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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
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( ) 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
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-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