bng phn phi xc sut ca X.b) Tnh P (2 X < 4).

BI GIIa) Lp bng PPXS ca X: Ta c: X = 1, 4. Gi Ai: bin c vin n th i trng mc tiu, i = 1, 4.Theo gi thit, {A1, A2, A3} l h c lp ton phn.Ta tnh cc xc sut:

P (X = 1) = P (A1) = 0, 7

P (X = 2) = P (A1.A2) = P (A1).P (A2) = 0, 3.0, 7 = 0, 21

P (X = 3) = P (A1.A2.A3) = P (A1).P (A2).P (A3) = 0, 3.0, 3.0, 7 = 0, 063

P (X = 4) = 1 P (X = 1) P (X = 2) P (X = 3) = 0, 027

Vy bng PPXS ca X l:

X 1 2 3 4p 0,7 0,21 0,063 0,027

b) Da vo bng PPXS: P (2 X < 4) = P (X = 2) + P (X = 3) = 0, 273

Bi 5. C 3 hp: mi hp ng 15 vin bi, trong hp th nht c 3 vinbi trng, hp th hai c 5 vin bi trng, hp th ba c 2 vin bi trng. Chnngu nhin mt hp, ri t hp chn ly ngu nhin ng thi 3 vin bi.Gi X l s vin bi trng trong 3 vin bi ly ra.

a) Lp bng phn phi xc sut ca X.b) Tnh cc xc sut P (0 < X < 3), P (1 X 2), P (X > 1).

BI GIIa) Lp bng PPXS ca X: Ta c: X = 0, 1, 2, 3. Gi Hi: bin c chn c hp th i, i = 1, 2, 3.= {H1,H2,H3} l h y v xung khcp dng CTXSTP:P (X = 0) = P (H1).P (X = 0|H1)+P (H2).P (X = 0|H2)+P (H3).P (X = 0|H3)

=1

3 C

03 .C

312

C315+

1

3 C

05 .C

310

C315+

1

3 C

02 .C

313

C315=

626

1365

P (X = 1) = P (H1).P (X = 1|H1)+P (H2).P (X = 1|H2)+P (H3).P (X = 1|H3)

42

=1

3 C

13 .C

212

C315+

1

3 C

15 .C

210

C315+

1

3 C

12 .C

213

C315=

193

455

P (X = 2) = P (H1).P (X = 2|H1)+P (H2).P (X = 2|H2)+P (H3).P (X = 2|H3)=

1

3 C

23 .C

112

C315+

1

3 C

25 .C

110

C315+

1

3 C

22 .C

113

C315=

149

1365

P (X = 3) = P (H1).P (X = 3|H1)+P (H2).P (X = 3|H2)+P (H3).P (X = 3|H3)=

1

3 C

33 .C

012

C315+

1

3 C

35 .C

010

C315+

1

3 0 = 11

1365

Vy bng PPXS ca X l:

X 0 1 2 3p 626

1365

193

455

149

1365

11

1365

b) Da vo bng PPXS, ta c:

P (0 < X < 3) = P (X = 1) + P (X = 2) =193

455+

149

1365=

8

15 0, 5333

P (1 X 2) = P (X = 0) + P (X = 1) + P (X = 2)=

626

1365+

193

455+

149

1365=

1354

1365 0, 9919

P (X > 1) = P (X = 2) + P (X = 3) =149

1365+

11

1365=

32

273 0, 1172

2.1.2 Hm mt ca LNN lin tc

Bi 1. Cho hm s f(x) =

3x2

8, nu x [0, 2]

0 , nu x / [0, 2]a) Chng minh rng f(x) l hm mt ca LNN X.

b) Tnh cc xc sut P (1 X 32) v P (1 < X < 3).

BI GIIa) Ta nhn thy f(x) 0,x nn

f(x) l hm mt xc sut +

f(x)dx = 1

43

V +

f(x)dx =

20

3x2

8dx =

[x3

8

]20

= 1 nn ta c pcm.

b) p dng cng thc P (a X b) = ba

f(x)dx ta c:

P (1 X 32) =

32

1

f(x)dx =

32

1

3x2

8dx =

[x3

8

] 32

1

=27

64 1

8=

19

64

P (1 X 3) = 31

f(x)dx =

21

3x2

8dx =

[x3

8

]21

= 1 18=

7

8

Bi 2. Chng minh rng hm s sau l hm mt xc sut ca LNN lin

tc X no : f(x) =

{ 1x2

, nu x 10 , nu x < 1

Khi , tnh: P (X 3); P (1 X 2); P (1 < X < 2); P (X > 5)

BI GII Nhn xt: f(x) 0,x

Ngoi ra: +

f(x)dx =

+1

1

x2dx =

[ 1x

]+1

= 0 (1) = 1

Vy f(x) l hm mt xc sut.

Tnh cc xc sut:

P (X 3) = 3

f(x)dx =

31

1

x2dx =

[ 1x

]31

=13 (1) = 2

3

P (1 X 2) = 21

f(x)dx =

21

1

x2dx =

[ 1x

]21

= 12 (1) = 1

2

P (1 X 2) = 21

f(x)dx =

21

1

x2dx =

[ 1x

]21

= 12 (1) = 1

2

P (X > 5) =

+5

f(x)dx =

+5

1

x2dx =

[ 1x

]+5

=1

5

Bi 3. Cho LNN X lin tc c hm mt xc sut xc nh bi:

f(x) =

{c(1 x2) , nu x [1, 1]

0 , nu x / [1, 1]

44

a) Xc nh hng s c.b) Tnh xc sut P (0, 5 X 0, 8)

BI GIIa) Theo gi thit, f(x) l hm mt xc sut nn ta c

+

f(x)dx = 1

M +

f(x)dx =

11

c(1x2)dx = 2 10

c(1x2)dx =[2c(x x

3

3

)]10

=4c

3

Cho ta 4c3

= 1 c = 34.

b) Ta c:

P (0, 5 X 0, 8) = 0,80,5

f(x)dx =

0,80,5

3

4(1 x2)dx

=

[3

4

(x x

3

3

)]0,80,5

=3263

4000= 0, 81575

Bi 4. Cho LNN X lin tc c hm mt xc sut l

f(x) =

{ax+ b , nu x (0, 1)

0 , nu x / (0, 1)

Gi s P (X > 12) =

1

3, hy xc nh hng s a, b.

BI GII Ta c f(x) l hm mt xc sut nn

+

f(x)dx = 1

M +

f(x)dx =

11

(ax+ b)dx =

[ax2

2+ bx

]10

=a

2+ b

Nn ta c phng trnh:a

2+ b = 1 hay a+ 2b = 2

Mt khc:

P (X >1

2) =

+1

2

f(x)dx =

11

2

(ax+ b)dx =

[ax2

2+ bx

]11

2

=3a

8+b

2

45

Theo gi thit P (X >1

2) =

1

3nn suy ra 9a + 12b = 8

Vy ta c h phng trnh

{2 + 2b = 2

9a+ 12b = 8

a = 4

3

b =5

3

Bi 5. Gi s tui th ca mt thit b in t l LNN lin tc X c hm

mt xc sut l f(x) ={

cecx , nu x 00 , nu x < 0

a) Xc nh hng s c.b) Tnh xc sut P (X 10).c) Nu P (X 10) = 1

2th gi tr ca c l bao nhiu?

BI GIIa) Nhn thy: f(x) 0,x

Mt khc, +

f(x)dx =

+0

cecxdx =

[ ecx

]+0

= 0 (1) = 1,c

Vy hm s cho l hm mt xc sut vi mi gi tr ca c.

b) P (X 10) = 10

f(x)dx =

100

cecxdx =

[ ecx

]100

= 1 e10c

c) Ta c: P (X 10) = 12 1 e10c = 1

2 e10c = 1

2 c = ln2

10

2.2 Hm phn phi xc sut

Cho LNN X, hm phn phi xc sut l hm F (x) = P (X < x)

Vi X l LNN ri rc c bng phn phi xc sut l

X x1 x2 . . . xnp p1 p2 . . . pn

th hm phn phi F (x) =xi

C th:

F (x) =

0 , nu x x1p1 , nu x1 < x x2p1 + p2 , nu x2 < x x3. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .p1 + p2 + . . .+ pn1 , nu xn1 < x xn1 , nu x xn

Nu X l LNN lin tc c hm mt xc sut l f(x) th hm phn phixc sut F (x) =

x

f(t)dt. C th:

Nu X c hm mt dng f(x) ={

(x) , khi x [a, b]0 , khi x / [a, b]

th F (x) =

0 khi x a xa

(t)dt khi a < x b1 khi x b

Nu X c hm mt dng f(x) ={

0 , khi x < a(x) , khi x a

th F (x) =

0 khi x < a xa

(t)dt khi x a

Tnh cht:

a) 0 F (x) 1, F () = 0, F (+) = 1.b) F (x) l hm khng gim.c) Nu X l LNN lin tc c hm mt xc sut f(x) th F (x) = f(x).d) Vi X l LNN ri rc th P (a X < b) = F (b) F (a).e) Vi X l LNN lin tc th

P (a < X < b) = P (a X < b) = P (a < X b)= P (a X b) = F (b) F (a)

Bi 1. Cho X l LNN ri rc c bng phn phi xc sut l

X 0 1 2p 0, 6 0, 3 0, 1

Tm hm phn phi xc sut ca X.

47

p s

F (x) =

0 , nu x 00, 6 , nu 0 < x 10, 9 , nu 1 < x 21 , nu x > 2

Bi 2. Cho X l LNN X ri rc, c bng phn phi xc sut nh sau:

X 2 0 1 2 3p 0, 1 0, 2 0, 1 0, 5 0, 1

a) Tm hm phn phi xc sut ca X.b) Tnh cc xc sut P (0 X < 3), P (2 < X 3).

BI GII

a) F (x) =

0 , nu x 20, 1 , nu 2 < x 00, 3 , nu 0 < x 10, 4 , nu 1 < x 20, 9 , nu 2 < x 31 , nu x > 3

b) Da vo bng PPXS:

P (0 X < 3) = P (X = 0) + P (X = 1) + P (X = 2) = 0, 2 + 0, 1 + 0, 5 = 0, 8

Cch 2: Da vo kt qu cu a):

P (0 X < 3) = F (3) F (0) = 0, 9 0, 1 = 0, 8

Tng t,

P (2 < X 3) = P (X = 0) + P (X = 1) + P (X = 2) + P (X = 3) = 0, 9

Bi 3. Tui th ca mt b phn trong mt dy chuyn sn xut l LNN X

(thng) c hm mt nh sau: f(x) =

2

25(x+ 10

)2 , nu x (0, 40)0 , nu x / (0, 40)

a) Tm hm phn phi xc sut ca X.

48

b) Tm xc sut tui th ca thit b nh hn 1 nm.

BI GIIa)

F (x) =

0 , nu x 0 x

f(t)dt , nu 0 < x 401 , nu > 40

=

0 , nu x 05x

4(x+ 10), nu 0 < x 40

1 , nu > 40

Tht vy, vi 0 < x 40 th x

f(t)dt =

x0

25

2(t + 10)2dt =

[ 25

2(x+ 10)

]x0

=25

2

( 110

1x+ 10

)=

25x

20(x + 10)

b) Ta cn tnh:

P (X < 12) = P (0 < X < 12) =

120

f(x)dx =

120

25

2(t+ 10)2dt

=

[ 25

2(x+ 10)

]120

=25

2

( 110

122

)=

15

22= 0, 68182

Cch 2: Da vo kt qu cu a):

P (X < 12) = P (0 < X < 12) = F (12) F (0) = 5x4(x+ 10)

x=12

0 = 1522

Bi 4. Cho LNN X c hm mt : f(x) =

{0 , nu x < 22

x2, nu x 2

a) Tm hm phn phi xc sut ca X.b) Tm P (3 < X < 5).

49

BI GIIa)

F (x) =

0 , nu x < 2 x

f(t)dt , nu x 2

=

{0 , nu x < 2

1 2x

, nu x 2

Tht vy, vi x 2 th x

f(t)dt =

x2

2

t2dt =

[ 2

t

]x2

= 2x( 2

2

)= 1 2

x

b) Ta cn tnh:

P (3 < X < 5) = 53

f(x)dx =

52

2

t2dt

=

[ 2x

]52

= 25( 2

2

)=

3

5= 0, 6

Cch 2: Da vo kt qu cu a):

P (3 < X < 5) = F (5) F (3) = 1 2x

x=5

0 = 35

Bi 5. Tui th ca ngi dn mt a phng l LNN X c hm phn

phi xc sut cho nh sau: F (x) ={

0 , khi x 01 e0,013x , khi x > 0

a) Tnh t l ngi dn th t 60 n 70 tui.b) Xc nh hm mt xc sut ca X.

BI GIIa) Ta cn tnh: P (60 X 70) :

P (60 X 70) = F (70) F (60) =(1 e0,013x

)x=70

(1 e0,013x

)x=60

= 1 e0,013.70 1 + e0,013.60 0, 0559

50

b) Ta c hm mt xc sut ca X l f(x) = F (x) Vi x < 0 th F (x) = 0 nn F (x) = 0 Vi x > 0 th F (x) = 1 e0,013x nn F (x) = 0, 013e0,013x Ti x = 0 ta c

F (x) = limx0

F (x) F (0)x 0 = limx0

1 e0,013xx

= limx0

e0,013x 10, 013x 0, 013 = 0, 013

Vy f(x) ={

0 , nu x < 00, 013e0,013x , nu x 0

2.3 Cc c trng bng s ca LNN

Mode v Median

Mode ca LNN X, k hiu ModX, l gi tr x0 sao cho: P (X = x0) max, nu X l LNN ri rc f(x0) max, nu X l LNN lin tc c hm mt xc sut f(x)Ch .ModX cn c gi l gi tr tin chc nht hay gi tr chc chnnht ca LNN X.

Median ca LNN X (cn gi l trung v ca X), k hiu MedX, l sthc m tha: P (X m) 1

2v P (X m) 1

2

Nu X l LNN lin tc th

MedX = m P (X m) = P (X m) = 12

K vng, Phng sai, lch chun

K vng ca LNN X, k hiu EX hoc MX, l mt s thc c xcnh nh sau:

Nu X l LNN ri rc c bng phn phi xc sut l

X x1 x2 . . . xnp p1 p2 . . . pn

th EX =ni=1

xipi = x1p1 + x2p2 + . . .+ xnpn

51

Nu X l LNN lin tc c hm mt xc sut f(x) th

EX =

+

xf(x)dx

Phng sai ca LNN X, k hiu DX hoc V arX, l mt s thc khngm c xc nh bi:

DX = E(XEX)2 = E(X2) (EX)2

Nu X l LNN ri rc c bng phn phi xc sut l

X x1 x2 . . . xnp p1 p2 . . . pn

th DX =ni=1

x2ipi (EX)

2 =

ni=1

x2ipi

( ni=1

xipi)2

Nu X l LNN lin tc c hm mt xc sut f(x) th

DX =

+

x2f(x)dx (EX)2

=

+

x2f(x)dx( +

xf(x)dx)2

ngha: EX l gi tr trung bnh ca LNN X v DX l phn tnca LNN X xung quanh gi tr trung bnh.

Tnh cht ca k vng v phng sai:a) Vi C l hng s th E(C) = C v D(C) = 0b) Vi l mt s thc th E(.X) = .EX v D(.X) = 2.DXc) Vi mi LNN X,Y th E(X + Y ) = EX + EYd) Nu X,Y l hai LNN c lp th D(X + Y ) = DX +DY

lch chun ca LNN X l gi tr (X) =DX ( rng (X) c

cng n v o vi X)

52

Bi 1. Cho LNN X ri rc c bng phn phi xc sut l:

X 1 0 2 4 5p 0, 15 0, 1 0,45 0, 05 0, 25

a) Tm ModX, MedX, EX, DX.b) Tm E(2X + 5), D(3X + 9).

BI GIIa) Da vo bng PPXS: ModX = 2; MedX = 2.

EX =

5i=1

xi.pi = 1.0, 15 + 0.0, 1 + 2.0, 45 + 4.0, 05 + 5.0, 25 = 115

= 2, 2

DX =

5i=1

x2i .pi (EX)2

= (1)2.0, 15 + 02.0, 1 + 22.0, 45 + 42.0, 05 + 52.0, 25 (2, 2)2 = 4, 16b) p dng tnh cht ca k vng v phng sai:

E(2X + 5) = E(2X) +E(2) = 2EX + 2 = 2.2, 2 + 2 = 9, 4

D(3X + 9) = D(3X) +D(9) = 9DX + 0 = 9.4, 16 = 37, 44

Bi 2. Cho LNN X ri rc c bng phn phi xc sut l:

X 1 2 3 4p 0, 15 a 0, 35 b

Tm gi tr ca hng s a, b EX = 2, 8.

BI GIIDa vo tnh cht ca bng PPXS, ta c phng trnh:

0, 15 + a+ 0, 35 + b = 1 hay a+ b = 0, 5

Li c: EX =4i=1

xi.pi = 1.0, 15 + 2a+ 3.0, 35 + 4b = 2a + 4b+ 1, 2

Theo gi thit, EX = 0, 2 nn suy ra 2a+ 4b+ 1, 2 = 2, 8 a+ 2b = 0, 8Do , ta c h phng trnh:

{a+ b = 0, 5a+ 2b = 0, 8

{

a = 0, 2b = 0, 3

Bi 3. Mt l hng c 8 sn phm, trong c 5 sn phm loi I v 3 snphm loi II. Ly ngu nhin ng thi 3 sn phm t l hng ny.

53

a) Tnh s sn phm loi I tin chc nht trong 3 sn phm ly ra.b) Tnh trung bnh s sn phm loi I trong 3 sn phm ly ra.c) Gi Y l s sn phm loi II trong 3 sn phm ly ra. Tnh EY,DY .

BI GIIGi X l s sn phm loi I trong 3 sn phm ly ra. Ta c: X = 0, 3

Ngoi ra, P (X = k) = Ck5 .C

3k3

C38vi k = 0, 3

Nn ta c bng PPXS ca X l:

X 0 1 2 3p 1

56

15

56

15

28

5

28

a) Da vo bng PPXS, ta c s sn phm tt tin chc nht trong 3 sn phmly ra l ModX = 2b) Trung bnh s sn phm loi I trong 3 sn phm ly ra l

EX =

3i=0

xi.pi =15

8= 1, 875

c) Ta tm bng PPXS ca Y : Ta c: Y = 0, 3

P (X = k) =Ck3 .C

3k5

C38vi k = 0, 3

Nn ta c bng PPXS ca Y l:

Y 0 1 2 3p 5

28

15

28

15

56

1

56

Khi ,

EY =3i=0

yi.pi =9

8= 1, 125

DY =

3i=0

y2i .pi (EY )2 =225

448 0, 5022

Bi 4. Cho LNN X lin tc c hm mt xc sut l

f(x) =

3x2

16, khi x (2, 2)

0 , khi x / (2, 2)

54

Tm k vng v phng sai ca X.

BI GII K vng: EX =

+

xf(x)dx =

22

x 3x2

16dx =

22

3x3

16dx = 0 (v hm

s di du tch phn l) Phng sai:

DX =

+

x2f(x)dx (EX)2 = 22

x2 3x2

16dx 0

=

22

3x4

16dx =

3

8

20

x4dx =

[3

8 x

5

5

]20

=12

5= 2, 4

Bi 5. Cho LNN X lin tc c hm mt xc sut l

f(x) =

{a(3x x3) , khi x [0,3]

0 , khi x / [0,3]

a) Xc nh hng s a.b) Tm ModX.c) Tm k vng v phng sai ca X.d) Tm k vng v phng sai ca Y = 3X + 5.

BI GIIa) f(x) l hm mt

+

f(x)dx = 1

Ta c: +

f(x)dx =

30

a(3x x3)dx =[a(3x2

2 x

4

4

)]30

=9a

4

Suy ra 9a4

= 1 a = 49.

b) ModX l s x0 sao cho f(x) t cc i ti x0 nn ModX [0,3] .

Vi x [0,3] : f(x) = 49(3x x3)

= f (x) = 49(3 3x2) = 0 x = 1

f (x) =8x3

V f (1) = 83> 0 v f (1) =

83

< 0

nn f(x) t cc i ti x = 1 hay ModX = 1

55

c) Ta c

EX =

+

xf(x)dx =

30

x 49(3x x3)dx =

30

4

9(3x2 x4)dx

=

[4

9

(x3 x

5

5

)]30

=83

15 0, 9238

DX =

+

x2f(x)dx (EX)2 = 30

x2 49(3x x3)dx

(83

15

)2

=

30

4

9(3x3 x5)dx 64

75=

[4

9

(3x44

x6

6

)]30

6475

=11

75 0, 1467

d) Cch 1: Tnh trc tip tng t cu c)

E(3X + 5) =

+

(3x+ 5)f(x)dx =

30

(3x+ 5) 49(3x x3)dx

=

30

4

9(3x4 5x3 + 9x2 + 15x)dx

=

[4

9

( 3x

5

5 5x

4

4+ 3x3 +

15x2

2

)]30

=83

5+ 5 7, 7713

D(3X + 5) =

+

(3x+ 5)2f(x)dx (EX)2

=

30

(3x+ 5)2 49(3x x3)dx

(83

5+ 5

)2=

30

4

9(9x5 30x4 + 2x3 + 90x2 + 75x)dx

(817

25+ 16

3

)

=

[4

9

( 3x

6

2 6x5 + x

4

2+ 30x3 +

75x2

2

)]30

(817

25+ 16

3

)=

(34 + 16

3

)(817

25+ 16

3

)=

33

25= 1, 32

Cch 2: p dng tnh cht ca k vng v phng sai

E(3X + 5) = E(3X) + E(5) = 3EX + 5 = 3 83

15+ 5 =

83

5+ 5 7, 7713

D(3X + 5) = D(3X) +D(5) = 9DX + 0 = 9 1175

=33

25= 1, 32

56

2.4 Cc quy lut phn phi xc sut c bit

2.4.1 Quy lut phn phi siu bi

Xt mt tng th gm N phn t, trong c M phn t loi A. Ly ngunhin ng thi n phn t t tng th ny. Gi X l s phn t loi A trongn phn t ly ra. Khi , ta ni: X tun theo quy lut phn phi siu bi.K hiu: X H(N,M, n). Cng thc tnh xc sut:

P (X = k) =CkM.C

nk

NM

CnNvi k = 0, n

Tnh cht: Cho X H(N,M,n)

a) K vng: EX = np vi p =M

N

b) Phng sai: DX = npqN nN 1 vi q = 1 p

Bi 1. Cho X H(20, 12, 5). Tnh cc xc sut:a) P (X = 4) b) P (X < 2) c) P (X > 1) d) P (X 3)

BI GIITa c: X H(20, 12, 5) = P (X = k) = C

k12.C

5k8

C520vi k = 0, 5

a) P (X = 4) = C412.C

18

C520=

165

646 0, 25542

b) P (X < 2) = P (X = 0) + P (X = 1) =C012.C

58

C520+C112.C

48

C520=

56

969 0, 05779

c) P (X > 1) = 1 P (X 1) = 1 P (X = 0) P (X = 1)= 1 C

012.C

58

C520 C

112.C

48

C520=

913

969 0, 94221

d) P (X 3) = P (X = 3) + P (X = 4) + P (X = 5)=

C312.C28

C520+C412.C

18

C520+C512.C

08

C520=

682

969 0, 70382

Bi 2. Cho X H(10, 4, 5). Tnh:a) P (X < 3) b) P (X 3) c) P (X > 3) d) P (X 3)e) EX f) DX g) E(3X 2) h) D(5X 3)

57

BI GIITa c: X H(10, 4, 5) = P (X = k) = C

k4 .C

5k6

C510vi k = 0, 4

a) P (X < 2) = P (X = 0) + P (X = 1) =C04 .C

56

C510+C14 .C

46

C510=

11

42 0, 2619

b) P (X 2) = 1 P (X > 2) = 1 P (X = 3) P (X = 4)= 1 C

34 .C

26

C510 C

44 .C

16

C510=

31

42 0, 7381

c) P (X > 3) = P (X = 4) =C44 .C

16

C510=

1

42 0, 0238

d) P (X 3) = P (X = 3) + P (X = 4) = C34 .C

26

C510+C44 .C

16

C510=

11

42 0, 2619

e) EX = np = 5 410

= 2

f) DX = npqN nN 1 = 5

4

10 610

69=

4

5= 0, 8

g) E(3X 2) = 3EX 2 = 3.2 2 = 4h) D(5X 3) = 25DX = 25.0, 8 = 20

Bi 3. Mt l hng c 100 sn phm, trong c 90 sn phm tt. Ly ngunhin ng thi 15 sn phm t l hng ny. Tm s sn phm tt trung bnhv phng sai ca s sn phm tt trong 15 sn phm c ly ra.

BI GIIGi X l s sn phm tt trong 15 sn phm ly ra, ta c: X H(100, 90, 15). S sn phm tt trung bnh: EX = np = 15 90

100=

27

2= 13, 5

Phng sai: DX = npqN nN 1 = 15

90

100 10100

8599

=51

44 1, 1591

Bi 4. C 20 chi tit my, trong c 15 chi tit my tt. Ly ngu nhinng thi 4 chi tit my. Gi X l s chi tit my tt trong s 4 chi tit myc ly ra.

a) Xc nh quy lut phn phi xc sut ca X.b) Tnh xc sut ly c 3 chi tit my tt.c) Tnh trung bnh s chi tit tt c ly ra v phng sai ca X.

BI GIIa) X tun theo phn phi siu bi: X H(20, 15, 4)

58

b) Ta c P (X = k) =Ck15.C

4k5

C420vi k = 0, 4

Do , P (X = 3) = C315.C

15

C420=

455

969 0, 4696

c) Trung bnh s chi tit tt c ly ra: EX = np = 4 1520

= 3

Phng sai: DX = npqN nN 1 = 4

15

20 520

1619

=12

19 0, 6316

Bi 5. Mt ng ch vn lan nhm 20 chu lan c hoa mu vo cngvi 100 chu lan c hoa mu tm (lan cha n hoa). Mt khch hng chn muangu nhin ng thi 15 chu t 120 chu lan ny.

a) Tnh xc sut khch hng mua c t 5 n 6 chu lan c hoa mu .b) Gi X l s chu lan c hoa mu m khch hng chn c. Tnh

trung bnh v phng sai ca X.

BI GIIGi X l s chu lan c hoa mu m khch hng mua c

Ta c X H(120, 20, 15) = P (X = k) = Ck20.C

15k100

C15120, k = 0, 15

a) P (5 X 6) = P (X = 5) + P (X = 6) = C520.C

10100

C15120+C620.C

9100

C15120 0, 07232

b) Trung bnh: EX = np = 15 20120

=5

2= 2, 5

Phng sai: DX = npqN nN 1 = 15

20

120 100120

105119

=125

68 1, 8382

2.4.2 Quy lut phn phi nh thc

Xt mt lc Bernoulli gm:

n php th c lp

Trong mi php th ch xy ra mt trong hai bin c A v A: P (A) = pGi X l s ln xy ra bin c A trong n php th. Ta ni rng: X tun theoquy lut phn phi nh thc. K hiu: X B(n, p).Cng thc tnh xc sut:

P (X = k) = Ckn.pk.qnk vi q = 1 p v k = 0, n

Tnh cht: Cho X B(n, p)

59

a) K vng: EX = npb) Phng sai: DX = npq vi q = 1 pc) np q ModX np+ p

Bi 1. Cho X B(10; 0, 7). Tnh:a) P (X = 5) b) P (X < 2) c) P (X 2) d) P (X > 2) e) P (X 2)f) EX g) DX h) E(9X + 4) i) D(4X + 9) j) ModX

BI GIITa c X B(10; 0, 7) = P (X = k) = Ck10.(0, 7)k.(0, 3)10k , k = 0, 10

a) P (X = 5) = C510.(0, 7)5.(0, 3)5 0, 10292

b) P (X < 2) = P (X = 0) + P (X = 1)= C010.(0, 7)

0.(0, 3)10 + C110.(0, 7)1.(0, 3)9 0, 00014

c) P (X 2) = P (X = 0) + P (X = 1) + P (X = 2) = C010.(0, 7)0.(0, 3)10+ C110.(0, 7)

1.(0, 3)9 + C210.(0, 7)2.(0, 3)8 0, 00159

d) P (X > 2) = 1 P (X 2) = 1 P (X = 0) P (X = 1) P (X = 2)= 1 C010.(0, 7)0.(0, 3)10 C110.(0, 7)1.(0, 3)9 C210.(0, 7)2.(0, 3)8 0, 99841

e) P (X 2) = 1 P (X < 2) = 1 P (X = 0) P (X = 1)= 1 C010.(0, 7)0.(0, 3)10 C110.(0, 7)1.(0, 3)9 0, 99986

f) EX = np = 10.0, 7 = 7g) DX = npq = 10.0, 7.0, 3 = 2, 1

h) E(9X + 4) = 9EX + 4 = 9.7 + 4 = 67i) D(4X + 9) = 16DX = 16.2, 1 = 33, 6

j) Ta c: np q ModX np+ p 10.0, 7 0, 3 ModX 10.0, 7 + 0, 7 6, 7 ModX 7, 7

Vy ModX = 7

Bi 2. Tung ba ln mt con xc xc. Tnh xc sut c:

a) Hai ln xut hin mt 1 chm.

b) t nht mt ln xut hin mt 1 chm.

BI GII

60

Gi X l s ln xut hin mt 1 chm trong ba ln tung con xc xcTa c X B(n, p) vi n = 3 v p = 1

6

Khi , P (X = k) = Ck3 (16

)k(56

)3k, k = 0, 3

a) Ta tnh P (X = 2) = C23 (16

)2(56

)1=

5

72 0, 0694

b) Tnh P (X 1) = 1 P (X < 1) = 1 P (X = 0)= 1 C03

(16

)0(56

)3=

91

216 0, 4213

Bi 3. Mt thi c 10 cu hi, mi cu c 4 phng n tr li trong chc 1 phng n ng. Sinh vin A tr li mt cch ngu nhin tt c cc cuhi. Gi X l s cu tr li ng trong 10 cu.

a) Xc nh quy lut phn phi ca X.b) Tnh xc sut sinh vin A tr li ng t 2 n 3 cu hi.c) Tnh xc sut sinh vin A tr li ng t nht mt cu hi.d) Tnh trung bnh s cu hi c tr li ng v phng sai ca X.e) Tnh s cu hi m sinh vin A c kh nng tr li ng ln nht.

BI GIIa) X tun theo phn phi nh thc: X B(10; 0, 25).b) V X B(10; 0, 25) nn P (X = k) = Ck10.(0, 25)k .(0, 75)10k , k = 0, 10Ta cn tnh: P (2 X 3)P (2 X 3) = P (X = 2) + P (X = 3)

= C210.(0, 25)2.(0, 75)8 + C310.(0, 25)

3.(0, 75)7 0, 531c) Ta cn tnh: P (X 1)P (X 1) = 1 P (X < 1) = 1 P (X = 0)

= 1 C010.(0, 25)0.(0, 75)10 0, 943d) EX = np = 10.0, 25 = 2, 5; DX = npq = 10.0, 25.0, 75 = 1, 875e) S cu hi m sinh vin c kh nng tr li ng nht chnh l ModXTa c: np q ModX np+ p 1, 75 ModX 2, 75Vy ModX = 2

Bi 4. Mt ngi nui 160 con g mi cng loi. Xc sut 1 con g trngtrong ngy l 0,8.

a) Tm xc sut ngi nui c c t nht 130 trng trong ngy.b) Gi s mi trng bn c 2200 ng, tin cho mi con g n trong ngy

l 1000 ng, tnh s tin li trung bnh thu c trong ngy.

61

BI GIIa) Gi X l s trng thu c trong ngy ( s con g trng trong ngy)Ta c: X B(160; 0, 8) nn

P (X = k) = Ck160.(0, 8)k.(0, 2)160k , k = 0, 160

Ta cn tnh: P (X 103)P (X 130) = P (X = 130) + P (X = 131) + . . .+ P (X = 160)

= C130160 .(0, 8)130.(0, 2)30 + C131160 .(0, 8)

131.(0, 2)29 + . . .+C160160 .(0, 8)

160.(0, 2)0 0, 39b) Gi Y l s tin thu c trong ngy. Ta cn tnh EYTa c: Y = 2200.X 1000.160 = 2200X 160000Do , EY = 2200EX 160000M X B(160; 0, 8) nn EX = np = 160.0, 8 = 128Vy EY = 2200.128 160000 = 121600 (ng)

Bi 5. Mt nh vn trng 121 cy mai vi xc sut n hoa ca mi cy maitrong dp tt l 0,75. Gi bn 1 cy mai n hoa l 500 ngn ng.

a) Tnh s cy mai trung bnh n hoa trong dp tt.b) Gi s nh vn bn ht nhng cy mai n hoa, tnh s tin trung bnh

m nh vn thu c.

BI GIIGi X l s cy mai n hoa trong dp tt. Ta c: X B(121; 0, 75)a) Ta tnh EX = np = 121.0, 75 = 90, 75 (cy)b) Gi Y l s tin nh vn thu c khi bn nhng cy mai. Ta c: Y = 500XDo , s tin trung bnh nh vn thu c l

EY = 500EX = 500.90, 75 = 45, 375 (triu ng)

2.4.3 Quy lut phn phi Poisson

i lng ngu nhin X c gi l c phn phi Poisson, k hiu: X P (),nu X nhn cc gi tr 0, 1, 2, . . . v xc sut c tnh bi:

P (X = k) =e.k

k!vi k = 0, 1, 2, . . .

trong l trung bnh s ln xut hin bin c no m ta quan tm.Lu rng: phn phi Poisson thng gn vi yu t thi gian.

Tnh cht:

62

a) EX = DX = b) 1 ModX

Bi 1. Cho X P (3). Tnh:a) P (X = 3) b) P (X < 2) c) P (X 3) d) P (X > 2)

BI GIIV X P (3) nn P (X = k) = e

3.3k

k!vi k = 0, 1, 2, . . .

a) P (X = 3) =e3.33

3!=

9

2e3 0, 224

b) P (X < 2) = P (X = 0) + P (X = 1) = e3.30

0!+e3.31

1!= 4e3 0, 1991

c) P (X 3) = P (X = 0) + P (X = 1) + P (X = 2) + P (X = 3)=

e3.30

0!+e3.31

1!+e3.32

2!+e3.33

3!= 13e3 0, 6472

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

3.30

0! e

3.31

1! e

3.32

2!= 1 17

2e3 0, 5768

Bi 2. Ti mt bn cng, trung bnh mi ngy c 5 tu cp bn. Tnh xc sut trong mt ngy:

a) C 3 tu cp bn.b) C t nht 4 tu cp bn.c) C ng 5 tu cp bn.d) C t 3 n 7 tu cp bn.

BI GIIGi X l s tu cp bn trong mt ngy th X P (5). Do ,

P (X = k) =e5.5k

k!vi k = 0, 1, 2, . . .

a) P (X = 3) =e5.53

3!=

125

6e5 0, 1404

b) P (X 4) = 1 P (X = 0) P (X = 1) P (X = 2) P (X = 3)= 1 e

5.50

0! e

5.51

1! e

5.52

2! e

5.53

3!=

118

3e5 0, 735

c) P (X = 5) =e5.55

5!=

625

24e5 0, 175

63

d) P (3 X 7) = P (X = 3)+P (X = 4)+P (X = 5)+P (X = 6)+P (X = 7)=

e5.53

3!+e5.54

4!+e5.55

5!+e5.56

6!+e5.57

7! 0, 742

Bi 3. Ti mt siu th, trung bnh c 5 pht th c 10 khch n quy tnhtin.

a) Tnh xc sut trong 1 pht c 3 khch n quy tnh tin.b) Tnh xc sut trong 1 pht c t 1 n 3 khch n quy tnh tin.c) Tnh s khch c kh nng n quy tnh tin ln nht trong 1 gi.

BI GIIa) Gi X l s khch n quy tnh tin trong 1 pht th X P (1) vi 1 ltrung bnh s khch n quy tnh tin trong 1 pht: 1 =

1.10

5= 2

Ta c: P (X = 3) =e1(1)3

3!=

e2.23

3!=

4

3e2 0, 18044

b) Ta cn tnh: P (1 X 3)P (1 X 3) = P (X = 1) + P (X = 2) + P (X = 3)

=e2.21

1!+e2.22

2!+e2.23

3!=

16

3e2 0, 722

c) Gi Y l s khch n quy tnh tin trong 1 gi th Y P (2) vi 2 l trungbnh s khch n quy tnh tin trong 1 gi ( 60 pht): 2 = 60.10

5= 120

S khch c kh nng n quy tnh tin ln nht trong 1 gi chnh l ModY

Ta c: 2 1 ModY 2 119 ModY 120Vy ModY = 119 hoc 120.

Bi 4. B phn qun l ca mt siu th cho bit: trong nhng ngy ngkhch, nhn vin tnh tin ca siu th trung bnh trong 5 pht tnh tin xongcho 2 khch. Tnh xc sut trong ngy ng khch, mt khch hng phich qu 5 pht ngi khch k trc c tnh tin xong.

BI GIIGi X l s khch c nhn vin tnh tin xong trong 5 pht. Ta c: X P (2)Ta cn tnh: P (X < 1) = P (X = 0) = e

2.20

0!= e2 0, 1353

Bi 5. Mt ngi c 4 xe t cho thu. Hng ngy, chi ph cho mi xe l 10usd(cho d xe c c thu hay khng). Gi cho thu mi xe l 70usd. Gi s yu

64

cu thu xe mi ngy l LNN c phn phi Poisson vi tham s = 2, 8 .Tnh s tin trung bnh ngi ny thu c trong mt ngy.

BI GIIGi X l s yu cu thu xe mi ngy.Theo gi thit, X P (2, 8) nn EX = 2, 8Gi Y l s tin thu c trong mt ngy.Ta c: Y = 70X 10.4 = 70X 40Vy s tin trung bnh ngi ny thu c trong mt ngy l

EY = 70EX 40 = 70.2, 8 40 = 156

2.4.4 Quy lut phn phi chun

LNN X lin tc c gi l c phn phi chun nu hm mt c dng:

f(x) =1

2

e1

2

(x

)2

K hiu: X N(, 2). Cng thc tnh xc sut:

P (a X b) = (b

)

(a

)Gi tr ca hm c tnh bng cch tra bng 1.

Tnh cht:

a) P (X = C) = 0 vi C l hng s bt k.b) EX = ModX = MedX = .c) DX = 2

Bi 1. Cho LNN X N(3, 4). Tnh:a) P (X > 1) b) P (0 < X < 1)c) P (|X 1| < 2) d) P (|X 1| > 1)

BI GIIV X N(3, 4) nn P (a X b) =

(b 32

)

(a 32

)a) P (X > 1) = P (1 < X < +) = (+)

(1 32

)= 0, 5 (1)

= 0, 5 + (1) = 0, 5 + 0, 3413 = 0, 8413

65

b) P (0 < X < 1) = (1 3

2

)

(0 32

)= (1) (1, 5)

= (1) + (1, 5) = 0, 3413 + 0, 4332 = 0, 0919

c) P (|X 1| < 2) = P (1 < X < 3) = (3 3

2

)

(1 32

)= (0) (2) = 0 + (2) = 0, 4772

d) V |X 1| > 1 X > 2 hoc X < 0 nn ta c:P (|X1| > 1) = P (X > 2)+P (X < 0) = P (2 < X < +)+P ( < X < 0)

= (+) (2 3

2

)+

(0 32

) ()

= (+) (0, 5) + (1, 5) + (+)= 2(+)+(0, 5)(1, 5) = 2.0, 5+0, 19150, 4332 = 0, 7583

Cch 2: P (|X 1| > 1) = 1 P (|X 1| 1) = 1 P (0 X 2)= 1

[(2 3

2

)

(0 32

)]= 1(0, 5) + (1, 5)

= 1 + (0, 5) (1, 5) = 1 + 0, 1915 0, 4332 = 0, 7583

Bi 2. Chiu cao ca mt loi cy ly g l i lng ngu nhin tun theoquy lut phn phi chun vi chiu cao trung bnh l 20m v lch chun l2,5m. Cy t tiu chun khai thc l cy c chiu cao ti thiu l 15m. Tnht l cy t tiu chun khai thc.

BI GIIGi X l chiu cao ca cy. Ta c: X N(, 2) vi = 20m; = 2, 5mT l cy t tiu chun khai thc chnh bng P (X 15):P (X 15) = P (15 X < +) = (+)

(15

)= (+)

(15 202, 5

)= 0, 5 (2)

= 0, 5 + (2) = 0, 5 + 0, 4772 = 0, 9772

Bi 3. Chiu cao ca cc sinh vin mt trng i hc l LNN c phnphi chun vi chiu cao trung bnh l 158cm v lch chun l 7,5cm. Nuchn ra 10% sinh vin c chiu cao cao nht th chiu cao ti thiu ca sinhvin trong nhm ny l bao nhiu?

BI GIIGi X l chiu cao ca sinh vin th X N(, ) vi = 158; = 7, 5

a l chiu cao ti thiu trong nhm sinh vin c chiu cao cao nht.Ta cn tm a sao cho P (X a) = 10% = 0, 1

66

Ta c: P (X a) = P (a X < +) = (+) (a

)= (+)

(a 1587, 5

)= 0, 5

(a 1587, 5

)Suy ra 0, 5

(a 1587, 5

)= 0, 1

(a 1587, 5

)= 0, 4

tra bng 1= (1, 29)

Do a 1587, 5

= 1, 29 hay a = 167, 675.

Bi 4. im thi Toeic ca sinh vin nm cui mt trng i hc l LNNX c phn phi chun vi gi tr trung bnh l 560 im v lch chun l 78im. Tnh:

a) T l sinh vin c im t 600 n 700 im.b) T l sinh vin c im thi trn 500 im.c) Gi s nh trng mun xc nh im Toeic ti thiu sinh vin c

th ra trng vi t l 80%, tnh im Toeic ti thiu ny.

BI GIIGi X l im thi Toeic ca sinh vin. Ta c: X N(, 2) vi = 560; = 78a) T l sinh vin c im t 600 n 700 im l:

P (600 < X < 700) = (700

)

(600

)=

(700 56078

)

(600 56078

)= (1, 79) (1, 28)

= 0, 4633 0, 3997 = 0, 0636 = 6, 36%

b) T l sinh vin c im thi trn 500 im l:P (X > 500) = P (500 < X < +) = (+)

(500

)= (+)

(500 56078

)= 0, 5 (0, 77)

= 0, 5 + (0, 77) = 0, 5 + 0, 2794 = 0, 7794 = 77, 94%

c) Gi a l mc im ti thiu m sinh vin cn phi t c.Ta cn tm a sao cho P (X a) = 80% = 0, 8V P (X a) = P (a X < +) = (+)

(a

)= (+)

(a 56078

)= 0, 5 +

(560 a78

)nn ta