1.5.2 xac suat cua mot bien co

5/24/2018 1.5.2 Xac Suat Cua Mot Bien Co

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Xc sut ca mt bin c

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XC SUT CA MT BIN C

I. TM TT L THUYT

Xt th nghim gieo qun xc sc 6 mt (c th gieo mt con, hai con hoc

nhiu qun xc sc) v xt schm xut hin, ta c cc khi nim sau y:

1. Php thngu nhin

Php thngu nhin l mt th nghim c kt qumang tnh cht ngu nhin

m ta khng thbit chc c kt qusxy ra nhng c thxc nh c

tp hp tt ccc kt quc thxy ra ca php th.

V d:Vic gieo qun xc sc l mt php thngu nhin.

2. Khng gian muTp hp tt ccc kt quc thxy ra ca php thngu nhin gi l khng

gian mu. Khng gian mu thng c k hiu l E hoc

V d:Nu gieo mt qun xc sc th khng gian mu E l {1, 2, 3, 4, 5, 6}

Nu gieo ln lt hai qun xc sc th khng gian mu E l

{(1;1), (1;2), (1;3), (1;4), (1;5), (1;6), (2;1), (2;2), (2;3), (2;4), (2;5), (2;6),

(3;1), (3;2), (3;3), (3;4), (3;5), (3;6), (4;1), (4;2), (4;3), (4;4), (4;5), (4;6), (5;1),

(5;2), (5;3), (5;4), (5;5), (5;6), (6;1), (6;2), (6;3), (6;4), (6;5), (6;6) }

3. Bin c

Mi tp hpcon ca khng gian mu l mt bin c. Mi phn tca bin c

A gi l mt kt quthun li cho A.

V d:Bin cgieo ln lt 2 qun xc sc c tng bng 5 l

{(1;4), (4;1), (2;3), (3;2)}

4. Cc loi bin c

4.1. Bin c s cp

Miphn tca khng gian mu l mt bin cscp.

V d:(1;2) l bin cscp


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Chng III. Thp, Xc sut v Sphc Trn Phng

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4.2. Bin c chc chn

Khng gian mu E cn gi l bin cchc chn, tc l bin clun lun xy ra

khi thc hin php thngu nhin.

V d:Bin cgieo 2 qun xc sc c tng ln hn hoc bng 2 v nhhn

hoc bng 12 l bin cchc chn.

4.3. Bin c khng th

Bin ckhng thl bin ckhng bao gixy ra khi thc hin php thngu

nhin. Bin ckhng thk hiu l .

V d:Bin cgieo 2 qun xc sc c tng ln hn 12 l bin ckhng th.

4.4. Bin c hp

Bin cA B l bin c"t nht c A hoc B xy ra" gi l hp ca hai bin

cA v B.

Bin c1 2

...k

A B A gi l hp ca kbin c1 2, , ...,

kA A A

V d:Gi A l bin cgieo ln lt 2 qun xc sc c tng ln hn 10 v

B l bin cgieo 2 qun xc sc c tng nhhn 4.

Khi bin c A B l {(6;5), (5;6), (6;6), (1;1), (1;2), (2;1)}

4.5. Bin c giao

Bin cA B l bin c"cA v B cng xy ra".

Bin c1 1

...k

A B A l bin c"1 2, , ...,

kA A A cng xy ra", gi l giao ca

bin c1 2, , ...,

kA A A .

V d:Gi A l bin cgieo 2 qun xc sc c tng ln hn 7 v B l bin

cgieo 2 qun xc sc c tng nhhn 10.

Khi bin c A B l {(2;6), (6;2), (3;5), (5;3) (4;4), (4;5), (5;4)}

4.6. Bin c xung khc

Hai bin cA v B gi l xung khc nu khi bin cny xy ra th bin ckia

khng xy ra, tc l 0A B .

V d:Bin cA gieo 2 qun xc sc c tng ln hn 10 v bin cB gieo 2

qun xc sc c tng nhhn 4 l hai bin cxung khc.


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Xc sut ca mt bin c

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4.7. Bin c i.Bin ci ca bin cA trong khng gian mu E, k hiu

A , l bin c"khng xy ra A".

V d:Bin cA gieo 2 qun xc sc c tng l mt schn, khi bin cA

l bin cgieo 2 qun xc sc c tng l mt sl.

4.8. Bin c c lp

Hai bin cA v B ca mt php thngu nhin gi l c lp vi nhau nu

sxy ra hay khng xy ra ca bin cny khng nh hng ti sxy ra hay

khng xy ra ca bin ckia.

V d:Khi gieo 2 qun xc sc, gi A, B l bin ctng ng qun xc sc

u tin v thhai nhn c mt 3. Khi A, B c lp vi nhau.

5. Tn sca mt bin c

Smln xut hin ca bin cA trong nln thc hin php thngu nhin gil tn sca bin cA ( )0 m n .

V d:Khi gieo 16 ln mt qun xc sc ta thy c 2 ln xut hin mt lc th

tn sca bin cqun xc sc xut hin mt lc l 2 trong 16 php th.

6. Tn sut ca mt bin c

Tsgia tn smca bin cA v sn ln thc hin php thngu nhin

gi l tn sut ca bin cA. K hiu mfn

= .

V d:Khi gieo 16 ln mt qun xc sc ta thy c 2 ln xut hin mt lc th

tn sut ca bin cqun xc sc xut hin mt lc l2

0,12516

f = = .

7. nh ngha xc sut

Xc sut ca bin cA l tsgia strng hp thun li cho A v tng s

trng hp c thxy ra trong php thngu nhin:

Nu bin cA c mphn ttrong khng gian mu E c nphn t ( )0 m n

th xc sut ca bin cA l :

Strng hp thun li cho A

Tng strng hp c thxy raP(A) =

Sphn tca A

Sphn tca E( ) AmP A

n E= = =


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8. Tnh cht

Cho mt th nghim ngu nhin c khng gian mu E v A, B l hai bin c.

Khi ( ) ( ) ( )0 1; 1; 0;P A P E P = = A v B xung khc ( ) 0A B =

9. Quy tc tnh xc sut

9.1. Quy tc cng xc sut

9.1.1. Bin cxung khc

Cho A v B l hai bin cxung khc. Ta c: P(A B) = P(A) + P(B)

Cho kbin cxung khc1 2, , ...,

kA A A . Ta c:

( ) ( ) ( ) ( )1 1 1 2... ...k kP A A A P A P A P A= + + +

9.1.2. Bin ci

Cho A l bin ci ca bin cA. Ta c: ( ) ( ) ( ) ( )1 1P A P A P A P A+ = =

9.2. Quy tc nhn xc sut:

9.2.1. Bin cc lp:

Cho A v B l hai bin cc lp vi nhau. Ta c: P(A B) = P(A).P(B)

Cho kbin c1 2, , ...,

kA A A c lp vi nhau. Ta c:

( ) ( ) ( ) ( )1 2 1 2... , ,...,k kP A A A P A P A P A=

9.2.2. Bin cxung khc:

Cho A v B l hai bin cxung khc.

Ta c (A B) lun khng xy ra, nn: P(AB) = 0

Ta c A v B xung khc th A v B khng c lp, nn:

( ) ( ) ( ).P A B P A P B vi ( ) 0P A > v ( ) 0P B >

9.3. Lin h gia quy tc cng xc sut v quy tc nhn xc sut

Cho A v B l hai bin cbt k. Ta c:

( ) ( ) ( ) ( )P A B P A P B p A B

= +

Ch : C sch k hiu giao ca hai bin cA v B l AB thay cho A B.

Giao ca kbin c1 2, , ...,

kA A A l

1 2...

kA A A .


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Xc sut ca mt bin c

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B. CC DNG BI TP CBN

DNG 1. TNH XC SUT CA MT BIN C

Phng php:Dng nh ngha xc sut ca mt bin c

Bi 1. Cho 9 qucn trng lng 1kg, 2kg, , 8kg, 9kg. Chn ngu nhin 3qu cn. Tnh xc sut tng trng lng 3 qucn c chn khng vtqu 8 kg.

Gii

Gi E l tp hp tt ccc cch chn 3 qucn trong 9 qucn 39

84E C= = .

Gi A l bin c"ly 3 qucn c tng trng lng khng vt qu 8kg".

Xt cc khnng xy ra: 1 + 2 + 3 = 6; 1 + 2 + 4 = 7; 1 + 2 + 5 = 8; 1 + 3 + 4 = 8.

Nhvy chc 4 cch chn ra 3 qucn c tng trng lng khng vt qu

8kg, tc l 4A = . Vy xc sut cn tm l ( ) 4 184 21P A = = .

Bi 2.Gieo ln lt ba ng xu. Gi A l bin cc t nht hai mt sp xut

hin lin tip; B l bin cc ba mt ging nhau.

1.Tnh xc sut ca A v B.

2.Tnh xc sut ca A B v ca A B .

Gii

1.Khng gian mu c 1 1 1 32 2 2. . 2 8C C C = = phn t:

{ }; ; ; ; ; ; ;E NNN NNS NSN NSS SNN SNS SSN SSS=

Bin c { ; ; ;A SSS SSN NSS SNS= ; bin cB = { ;NNN SSS

Xc sut ca A: ( ) 4 18 2

P A = = ; xc sut ca B: P(B) 2 18 4

= =

2.Ta c: { }; ; ; ;A B SSS SSN NSS SNS NNN= v { }A B SSS=

Xc sut ca ( ) 5:8

A B P A B = ; Xc sut ca ( ) 1:8

A B P A B =

Bi 3.Gieo ln lt hai qun xc sc. Tnh xc sut ca cc bin csau y:

1.A: "Tng schm trn hai mt xut hin ca hai qun xc sc 6". 2.B: "C ng mt qun xc sc xut hin schm l sl".

3.C: "Schm xut hin trn hai qun xc sc hn km nhau 2".


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Gii

Khng gian mu ( ) ( ) ( ) ( ){1,1 ; 1, 2 ; 1,3 ; ... 6,6E= : c 6.6 =36 phn t.

1.Bin c: ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ){1,1 ; 1, 2 ; 1,3 ; 1, 4 ; (1,5); 2, 1 ; 2, 2 ; 2,3 ; (2, 4); 3,1 ;A =

( ) ( )3,2 ;(3,3); 4,1 ;(4,2);(5,1) c 15 phn t. Xc sut ca A: ( ) 15 5

36 12P A = = .

2.Bin c: ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ){1,2 ; 1,4 ; 1,6 ; 2, 1 ; 2,3 ; 2,5 ; 3,2 ; 3,4 ; 3,6 ;B=

( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )}4,1 ; 4,3 ; 4,5 ; 5,2 ; 5,4 ; 5,6 ; 6, 1 ; 6,3 ; 6,5 c 18 phn t

Xc sut ca B: ( ) 18 136 2

P B = = .

3.Bin c ( ) ( ) ( ) ( ) ( ) ( ){ }1, 3 ; 3, 1 ; 2, 4 ; 4, 2 ; 3, 5 ; 5, 3 ; (4,6);(6, 4)C= c 8 phn t

Xc sut ca C: ( ) 8 236 9

P C = = .

Ch :Nu gieo ng thi 2 qun xc sc th hai khnng xy ra (p;q) v(q;p) vi p q l nhnhau nn khng gian mu E chc 21 phn t, bin cAc 9 phn t, bin cB c 9 phn tv bin cC c 4 phn t.

Bi 4.Gieo ln lt mt ng xu v mt qun xc sc.

1.Tnh xc sut ca mt bin cA c mt mt sp v mt qun xc sc xuthin l mt schn.

2.Tnh xc sut ca bin cB c mt qun xc sc xut hin l mt snguyn t.

3.Tnh xc sut ca bin cC c mt mt nga v mt qun xc sc xut hinl mt sl.

4.Tnh xc sut ca A B , ca A B v ca A B C .

Gii

1.Khng gian mu c 1 12 6

12C C = phn t:

{ 1; 2; 3; 4; 5; 6; 1; 2; 3; 4; 5; 6E N N N N N N S S S S S S=

Bin c { 2; 4; 6A S S S= . Xc sut ca A: ( ) 3 112 4

P A = = .

2.Bin c { }2; 3; 5; 2; 3; 5B N N N S S S= . Xc sut ca B: ( ) 6 112 2

P B = = .

3.Bin c { }1; 3; 5C N N N =

. Xc sut ca C: ( ) 3 1

12 4P C = =

.4.Bin c { }2; 3; 4; 5; 6; 2; 3; 5A B S S S S S N N N=

Xc sut ca ( ) 8 2:12 3

A B P A B = = .


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Xc sut ca mt bin c

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Bin c { }2A B S= . Xc sut ca ( ) 1:2

A B P A B =

Bin cA B C= . Xc sut ( )P A B C = .

Bi 5.Mt bnh ng 5 vin bi xanh, 7 vin bi v 4 vin bi vng. Ly ngunhiu 4 vin bi.

1.Tnh xc sut c 1 vin bi xanh v 3 vin bi vng.

2.Tnh xc sut c 3 mu.

3.Tnh xc sut c 4 vin bi cng mu.

Gii

Khng gian mu c 416

1820C = phn t.

1.Bin cA ly c 1 vin bi xanh v 3 vin bi vng c 1 35 4. 20C C = phn t.

Vy xc sut l: ( ) 120

1820 91

P A = = .

2.Bin cB ly c 3 mu 1 1 2 1 2 1 2 1 15 7 4 5 7 4 5 7 4. . . . . . 910C C C C C C C C C + + = phn t

Vy xc sut l ( ) 910 11820 2

P B = = .

3.Bin cC ly c 4 vin bi cng mu c: 4 4 45 7 4

41C C C+ + = phn t.

Vy xc sut l: ( ) 411820

P C = .

Bi 6.Mt hp ng 5 tbc 50000v 10 tbc 20000.

1.Ly ngu nhin 2 tbc. Tnh xc sut c 70000.

2.Ly ngu nhin 4 tbac. Tnh xc sut c 140000.Gii

1.Gi A l bin cly c 150000. Ta c:

700000= 50000+ 20000nn 2 tbc phi khc nhau.

A c 1 15 10. 5.10 50C C = = phn t. Khng gian mu c 2

15 105C = phn t.

Vy xc sut l: ( ) 50 10105 21

P A = = .

2.Gi B l bin cly c 140000. Ta c: 140000= 100000+ 40000

Nn B c 2 tbc 50000v 2 tbc 20000.

B c: 2 25 10. 10.45 450C C = = phn t. Khng gian mu c 4

15 1365C = phn t.

Vy xc sut l ( ) 450 301365 91

P B = = .


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Bi 7.on tu thng nht S1c 4 toa i Hu Nng. Ti ga Thanh Hac 6 hnh khch mua v i 4 toa ny. Mi hnh khch c lp vi nhau chnngu nhin mt toa. Tnh xc sut mi toa c t nht mt hnh khch mua v.

Gii

Gi E l tp hp tt ccc dy s ( )1 2 3 4 5 6, , , , ,x x x x x x , trong ix l stoam khch thimua v vi

ix {1, 2, 3, 4 . Khi 6E 4= .

Gi A l bin c"mi toa u c hnh khch mua v". Ta tnh sphn tca

bin cA bng cch phn chia tp A thnh hai tp con1

A v2

A sau y:

i1

A l tp hp cc cch mua v tu sao cho 1 toa c 3 ngi, v 3 toa cn li

mi toa c mt ngi, khi ta c cc khnng sau:

+ C 3 ngi mua v toa 1 + C 3 ngi mua v toa 2

+ C 3 ngi mua v toa 3 + C 3 ngi mua v toa 4

Gisxt khnng c 3 ngi ln toa 1: iu ny c ngha trong dy1 2 3 4 5 6, , , , ,x x x x x x s1 xut hin 3 ln, mi s2, 3, 4 xut hin 1 ln.

Khi scc khnng xy ra l 36.3! 20.6 120C = =

Do 4 khnng 3 ngi mua v 1 toa c vai tr nhnhau nn sphn tca

bin cA1l 1 120.4 480A = =

i 2A l tp hp cc cch mua v sao cho c 2 toa c 2 ngi v 2 toa c 1 ngi.

Khi ta c cc khnng sau:

+ C 2 ngi mua v toa 1 v 2 ngi mua v toa 2






Xt i din khnng c 2 ngi mua v toa 1 v 2 ngi mua v toa 2

iu ny c ngha trong dy1 2 3 4 5 6, , , , ,x x x x x x s1 v 2 xut hin 2 ln, s

3 v 4 xut hin 1 ln. Khi skhnng xy ra l: 2 26 4. .2! 15.6.2 180C C = =

Suy ra 2 6.180 1080A = = v 1 2 1320A A A= + = nn ( ) 1320 165

4096 512

AP A

E= = = .


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Xc sut ca mt bin c

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DNG 2: XC SUT CA NHIU BIN C

Phng php:Sdng cng thc cng xc sut v cng thc nhn xc sut:

( ) ( ) 1P A P A+ =

( ) ( ) ( ) ( )P A B P A P B P A B= + (A, B bt k)

( ) ( ) ( )P A B P A P B= + (A, B xung khc)

( ) ( ) ( ).P A B P A P B= (A, B c lp).

Bi 1.Lp 12C c 30 hc sinh, trong c 5 em gii, 17 em kh v 8 emtrung bnh. Chn ngu nhin 3 em. Tnh xc sut :

1.C 3 em gii. 2.C t nht 1 em gii. 3.Khng c em trung bnh.

Gii

1. ( )3

53

30

10 14064060

CP AC

= = = .

2.Gi B l bin cchn c t nht mt em gii th B l bin cchn khng

c hc sinh gii. Do ( )3

25

3

30

2300 115

4060 203

CP B

C= = = nn ( ) ( ) 881

203P B P B= =

3.C 322C cch chn 3 hc sinh khng trung bnh nn ( )3

22

3

30

1540 11

4060 29

CP C

C= = = .

Bi 2.Cng ty Samsung pht hnh 25 v khuyn mi trong c 5 v trng

thng. Mt i l c phn phi 3 v. Tnh xc sut i l c:

1.Mt v trng. 2.t nht mt v trng.

Gii

1.Gi A l bin cc 1 v trng. Xc sut ca A l: ( )1 2

5 20

3

25

. 1946

C CP A

C= = .

2.Gi B l bin ckhng c v no trng th ( )3

20

3

25

57115

CP B

C= = .

Bin cB l bin cc t nht mt v trng th ( ) ( ) 581 115P B P B= = .

Bi 3.Mt bnh ng 5 vin bi xanh v 4 vin bi mu . Ly ln lt 2 vinbi. Tnh xc sut c vin bi thhai mu .


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Gii

Gi A l bin cly c vin bi xanh ln thnht; B l bin cly c vin

bi ln thnht; C l bin cly c vin bi ln thhai.

Bin cly c vin bi ln thhai l tp hp ca hai bin c: A C v

BC, trong A v C; B v C l hai cp bin cc lp.

Ta c: ( ) 5 549 8 18

P A C = = v ( ) 3 349 8 18

P B C = =

Vy xc sut phi tnh l: 5 3 418 18 9

P= + = .

Bi 4.Hai xthA v B cng nhm bn mt con th. Xc sut xthA

bn trng l 27

; xc sut xthB bn trng l 18

. Tnh xc sut :

1.Chai xthu bn trng. 2.Chmt trong hai ngi bn trng.

3.t nht mt trong hai ngi bn trng. 4.Chai xthu bn trt.

Gii

1.Gi A l bin cxthA bn trng. Xc sut l ( ) 27

P A = ; B l bin cx

thB bn trng. Xc sut l ( ) 18

P B = . Hai bin cA v B c lp.

Vy xc sut hai xthcng bn trng l ( ) ( ) ( ) 2 1 1.7 8 28

P A B P A P B= = =

2.Gi A l bin cxthA bn trt, ta c ( ) ( ) 517

P A P A= = ;

B l bin cxthB bn trt, ta c ( ) ( ) 71 8P B P B= = .

Bin cmt ngi bn trng l hp ca hai bin cA B v A B .

Vy xc sut mt ngi bn trng l:

( ) ( ) 5 7 5 191 2 147 8 7 8 56 56 56

P P A B P A B= + = + = + = .

3. l bin cA B : ( ) ( ) ( ) ( ) 32 1 17 8 28 8

P A B P A P B P A B= + = + = .

4.Xc sut phi tm l ( )P A B . M A v B c lp nn A v B c lp.

Do : ( ) ( ) ( ) 5 7 5.7 8 8

P A B P A P B= = =

Cch khc: ( ) ( ) ( ) 3 51 18 8

P A B P A B P A B= = = =


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Xc sut ca mt bin c

275

Bi 5.Hai ngi A, B cng bn mt con chim. Xc sut ca A bn trng l 0,7ca 2 ngi cng bn trng 0,42 v ca con chim bbn trng l 0,88.

1.Tnh xc sut B bn trng?

2.Kim chng rng hai bin cA v B l c lp.

Gii1.Gi A l bin cngi A bn trng

A B l bin chai ngi cng bn trng, th A B l bin ccon chim bbn

trng ( ) ( ) ( ) ( ) ( ) ( )0,88 0,7 0, 42 0,6P A B P A P B P A B P B P B= + = + =

2.Ta c P(A).P(B) = 0,7.0,6 = 0,42 = P(A B). Vy A v B c lp nhau.

Bi 6. Ba ng v ba b ngi trn mt dy 6 gh.

1.Tnh xc sut ngi cng phi ngi gn nhau.

2.Tnh xc sut ba b ngi gn nhau.

3.Tnh xc sut hngi nam nxen knhau

Gii

1.K hiu tt ng l M v b l W. Khng gian mu E c 6! = 720 phn t.

C 2 cch xp ngi cng phi ngi gn nhau: MMMWWW, WWWMMM

C 3! = 6 cch ngi ca 3 ng v c 3! = 6 cch ngi ca 3 b.

Vy xc sut phi tnh l: 2.3!3! 2.6.6 16! 720 10

P= = =

2.C 4 cch sp xp ba b ngi gn nhau: MMMWWW, MMWWWM,MWWWMM,WWWMMM. C 3! = 6 cch sp xp 3 ng v c 3! = 6 cch

sp xp 3 b. Vy xc sut phi tnh l:4.3!3! 1

6! 5P= =

3.C 2 cch sp xp ba ng v ba b ngi xen knhau: MWMWMW,WMWMWM. C 3! = 6 cch sp xp ba ng v c 3! = 6 cch sp xp ba b.

Vy xc sut phi tnh l: 2.3!3! 16! 10

P= =

Bi 7.Mt hp ng 4 bi vng, 3 bi xanh, 2 bi trng v 1 bi , cc bi ny chkhc nhau vmu sc. Ly ngu nhin 3 bi cng mt lc. Tnh xc sut c 3vin bi khc mu trong phi c bi vng.

Gii

Gi 1A l bin cly c 1 bi vng, 1 bi xanh, 1 bi trng: 1 1 14 3 2C C C

Gi2

A l bin cly c 1 bi vng, 1 bi xanh, 1 bi : 1 1 14 3 1

C C C

Gi3

A l bin cly c 1 bi vng, 1 bi trng, 1 bi : 1 1 14 2 1

C C C


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276

Ta c:1 2 3

A A A A= vi1 2 3, ,A A A xung khc nhau, nn

( ) ( ) ( ) ( )1 1 1 1 1 1 1 1 1

4 3 2 4 3 1 4 2 1

1 2 3 3

10

1130

C C C C C C C C C P A P A P A P A

C

+ += + + = =

Bi 8. a.Gieo mt qun xc sc lin tip 6 ln. Tnh xc sut t nht c mt

ln ra mt "lc".

b.Gieo mt cp hai qun xc sc 2009 ln. Tnh xc sut t nht c mt lnchai con u ra mt "lc".

Gii

a.Gii

A l bin c"ln thixut hin mt lc". 1, 6i= . Ta c ( ) 16i

P A =

1,6i = . Khi bin cii

A "ln thikhng xut hin mt lc" c xc

sut ( ) 5 , 1,66i

P A i= =

Gi A l bin c"t nht c mt ln ra mt lc", th bin ci A l bin c"c6 ln u khng xut hin mt lc".

R rng1 2 3 4 5 6. . . . .A A A A A A A= . R rng cc bin c

iA l c lp.

Theo quy tc nhn, ta c ( ) ( ) ( ) ( ) ( )6

1 2 6

5...6

P A P A P A P A= = .

Theo cng thc xc sut ca bin ci, ta c ( ) ( ) ( )6

51 16

P A P A= =

Vy xc sut c t nht mt ln ra lc l ( )

65

1 6

.

b.Nui

A l bin c"Ln thichai mt u ra lc" th ( ) ( )2

1 1 , 1,246 36i

P A i= = =

Khi bin cii

A , "ln thi khng xut hin chai mt lc", c xc sut

l ( ) ( ) 3511 1 , 1, 2436 36i i

P A P A i= = = =

Gi A l bin c"C t nht mt ln chai mt u ra mt lc", th A l binc" c2009 ln, khng c ln no ra chai mt lc".

Theo quy tc nhn ( ) ( ) ( ) ( ) ( )

2009

1 2 2009

35... 36P A P A P A P A= =

Theo cng thc tnh xc sut ca bin ci, th ( ) ( ) ( )2009

351 136

P A P A= =


13/14

Xc sut ca mt bin c

277

Bi 9. Gieo ng thi 3 qun xc sc. Anh l ngi thng cuc nu c xuthin t nht "2 mt lc". Tnh xc sut trong 5 vn chi anh thng t nht l ba vn.

Gii

Gii

A l bin c"Ln thithng". R rng khi tung 1 qun xc sc, xc sut

xut hin mt lc l 16 (v do xc sut khng xut hin mt lc l56 ).

Ln thithng khi

Hoc l c3 qun xc sc ra mt lc. iu ny xy ra vi xc sut ( )3

1 16 216

=

Theo quy tc cng xc sut, ta c ( ) 1 15 2 , 1,7216 27i

P A i+= = =

Vy bin cii

A "Ln thithua" l ( ) ( ) 25127i i

P A P A= =

Gi A l bin c"t nht thng ba vn".

R rng t nht thng ba vn, ta cn c

Cnm vn u thng. iu ny xy ra vi xc sut ( )5

227

.

C 4 vn thng, 1 vn thua th xc sut l ( ) ( ) ( ) ( )4 4

4

5

25 252 2. 5.27 27 27 27

C =

C 3 vn thng, 2 vn thua th xc sut l ( ) ( ) ( ) ( )2 23 3

3

5

25 252 2. 10.27 27 27 27

C =

Theo quy tc cng, ta c ( ) ( ) ( ) ( ) ( ) ( )5 4 3

5

25 25 520322 2 25 1027 27 27 27 27 27

P A = + + =

Bi 10.Bn lin tip vo mc tiu cho n khi c mt vin n u tin trngmc tiu th ngng bn. Tm xc sut sao cho phi bn n vin n th3, bitrng xc sut trng mc tiu ca mi ln bn l nhnhau v bng 0,4.

Gii

Gi A i l bin c"bn n vin th i mi trng mc tiu"

A l bin c"bn n vin th3 th ngng li", suy ra A = 1 2 3A A A

Cc bin c 1 2 3A , A , A khng c lp nhau v vic xy ra bin c A i snh

hng n khnng xy ra 1A i+ : ( ) ( )1 1A / A 0; A / A 0, 4i i i iP P+ += =

( ) ( ) ( ) ( ) ( ) ( ) ( )1 2 1 3 1 2 1 2 1 3 1 2. / . / 1 1 / / P A P A P A A P A A A P A P A A P A A A = =

Mt khc 2 1A A nn 1 2 2A A A= suy ra ( ) ( ) ( )1 0,4 1 0,4 .0,4 0,144P A = =


14/14


278

Bi 11. Mt cp trsinh i c thdo cng mt trng (sinh i tht), hay do

hai trng khc nhau sinh ra (sinh i gi). Cc cp sinh i tht lun lun c

cng mt gii tnh; cc cp sinh i gith gii tnh ca mi a trc lp

vi nhau v c xc sut l 0,5 con trai. Thng k cho thy 34% cp sinh i

u l trai, 30% cp sinh i l gi v 36% cp sinh i c gii tnh khc nhau.

1.Tnh xc sut cp sinh i tht.

2.Giskhi chn ngu nhin mt cp sinh i th c cp c cng gii tnh.

Tnh xc sut cp sinh i l cp sinh i tht.

Gii

1.Gi A l bin c"cp sinh i cng gii";1

B l bin c"cp sinh i tht"

v2

B l bin c"cp sinh i gi". Tgithit ( ) 0,34 0,30 0,64P A = + =

Nu c cp sinh i tht th chng lun c cng gii tnh nn ta c1

1APB

=

Nu c cp sinh i gith gii tnh ca mi a trc lp vi nhau v c

xc sut l 0,5 con trai nn ta c2

12

APB

=

.

t ( )1P B x , th ( )2 1P B x= . Ta c: ( ) ( ) ( )1 21 2

A AP A P P B P P BB B

= +

( )10,64 1 1, 28 2 1 0, 282

x x x x x = + = + =

Vy xc sut c c cp sinh i tht l 0,28.

2.Trong cc cp sinh i c cng gii tnh th xc sut xy ra cp sinh i tht

l( )

( )

11 P B AB

PA P A

=

. V mt bin csinh i tht cng l bin csinh i cng

gii nn ( ) ( )1 1 0,28P B A P B= = . T suy ra1 0,28 7 0,4375

0,64 16

BP

A

= = =

Nhn xt. C thsdng bin i sau y:

( ) ( ) ( )

( ) ( )

1

1 1 11 1

P ABAP B P P B P ABB P B

= =

( ) ( )( )

1 11 / 0,28.1

0,43750,64

P B P A BBP

A P A

= = =

.

Bi 12. Mt chung g c 9 con mi v 1 con trng. Chung g kia c 1 con

mi v 5 con trng. Tmi chung ta bt ra ngu nhin mt con lm tht. Cc

con g cn li c dn vo mt chung thba. Tchung thba ny li bt

ngu nhin mt con g. Tnh xc sut ta bt c g trng tchung thba.

1.5.2 xac suat cua mot bien co

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