CHNG I. I CNG V XC SUT1:Bin c v quan h gia cc bin c

1. Php th v bin c.

2. Phn loi bin c : gm 3 loi

- Bin c chc chn:

- Bin c khng th c hay khng th xy ra:

- Bin c ngu nhin: A, B, C

- Bin c ngu nhin: A, B, C

3. So snh cc bin c.

nh ngha 1.1: (A nm trong B hay A ko theo B) nu A xy ra th B xy ra.Vy

A B

A BA B

B A

1

nh ngha 1.2: A c gi l bin c s cp , .B A B A

4. Cc php ton trn bin c (hnh 1.1 v 1.2 ):

xy ra khi v ch khi A xy ra v B xy ra.

xy ra khi v ch khi A xy ra hoc B xy ra.A B A B

.A B A B

xy ra khi v ch khi A xy ra v B khng xy ra.

xy ra khi v ch khi A khng xy ra.

A B

A A

2

Hnh 1.1 Hnh 1.2

3

Cc php ton ca bin c c tnh cht ging cc php ton ca tp hp, trong c cc tnh cht i ngu:

Ngn ng biu din: tng = c t nht mt ;tch = tt c u.

(A = c t nht 1 phn t c tnh cht x) suy ra (khng A = tt

,i i i ii ii i

A A A A

4

(A = c t nht 1 phn t c tnh cht x) suy ra (khng A = tt c u khng c tnh cht x).

V d 1.1: (A = c t nht 1 ngi khng b ln) suy ra( khng A = tt c u ln).

nh ngha 1.3: bin c A v B c gi l xung khc vi nhau nu

.A B

2: Cc nh ngha xc sut.

1. nh ngha c in v xc sut

nh ngha 2.1: gi s trong mi php th cc kt cc l ng kh nng v c tt c n kt cc nh vy. K hiu m l s cc kt cc thun li cho bin c A. Khi y xc sut ca bin c A l:

( )m

A

V d 2.1: Trong 1 hp c 6 bi trng, 4 bi en.Ly ngu nhin ra 5 bi. Tnh xc sut ly c ng 3 bi trng.

Gii ( phn phi siu bi)

( )An

3 26 4

510

.C C

C

5

Ch : ly 1 lc 5 bi ging ly ln lt 5 bi khng hon li

V d 2.2: C 10 ngi ln ngu nhin 5 toa tu. Tnh xc sut toa th nht khng c ngi ln:

2. nh ngha hnh hc v xc sut:

nh ngha 2.2: Gi s trong mi php th cc kt cc l ng

kh nng v c biu din bng cc im hnh hc trn min

1 0

1 0

4

5

.kh nng v c biu din bng cc im hnh hc trn min

K hiu D l min biu din cc kt cc thun li cho bin c A. Khi y xc sut ca bin c A l:

( o l di,din tch

hoc th tch)

.

6

o D( )

o

oP A

o

V d 2.3: Chia on AB c nh ngu nhin thnh 3 on. Tnh xc sut 3 on lp thnh 3 cnh ca 1 tam gic.

Gii: Gi di on th 1,2 l x,y.Khi y on th 3 l l-x-y

0 , 0x y

x y l

l

7

2

1( )

2 4

2

lx y

x y l x yl

D x l x y y y A

y l x y xl

x

HNH 2.1

8

V d 2.4: Nm ln mt phng c k nhng ng thng song

song cch nhau 1 khong l 2a mt cy kim c di 2t

HNH 2.2

10

HNH 2.3

11

Cc tnh cht ca xc sut : xem sch gio khoa

3. nh ngha xc sut theo tin

nh ngha 2.3: K hiu l tp hp cc bin c trong 1 php th. Ta gi xc sut l 1 quy tc t mi bin c A vi 1 s P(A) tha mn cc tin :

(I)

(II)

0 1P A ( ) 1, 0P P

(II)

(III) Vi mi dy bin c i mt xung khc,ta c:

H qu :

4.nh ngha xc sut theo thng k:xem sch gio khoa

( ) 1, 0P P

1 1

i ii i

A A

12

( ) 1 ( )P A P A

3: Cc nh l xc sut

1: nh l cng xc sut

nh l 3.1(hnh 3.1): P(A+B) = P(A) + P(B) P(AB)

V d 3.1: C 10 ngi ln ngu nhin 5 toa tu. Tnh xc

sut toa th nht hoc toa th hai khng c ngi ln.

13

A l bin c toa th 1 khng c ngi ln,

B l bin c toa th 2 khng c ngi ln. Ta c :

1 0 1 0 1 0

1 0 1 0 1 0

( ) ( ) ( ) ( )

4 4 3

5 5 5

A B P A P B P A B

HNH 3.1

14

nh l 3.1

1 1 21 1

... ( 1) ( ... )n n

ni i i j i j k n

i i i j i j k

A A AA AAA P AA A

V d 3.2: C k ngi ln ngu nhin n toa tu (k>n). Tnh xc sut tt c cc toa u c ngi ln

Bi gii A - tt c cc toa u c ngi ln

Ch : v phi trong tng th 1 c s hng, trong tng th 2 c s hng,, trong tng th k c s hng,, trong tng th n c s hng.

1nC

nnC

knC

2nC

15

Bi gii A - tt c cc toa u c ngi ln

- c t nht 1 toa khng c ngi ln.

- toa th i khng c ngi ln, i =1, 2,n

1

n

ii

A

iA

V cc toa tu c vai tr nh nhau nn p dng cng thc cng xc sut ta c :

1 2 31 1 2 1 2 3

11 2

1 2 3 1

. . .

... ( 1) ( ... )

1 2 3 1... 1 . 0

n n n

nn

k k k kn n

n n n nk k k k

C A C A A C A A A

P A A A

n n nC C C C

n n n n

1 1 21 1

... ( 1) ( ... )n n

ni i i j i j k n

i i i j i j k

A A AA AAA P AA A

... 1 . 0n n n nk k k kC C C Cn n n n

16

1

V d 3.3: C n bc th b ngu nhin vo n phong b c sn a ch.a)Tnh xc sut c t nht 1 bc th ng a ch.b) Tnh xc sut ch c ng 1 bc th ng a ch

Bi gii A - C t nht 1 bc ng.

- Bc th i ng

V cc bc th c vai tr nh nhau nn p dng cng thc cng xc sut ta c :

i 1

n

ii

A A

1 1 21 1

... ( 1) ( ... )n n

ni i i j i j k n

i i i j i j k

A A AA AAA P AA A

1 2 31 1 2 1 2 3

11 2

1 2 3 1

1 1

. . .

... ( 1) ( ... )

1 ! 2 ! 3 ! 1!... 1 .

! ! ! !

1 1 1 1 11 . 1 ... 1 .

! 2 ! 3 ! 4 ! !

n n n

nn

n nn n n n

n n

C A C A A C A A A

P A A A

n n nC C C C

n n n n

n n

17

-Khng c bc no ng a ch trong n bc th

-Ch c ng 1 bc ng a ch trong n bc th

nB

nC

1 1 1 1

( ) 1 ( ) ... 1 .2! 3! 4! !

n

nP B P An

nB A

18

( ) 1 ( ) ... 1 .2! 3! 4! !

nP B P An

1 1

1

( ) . ( ). ( )

1 1 1 1... 1 .

2! 3! 4! ( 1)!

n n

n

P C n P A P B

n

2. nh l nhn xc sut

nh ngha 3.2: Xc sut ca bin c B khi bit rng bin c A xy ra c gi l xc sut ca B vi iu kin A v k hiu l P(B/A).

Ch : bin c A c th xy ra trc, ng thi hoc sau B

Ngn ng biu din: P(B/A) = xc sut B bit (nu)A hoc Cho Ngn ng biu din: P(B/A) = xc sut B bit (nu)A hoc Cho A tnh xc sut B.

nh l 3.2: P(AB)=P(A).P(B/A)=P(B).P(A/B)

H qu:

1 2 1 2 1 3 1 2 1 2 1. ... . / . / ... / ...n n n

. //

19

nh ngha 3.3: Hai bin c A,B c gi l c lp vi nhau nu xc sut ca bin c ny khng ph thuc vo vic bin c kia xy ra hay cha trong 1 php th.

nh ngha 3.4: Mt h cc bin c c gi l c lp ton phn nu mi bin c ca h c lp vi 1 t hp bt k ca cc bin c cn li.

nh l 3.3: A, B c lp khi v ch khi P(AB)=P(A).P(B)

20

nh l 3.4: Gi s l c lp ton phn. Khi y ta c:

, 1,i i n

1 1

1 1

1 . ( )

2 . ( ) 1

n n

i ii i

n n

i ii i

A

A

Ch : Trong trng hp c lp khng nn dng cng thccng xc sut m nn dng cng thc nhn xc sut.

V d 3.3: 1 mng gm n chi tit mc ni tip.Xc sut hng ca chi tit th i l . Tnh xc sut mng hng.

Gii: - bin c chi tit th i hng

A - bin c mng hng

Vy xc sut mng hng l:

i

1

n

ii

iP

Vy xc sut mng hng l:

Ch :

1 21

1

1 1 1 1 ... 1n n

i i ni

i

21

1 2( ) 1 1 ... 1 nP A

V d 3.4: Tung 3 con xc xc cn i,ng cht. Tnh xc sut

:

1. Tng s chm bng 9 bit c t nht 1 mt 1 chm

2. C t nht mt mt 1 chm bit s chm khc nhau tng i mt.

Gii:

1. Gi A l c t nht 1 mt 1 chm.

22

1. Gi A l c t nht 1 mt 1 chm.

B l tng s chm bng 9

C l cc s chm khc nhau tng i mt

3 3

3

3

6 5

6

1 5

6

3

3 3 3

15 6 15/ .

6 6 5 91

S cch c t nht mt mt 1 chm v tng bng 9:

1+2+6 suy ra c 3! cch

1+3+5 suy ra c 3! cch

1+4+4 suy ra c 3 cch

Suy ra c 15 cch c t nht mt mt 1 chm v tng bng 9

2.

23

2.

3

3

6 .5 .4

6

3 .5 .4

6

C

C

( ) 1

/( ) 2

P ACC

P C

V d 3.5: T 1 hp c 10 bi trng , 6 bi en ,ngi ta ly lnlt khng hon li tng bi cho n khi c 5 bi en thi dngli.Tnh xc sut ln th 3 ly c bi trng nu bit rng a dng li ln th 9.

Gii:

Gi A l dng li ln th 9, B l ln th 3 ly c bi trng

24

: 4 4Am T

4748

( / ) AB

A

CmP B A

m C

: 3 4ABm T

3. Cng thc xc sut y v cng thc Bayes:

nh ngha 3.5: H c gi l h y , nu trong mi php th nht nh 1 v ch 1 trong cc bin c Hi xy ra.

nh l 3.4: Gi s l h y . Ta c:

, 1,iH i n

, 1,iH i n

(cng thc y ).

(cng thc Bayess)

( )

1

( ). ( / )

iP AHn

i ii

A P H P A H

. // , 1,i i ii

H H HH i n

25

Ch :

1.

2.

1

/ / /n

i ii

H H

/

2.

Vi:

/

1

/n

i ii

H H

26

V d 3.5: C 2 hp bi cng c, hp 1 cha 4 bi trng v 6 bi xanh, hp 2 cha 5 bi trng v 7 bi xanh.Ly ngu nhin 1 hp, t hp ly ngu nhin1 bi th c bi trng. Tm xc sut vin bi tip theo, cng ly t hp trn ra l bi trng.

Gii:. Ly ngu nhin 1 hp: H1 ly c hp 1

H2 ly c hp 2

1/ 2H H

Hp 1: 4t + 6x , Hp 2: 5t + 7x

A- bin c ly c bi trng ln 1

B- bin c ly c bi trng ln 2

1 2 1/ 2H H

/

27

Cch 1:

1 1 2 2

1 11

/ /

1 4 1 5. .

2 10 2 12

1 4./ 2 10/( )

H H H H

H HH

P A

1

2 22

1 1 2 2

3/9 4/11

/( )

1 5./ 2 12/( )

/ / . / / . /

HP A

H HH

P A

H H H H

28

Cch 2:

1 1 2 2/ /1 4 1 5

. .2 10 2 12

H H H H

/

1 1 2 2

1 1 1 2 2 2

2 10 2 12

. / . /

. / . ( / ) . / . ( / )

1 4 3 1 5 4. . . .

2 10 9 2 12 11

H H H H

H H P B AH H H P B AH

29

Ch

Nu sau ln 1 ly c bi trng ta tr bi vo hp ri mi ly tip ln 2 th li gii thay i nh sau:

3 4;

9 1 0

4 5

1 1 1 2

P(B)=P(A), trong c 2 bi ton.

Nu cu hi l :Gi s ln 1 ly c bi trng tnh xc sut bi ly c hp 1, th p s l:

1 1 1 2

1( / )P H A

30

V d 3.6: C 1 tin tc in bo to thnh t cc tn hiu(.)v (-). Qua thng k cho bit l do tp m, bnh qun 2/5 tn hiu(.) v 1/3 tn hiu(-) b mo. Bit rng t s cc tn hiu chm v vch trong tin truyn i l 5:3. Tnh chm v vch trong tin truyn i l 5:3. Tnh xc sut sao cho nhn ng tn hiu truyn i nu nhn c chm.

31

Gii :H1 l bin c truyn i chm,

H2 l bin c truyn i vch.

Gi A l bin c nhn c chm .

1 1 2 2. / /5 3 3 1 1

H H H H

1 2

5 3( ) , ( )

8 8P H P H

1 1

1

5 3 3 1 1. .

8 5 8 3 2

5 3./ 38 5/1 42

H HH

32

4. Cng thc Bernoulli:

nh l 3.5: Gi s trong mi php th 1 bin c A c th xut hin vi xc sut p (khi A xut hin ta quy c l thnh cng). Thc hin n php th ging nhau nh vy. Khi y xc sut c ng k ln thnh cng l :

(Phn phi nh thc)

Ch 1 : t nay tr i ta k hiu q=1-p

Ch 2: Thc ra cng thc c dng

, , . . , 0,1,...,k k n knn k p C p q k n

Ch 2: Thc ra cng thc c dng

nh ngha 3.6: K hiu k0 l s sao cho:

Khi y k0 c gi l s ln thnh cng c nhiu kh nng xut hin nht(tc l ng vi xc sut ln nht)

0, , , , , 0n k p Max n k p k n

33

nh l 3.6: hoc 0 1k n p 0 1 1k n p

, , . . . , 0,1,...,k k n k n kn n kn k p C p C q k n

Ch :

V d 3.7: Tung cng lc 20 con xc xc.

1. Tnh xc sut c ng 4 mt lc xut hin.

2. Tnh s mt lc c nhiu kh nng xut hin nht.

3. Tnh xc sut c t nht 1 mt lc.

4. Tnh s con sc sc t nht cn tung xc sut c

, ,1/ 2 . 0,5nk

nn k C

t nht 1 mt lc khng nh hn 0,99.

Gii:

4 16420

0 0

1) 20,4,1/ 6 1/ 6 . 5 / 6

2) 20 1 / 6 3 2

C

k k

34

203) 1 (5 / 6) 0,9739P

4)1 (5 / 6) 0,99 (5 / 6) 0,01 26n n n

V d 3.8:Trong 1 hp c N bi trong c M bi trng cn li l en. Ly ngu nhin ln lt tng bi c hon li ra n bi. Khi y xc sut ly c ng k bi trng c tnh bng cng thc Bernoulli ni trn vi p = M/N (phn phi nh thc) :

V d 3.9:Trong 1 hp c N bi trong c M bi trng cn li l en. Ly ngu nhin ln lt tng bi khng hon li ra n bi.

, , . . 1 , 0,1,...,k n k

k n kn n k

M M Mn k C C k n

N N N

en. Ly ngu nhin ln lt tng bi khng hon li ra n bi. Khi y xc sut ly c ng k bi trng l

(Phn phi siu bi)

Ch : y N-M l s bi khng trng.

Ch : Ly bi : + Khng hon li l siu bi

+ C hon li l nh thc.

35

. , 0 ,k n kM N M

nN

C Ck n

C