Sinh vin kinh t quc dn Sa mt s bi tp chng 1 1.4 a) Gi H1 l bin c
qu th 1 l trng H2 l bin c qu th 2 l en A l bin c qu th 2 l trng H1
v H2 l mt nhm y cc bin c nn theo cng thc XS y c:
PA=PH1PAH1+P(H2)P(A|H2) =aa+b a-1a+b-1+ ba+b aa+b-1 = aa+b b) Gi B
l bin c qu cui cng l trng S kt cc duy nht ng kh nng ca vic ly ra
a+b qu cu l m=(a+b)! C a cch chn qu cui cng l trng v (a+b-1)! cch
ly nhng qu cn li n=a*(a+b-1)! Vy P(B) = m/n = a(a+b-1)!(a+b)! =
aa+b 1.91 Gi Hi l bin c lc u hp c I sn phm tt ( i=0,n ) Do mi gi
thit v trng thi cu thnh ban u ca hp l ng xc xut nn P(Hi)= 1n+1 Gi A
l bin c ly c sn phm tt P(A/Hi) = i+1n+1 H0, H1,, Hn l mt nhm y cc
bin c nn Theo cng thc xc xut y : P(A) = i=0nP(Hi)P(A|Hi) =1n+1 (
1n+1 + 2n+1++n+1n+1 ) =n+22(n+1) 1.27 a) V mt nm c 366 ngy khc nhau
( tnh c 29/2) nn s kt cc duy nht ng kh nng v ngy sinh ca 3 ngi l
chnh hp lp chp 3 ca 366 phn t: n = 366^3 Gi A l bin c 3 ngi c ngy
sinh khac nhau S kt cc thun li cho A l: A3663 P(A) = m/n 0.992 b)
Gi B l bin c 3 ngi c ngy sinh trng nhau => m = 366 => P(B) =
366/(366^3) 7.510^- 6 1.28 Ta c: 9 sn phm ch v np 6 sn phm ch st vo
3 sn phm ch m ming a. S sn phm b khuyt tt l: 9+ 6+3+6+2+4+1= 31 Tng
s sn phm l 100 nn: Pa = 0.31 b. m = 6 => Pb = 0.06 c. Gi H1 l
bin c sn phm b v np H2 l bin c sn phm b st vi C l bin c sn phm b st
vi bit rng n b v np => C=H2/H1
=>P(C) = P(H2/H1) = 7/20 = 0.35 ( y l trng hp xc xut c iu kin
) 1.32 ( cch khc) S kt cc duy nht ng kh nng 10 ngi vo 3 quy l chnh
hp lp chp 10 ca 3 v = 3^10 Gi A l bin c c 3 ngi vo quy 1 => s kt
cc thun li cho A l :C103.27 Vy P(A) = C10327310 1.34 Gi Ak l bin c
ngi th k bn trng mc tiu ( k=1,3 ) a) A1A2 A3 l bin c ch c ngi thc
nht bn trng mc tiu b) v c) cc bn trc lm ng d) A1 A2 A3 l bin c c
ngi bn trng mc tiu 1.43 Gi A l bin c sau khi gia cng chi tit c
khuyt tt Xc xut chi tit cng on thc i khng c khuyt tt l 1- Pi (
i=1,k ) Vy P(A) = 1- P(A ) = 1- i=1k(1-Pi) 1.44 ( cch khc d hiu hn
) Sau khi ly ra k qu ri b li vo hp th cn n-k bng mi S cch ly tip k
qu bng mi l: Cn-kk Tng s cch ly tip k qu bng l: Cnk => P=
Cn-kkCnk 1.81 a) chia hp thnh 3 phn bng nhau, ta c : C9
3C63C33=1680 cch Gi A l bin c c 3 qu cu trong 1 phn => s kt cc
thun li cho A l: 3.C63C33=60 Vy P(A)= 60/1680 0.036 b) Gi B l bin c
mi phn c 1 qu cu => s kt cc thun li cho B l: 3!.C62C42C22=540 Vy
P(B)= 540/1680 0.321
Chng I: BIN C V XC SUT CA BIN C Bi 1.1: a) Gi A l bin c xut hin
mt su chm khi gieo con xc xc. S kt cc ng kh nng n = 6. S kt cc thun
li cho bin c A l m =1. Vy: P(A)= = b) Gi B l bin c mt c s chn chm
xut hin.
S kt cc thun li cho B l n = 3. Vy: P(B) = = = 0.5 Bi 1.2: a) Gi
A l bin c ly ra tm ba c xut hin ch s 5. khi l bin c khng xut hin ch
s 5. V s kt cc ng kh nng l 100, trong khi s kt cc thun li cho A l
19, nn s kt cc thun li cho l 81. Vy P ( ) = 0.81. b) t 1 n 100 c 50
s chn nn c 50 s chia ht cho 2. C 20 s chia ht cho 5, trong 10 s va
chia ht cho 5 va chia ht cho 2. Do vy s kt cc thun li cho bin c ly
ln ba c s hoc chia ht cho 2, hoc chia ht cho 5, hoc chia ht cho c 2
v 5 l 50 +20-10 = 60. Vy P(A)= =0.6.
Bi 1.3: a) A = qu cu th nht l trng S kt cc duy nht ng kh nng l
tt c cc phng php ly c 1 qu cu ra khi (a+b) qu cu. Vy n = a+b. S kt
cc thun li ly ra qu cu th nht mu trng l a. Vy xc sut P(A) = b) Nu
qu th nht trng th chn qu th 2 s cn a+b-1 kt cc ng kh nng. S kt cc
thun li qu th 2 mu trng l a-1 Vy xc sut P(B) = c) tng t cu b), v qu
th hai l trng nn s kt cc ng kh nng khi chn qu th nht l a+b-1 trong
khi s kt qu thun li l a-1. Vy P(C) = Bi 1.4.
a) S kt qu ng kh nng thc ra hon v ca a + b qu cu nn m = (a+b)!
nu qu cu th 2 l trng th s kt qu thun li cho bin c ny l chnh hp chp
a+b-1 phn t ca a+b phn t. Vy xc sut P(A) = . b) gi B l bin c qu cu
cui cng l trng. Khi tng t cu a), ta cng c xc sut P(B) =
Bi 1.5. 1 2 Sp (S) Nga (N) Sp (S) SS NS Nga (N) SN NN
a) Da vo bng trn, c th thy s kt cc ng kh nng l 4. S kt cc thun
li cho bin c A = Hai mt cng sp xut hin l 1. Vy P (A) = = 0,25 b) S
kt cc thun li cho bin c B = Mt sp mt nga l 2. Vy xc sut P(B) = =
0,5. c) S kt cc thun li cho bin c C = C t nht 1 mt sp l 3. Vy P(C)
= 0,75. Bi 1.6. Gieo ng thi 2 con xc xc th s kt cc ng kh nng l
6.6=36. a) c 6 kt cc thun li cho bin c A=hai mt c tng s chm bng 7 l
cc cp 1&6, 2&5, 3&4, 4&3, 5&2, 6&1 nn P(A)
= = b) B = hai mt c tng s chm nh hn 8. Bi =hai mt c s chm nhi hn i,
vi i=2,3,7 (nh nht l 2) B2 c 1 kt cc thun li B3 c 2 kt cc thun li
B4 c 3 kt cc thun li B5 c 4 kt cc thun li B6 c 5 kt cc thun li B7 c
6 kt cc thun li Vy s kt cc thun li ca bin c B l 21. Nn P(B) = = c)
D = hai mt c t nht 1 mt 6 chm khi s kt cc thun li cho D l 11. Vy
P(D) = . Bi 1.7.
S cch tr m c th xy ra l A =6. d) c 3 ngi cng c tr ng m th ch c 1
kt cc thun li nn P(D)= c) Khng th c kh nng ch c ng 2 ngi c tr ng m,
v chc chn ngi th 3 cng s ng m, nn P( C) = 0 b) s kt cc thun li c ng
1 ngi c tr ng m l 3 Vy P(B) = = 0,5. a) xc sut c 3 ngi b tr sai m
l: P(A) = 1 - - 0,5 = Bi 1.8. Ta c biu tp hp nh sau:
a) S hc sinh hc t nht 1 ngoi ng trn l 50% + 10% + 15% + 5% =
80%. Vy xc sut ca bin c ny l P(A) = 0,8. b) S hc sinh ch hc ting
Anh v ting c l 10%. Vy P(B) = 0,1. c) S hc sinh ch hc ting Php l
15% Vy xc sut l P( C) = 0,15. Bi 1.9. S kt cc ng kh nng l chnh hp
chp 3 ca 10 s t nhin, nn m = A = 720.
Ch c 1 kt cc thun li cho vic gi in ng s in thoi, vy xc sut
P=
Bi 1.10. S kt cc ng kh nng khi thc hin php th ly ngu nhin 3 sn
phm l t hp chp 3 ca 15 phn t, vy n = C = 455 a) A = 3 chi tit ly ra
t tiu chun th s kt cc thun li cho A l C =120 P(A) = = 0,264 b) B =
ch c 2 chi tit t tiu chun th s kt cc thun li cho B l C .5 = 225. P
(B) = = 0,495
Bi 1.11: S kt qu ng kh nng: P(6) = 6! A= Xp c ch NGHNH Ch N c 2
cch chn Ch H c 2 cch chn Ch G, , N mi ch c 1 cch chn S kt qu ng kh
nng xy ra A l: m = 2.2.1.1.1 = 4 P(A) = = Bi 1.12 a) Mi khch u c kh
nng ra 6 tng cn li ca ta nh. Do s kt cc ng kh nng n = 63 =216 A =
Tt c cng ra tng 4, m=1 P(A) = b) B= Tt c cng ra 1 tng Tt c u c kh
nng ra 6 tng cn li ca ta nh. Do : P(B) = 6P(A) = = c) C = Mi ngi ra
1 tng khc nhau P(C) = = Bi 1.13 S kh nng c th xy ra: P(12) =12! A =
Cc tp c xp th t t phi sang tri hc t tri sang phi nn m = 2 P(A) = Bi
1.14 S kh nng c th xy ra: a) A= Ly c 3 qun t m= =4 P(A) = = b) B=
Ly c 1 qun t P(B) = = = Bi 1.15 Chia ngu nhin l hng thnh 2 phn bng
nhau tc l ly ngu nhin 5 sn phm t 10 sn phm . Do s kh nng c th xy ra
l: n = = 252 Mi phn u c s chnh phm nh nhau tc l mi phn c 3 chnh
phm, 2 ph phm. Do , m = = 120 A = Mi phn c s chnh phm nh nhau =
22100
P(A) = = Bi 1.16 Mi v tr u c th nhn gi tr t 0 n 10 S kh nng c th
xy ra l: n = 105 a) A= C 5 ch s khc nhau P(A) = = 0.3024 b) B= C 5
ch s u l mi v tr u c th nhn gi tr 1, 3, 5, 7 ,9. Do : m= 55 P(B) =
=0.03125 Bi 1.17 S kh nng c th xy ra: P(5) = 120 a) A= C ngi chnh
gia m=1.P(4) = 24 P(A) = = 0.2 b) B= A v B ngi 2 u gh P(B) = = 0.1
Bi 1.18 S kh nng c th xy ra: A= Ly c 1 qu c s hiu nh hn k v 1 qu c
s hiu ln hn k m= P(A) = = bi 1.19 S kt qu ng kh nng l:n= 6n A = Tng
s chm l n+1 S kt qu thun li cho A l m=n P(A) = Bi 1.20 f= 0,85
n=200 k = f.n = 0,85 . 200 =170 Bi 1.21 Gi A l bin c Sinh c con
trai. Theo bi ra ta c, xc sut sinh c con trai l :
P ( A) =
45600 = 0,517 88200
Bi 22: Ln tung 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. 12. 13. 14.
15. 16. 17. 18. 19. 20. 21. 22. 23. 24. 25. 26. 27. 28. 29. 30. 31.
32. 33. 34. 35. 36. 37. 38. 39.
Kt qu mt xut hin 2 1 5 5 6 2 3 4 5 1 3 3 5 6 4 1 4 6 2 2 3 1 4 5
5 4 4 1 6 1 1 2 5 5 6 6 3 3 4
40. 41. 42. 43. 44. 45. 46. 47. 48. 49. 50. Mt 1 chm 2 chm 3 chm
4 chm 5 chm 6 chm
4 4 2 2 3 4 6 1 5 2 5 S ln xut hin 8 9 8 8 7 10 Tn sut 0,16 0,18
0,16 0,16 0,14 0,2
th cc tn sut ny s cng gn dng ng thng khi tung 1 triu ln. Bi 1.23
Gi A l bin c sn phm chn ra l chnh phm. Gi A1 v A2 l bin c sn phm b
mt ln lt l chnh phm v ph phm. Vy yu cu ca bi ton l tnh xc sut c iu
kin P(A1/A). Ta c A1, A2 l h y v xung khc vi nhau tng i mt. Ta c:a
a+b b P ( A2) = a+b P ( A1) =
Theo cng thc Bayes ta c:P ( A1 / A) = [P ( A1).P ( A / A1)] P(
A) a 1 P ( A / A1) = a + b 1 a P ( A / A2) = a + b 1 a (a 1)
Vi
P(A)= P(A1).P(A/A1) + P(A2).P(A/A2) = (a + b)(a + b 1) + (a +
b)(a + b 1)a.b
Vy: P(A1/A)= (a-1)(a+b-1).
Bi 1.24 Theo bng s liu ta c, tng s nhn vin trong cng ty l : 120
+ 170 + 260 + 420 + 400 + 230 = 1600 ( nhn vin) Khi ly ngu nhin mt
ngi ca cng ty th a. Xc sut c mt nhn vin t 40 tui tr xung l :120 +
170 + 260 + 420 = 0, 61 1600 400 = 0, 25 1600
b. Xc sut c mt nam nhn vin trn 40 l : c. Xc sut c mt n nhn vin t
40 tui tr xung l :170 + 420 = 0,37 1600
Bi 1.25 Gi A l bin c 3 bng in c ly ra trong hp c 4 bng hng u tt
S kt hp ng kh nng xy ra l s t hp chp 3 t 12 phn t. Nh 3 vy ta c : n
= C12 = 220 S kt cc thun li cho A xy ra bng s t hp chp 3 ( bng in
tt) t 8( trong 1 hp c 4 bng in b hng). Vy m= C83 = 56 Do xc sut mt
hp bng n c chp nhn trong c 4 bng b hng l :P ( A) = 56 = 0, 254
220
Bi 1.26 Ta c n=8 kt cc kh nng l GGG, GGT, GTG, TTT, TGG, TGT,
TTG. a. Gi A l bin c Gia nh c hai con gi. C 3 kt qu thun li cho A
nn ta c :P ( A) = 3 8
b. Gi B l bin c Gia nh c t nht 2 con gi . S kt qu thun li cho B
l 4 nn ta c :P( B) = 4 1 = 8 2
c. Gi C l bin c Gia nh c hai con gi bit rng a u lng l con gi. S
kt qu thun li cho C l 2 nn ta c :P (C ) = 1 4
d. Gi D l bin c Gia nh c t nht 2 a con gi bit rng gia nh c t nht
1 a con gi. Nu gia nh c t nht 1 a con gi th s kt cc ng kh nng l 7.
S kt cc thun li cho D l 4 nn ta c :P( D) = 4 7
Bi 1.27 Gi A l bin c C 3 ngi c ngy sinh nht trng nhau S kt cc ng
kh nng l t hp chp 3 ca 30 nn ta c :3 n = C30 = 4060
S kt cc c li cho bin c A l 30( v 1 thng c 30 ngy) nn ta cP ( A)
= 30 = 0, 008 4060
Gi B l bin c C 3 ngi c ngy sinh nht khc nhau Do A v B l 2 bin c
i nhau nn ta c P(B) = 1 P(A) = 0.992 Bi 1.28 Phn tch t d liu bi ta
c : 7 sn phm ch b v np 4 sn phm ch b v vi 1 sn phm ch b m ming 5 sn
phm va b m ming va b v np 3 sn phm va b st vi va m ming 7 sn phm va
b st vi va b v np 1 sn phm c tt c cc khuyt im trn a. Gi A l bin c
sn phm c khuyt tt S sn phm b khuyt tt l : 7+4+1+5+3+7+1=28 Vy :P (
A) = 28 = 0, 28 100
b. Gi B l bin c sn phm ch b st vi S sn phm ch b st vi l 4. Nh vy
:P( B) = 4 = 0, 04 100
c. Gi C l bin c sn phm b st vi bit rng n v np Gi D l bin c sn
phm va b st vi va b v np Gi E l bin c sn phm b c 3 khuyt tt Nh vy
C=D+E. Do D v E c lp vi nhau nn ta c : P(C)= P(D) + P( E) Vi
P(D)=0,07 v P(E)= 0,01 nn ta c P(C)= 0,08 Bi 1.29 Gi A l bin c Khng
c ngy no c qu 1 v tai nn lao ng S kt cc ng kh nng l s chnh hp lp
chp 6 t 92 phn t6 n = B92
S kt cc thun li l s chnh hp chp 6 t 92 phn t6 m = A92
Vy : P(A) =
6 A92 = 0,85 6 B92
Bi 1.30 a. C n ngi xp thnh hang ngang th s c n! cch xp Gi A l
bin c m ngi trng tn ng cnh nhau khi h xp hng ngang. Nu coi m ngi ng
trng tn cnh nhau ny lag 1 ngi th ta c
(n m +1 ) ! cch xp. C m! cch xp cho m ngi trng tn . Vy xc sut m
ngi trng tn ng cnh nhau khi h xp hng l:P ( A) = m !(n m + 1)!
n!
b. C n ngi xp thnh vng trn s c ( n-1) cch xp. Gi B l bin c c m
ngi trng tn ng cnh nhau khi h xp thnh vng trn. Nu coi m ngi trng tn
ng cnh nhau ny l 1 ngi th khi xp n ngi thnh vng trn ta c (n-m)! cch
xp. S kt qu thun li cho B l m!(n-m)! Vy ta c :P ( B) = m !(n m)! (n
1)!
Bi 1.31: Ta c 4=4+0+0=3+1+0=2+1+1=2+2+0 a.Gi M l bin c ch A nh v
3 chn v ch B nh v 1 chn S trng hp duy nht ng kh nng: n=3+
C23.2!.C34+3.C24.2!+C23.C24=81 P(M)=C34/81=4/81
b.Pb=C23.2!.C34=8/27 c.Pc=3/81 Bi 1.32. Xc sut s ngi n mi quy l nh
nhau v bng 1/3 Gi A l bin c c 3 ngi n quy 1 Lc Becnulli: n=10, k=3
Vy Bi 1.33. Gi A l bin c chi tit ly ra thuc loi I B l bin c chi tit
ly ra thuc loi II C l bin c chi tit ly ra thuc loi III Ta c : a.
A+B l bin c chi tit ly ra khng thuc loi III b. AB+C l bin c chi tit
ly ra thuc loi III hoc l va thuc loi I, va thuc loi II hoc va l loi
I, loi II v loi III c. l bin c chi tit ly ra l chi tit loi III nhng
khng thuc loi I hoc loi II hay thuc c loi I v loi II
d. AC l bin c chi tit ly ra va thuc loi I va thuc loi III Bi
1.34. Gi A l bin c ngi th k bn trng bia (k=1,2,3) Ta c: a. l bin c
ch c ngi th nht bn trng mc tiu b. mc tiu c. mc tiu d. l bin c c ngi
bn trng mc tiu l bin c ch c 2 ngi bn trng l bin c ch c mt ngi bn
trng
Bi 1.35. a) A=A1A2A3A4A5A6A7A8A9A10 b)
A=A1+A2+A3+A4+A5+A6+A7+A8+A9+A10 c) A= d) A=1 1 2 2 3 3 5 5 6
A7A8A9A106
A7
8
A9
10
Bi 1.36. Gi A l bin c Sinh con gi B l bin c sinh con c trng lng
hn 3kg Ta c: A+B = sinh con gi hoc con nng hn 3kg A.B = Sinh con gi
nng hn 3kg Bi 1.37. A l bin c cng ty thng thu d n th nht B l bin c
cng ty thng thu d n th hai Tng A+B l bin c: cng ty thng thu t nht
mt trong hai d n Tch A.B l bin c: cng ty thng thu ng thi c hai d n
Bi 1.38. Gi A1 l bin c sn phm ly ra thuc loi I A2 l bin c sn phm ly
ra thuc loi II A l bin c sn phm ly ra thuc loi I hoc loi II th V
A1, A2 xung khc nn Bi 1.39.
Gi A1,A2,A3 ln lt l bin c m sn phm ca nh my i qua phng kim tra s
1,2,3 l ph phm A l bin c sn phm nhp kho l ph phm Ta c:
P(A1)=1-0.8=0.2 P(A2)=0.1 P(A3)=0.01 V 3 phng kim tra hot ng c lp
ln A1, A2, A3 l cc bin c c lp Ta c
P(A)=P(A1.A2.A3)=P(A1).P(A2).P(A3)=0.1x0.2x0.01=0.0002 Bi 1.40. Xc
sut khi o mt i lng vt l phm sai s vt qu tiu chun cho php l 0.4. Thc
hin 3 ln o c lp. Tm xc sut sao cho c ng mt ln o sai s vt qu tiu
chun cho php. Gii: Gi A l bin c php o i lng vt l vt qu tiu chun cho
php. P(A) = 0,4 P( A )= 0.6 Vic thc hin cc ln o l c lp p dng cng
thc Bernoulli, xc sut A xut hin ng mt ln trong 3 php th l:
P 3 (1) = C
1 2 1 3 0.4 0.6 = 0.432
Bi 1.41: Gi A l bin c: Hai bi ly ra cng mu trng. B l bin c: Hai
bi ly ra cng mu . C l bin c: Hai bi ly ra cng mu xanh. S cch chn
ngu nhin mi hp 1 vin bi l P= cch. Khi xc sut ly ra 2 bi cng mu t 2
hp khc nhau l: P= P(A)+P(B)+P(C) = Vy P = Bi 1.42: a. Gi A l bin c:
Ngi th nht bn trng mc tiu P(A) = 0,8 Gi B l bin c: Ngi th hai bn
trng mc tiu P(B) = 0,9207 6251 1 1 1 1 1 C 3 .C10 + C 7 .C 6 + C15
.C 9 207 = 625 625
(7 + 3 +15 ).( 10 + 6 + 9) =
625
Do ch c mt ngi bn trng mc tiu. Suy ra nu A bn trng th B ko bn
trng v ngc li Vy xc sut mt duy nht 1 ngi bn trng mc tiu l: P 1 =
P(A).P( B ) + P( A ).P(B) = 0,8.0,1 + 0,9.0,2 = 0,26. Vy P1 = 0,26
b. Gi P2 l xc sut c ngi bn trng mc tiu. C ngi bn trng mc tiu khi
hoc ngi th nht bn trng, hoc ngi th 2 bn trng, hoc c 2 cng bn trng.
P2 = 0,8.0,1 + 0,9.0,2 + 0,8.0,9 = 0,98
Vy P2 = 0,98 c. Gi P3 l xc sut c 2 cng bn trt. C 2 ngi cng bn
trt khi khng c ai bn trng. Suy ra P3= 1 P2 =1 - 0,98 = 0,02. Vy
P3=0,02 Bi 1.43: Gi A l bin c "sau khi gia cng xong chi tit c khuyt
tt " Ai la bin c "gy ra khuyt tt cng on th i" P(Ai)=Pi suy ra P( A
) =1 Pi (i=1,2,k)i
A=
i =0
k
Aii
P(A) = 1 - P( A )= 1 - ( 1-Pi )k. Bi 1.44: Gi Ai l bin c qu bng
th i l qu bng mi. Sau khi ly k qu bng ra chi v b li hp th trong hp
ch cn n-k qu bng mi. Ly ln lt tng qu. Qu th nht c n cch ly nhng ch
c n-k cch ly ra qu mi hay cch khc s kt cc ng kh nng ca bin c A1 l n
v s kt cc thun li l n-k, vy P(A1) = Ly qu th 2 th s kt cc ng kh nng
l n-1 v s kt cc thun li l n-k-1 nn P(A2) =
Tng t P(A3) = P(Ak) = Xc sut k qu bng ly ra chi u l mi bng tch
xc sut qu th 1, th 2, th 3, th k u l qu mi. Hay: P =
P(A1).P(A2)P(Ak) = . . . = T s bng Mu s bng Vy P = [(n-k)!]2 /
[n!(n-2k)!] Bi 1.45: a) Gi Ai l bin c ln i khng thu c tn hiu th Pi
= 0,6. Bin c ngun khng nhn c thng tin l tch ca 3 bin c c lp A1, A2,
A3. P(A) = 0,6 . 0,6 .0,6 = 0,216. Bin c i ca A l ngun thu c thng
tin c xc sut l: Pa = 1 - P(A) = 0,784. Vy Pa = 0,784. b) Nu mun xc
sut thu c thng tin ln 0,9 th bin c ngun khng thu c tn hiu phi c xc
sut l 0,1. S ln phi pht l n sao cho 0,6n = 0,1 => n =log Vy phi
pht t nht 5 ln. Bi 1.46: Gi xc sut bn trng ch ca ngi th nht l a.
Theo bi ra ta c: 0,2.a + 0,8.(1-a) = 0,38 0,6a = 0,420, 6
0,1 = 4,5.
a = 0,7
Vy xc sut bn trng ca ngi th nht l P = 0,7. Bi 1.47: a. Do ly ngu
nhin 2 sn phm trong 10 sn phm v c hon li nn ta c 100 cch chn 2 sn
phm. Gi A l bin c: Sn phm ly ra ln th nht l ph phm P(A) =1 C2 =
0,02 100
Gi B l bin c: Sn phm ly ra ln 2 l ph phm. P(B) =
1 C2 = 0,02 100
Xc sut 2 ln ly ra u c ph phm l P = 0,02 + 0,02 = 0,04
Vy P= 0,04 b. Do ly ra ngu nhin 2 sn phm trong 10 sn phm v khng
hon li nn ta c 10.9 = 90 cch chn sn phm.1 Gi A l bin c: C 2 sn phm
u l ph phm. Ta c C 2 .C11 = 2 cch
chn ph phm. P(A) =2 = 0,022 90
Vy xc sut c 2 sn phm ly ra u l ph phm l P(A) = 0,022 Bi 1.48: Gi
A1 l bin c van 1 b hng P1 = 0,1. A2 l bin c van 2 b hng th P2= 0,2.
Bin c ni hi hot ng mt an ton l tch ca hai bin c c lp A 1 v A2 nn P
= 0,1.0,2 = 0,02 Bin c ni hi hot ng an ton l P = 1 - 0,02 = 0,98.
Bi 1.49: Gi A l bin c: Bn n vin th 6 mi trng ch. A1 l bin c: Vin th
nht bn trng ch. P(A1) = 0,2 A2 l bin c : Vin th 2 bn trng ch. P(A2)
= 0,2 ..... A6 l bin c : Vin th 6 bn trng ch. P(A6) = 0,2
Theo u bi ra, bn lin tip vo mt mc tiu cho n khi vin n u tin trng
mc tiu th dng. Do bn n vin th 6 th 5 vin u phi bn trt, vin th 6 bn
trng mc tiu. Mt khc cc ln bn c lp nhau nn cc bin c A1,
A2,A3,A4,A5,A6 l cc bin c c lp. Vy xc sut bn n vin th 6 mi trng ch
l: P(A) = P( A1 ) .P( A 2 ) . P( A 3 ).P( A4 ).P( A ).P(A6)=
0,85.0,2=0,0655365
Bi 1.50: Gi A1 l bin c: ln th th nht khng m c ca kho A2 l bin c:
ln th th hai khng m c ca kho A3 l bin c: ln th th ba khng m c ca
kho A4 l bin c: ln th th t m c ca kho A l bin c: m c ca kho ln th 4
Theo u bi, th kho th ngu nhin tng cha mt, chic no c th th khng th
li. Do A1,A2,A3,A4 l bin c ph thuc. Xt bin c A1, chm cha kha c 9
cha trong ch mt cha m c, 8 cha cn li khng m c. Ln th th nht khng m
c. Vy bin c A1 c xc sut: P(A1)= 8/9. Xt bin c A2, sau khi th ln mt,
cn 8 chic cha kha trong 1 chic m c v 7 chic khng m c. Ln th th hai
khng m c. Vy bin c A2 c xc sut: P(A2/A1)=7/8. Tng t, xt bin c A3,
sau khi th ln hai, cn 7 chic cha kha trong 1 chic m c v 6 chic khng
m c. Ln th th ba khng m c. Vy bin c A3 c xc sut: P(A3/A1A2)=6/7. Xt
bin c A4, sau khi th ln ba, cn 6 chic cha kha trong 1 chic m c v 5
chic khng m c. Ln th th t m c. Vy bin c A 4 c xc sut:
P(A4/A1A2A3)=1/6. Vy xc sut m c ca kho ln th 4 l
P(A)=P(A1).P(A2/A1).P(A3/A1A2).P(A4/A1A2A3)=8/9.7/8.6/7.1/6=1/9 Kt
lun xc sut m c ca kho ln th 4 l 1/9.
Bi 1.51: bng cch v s ven ra, ta c hnh
15%
24% 10%
Vy xc sut chn ngu nhin 1 khch hng bit thng tin v sn phm ca cng
ty l : P=15%+10%+24%=49% 1.51. Gi A l bin c khch hng nm c thng tin
qua v tuyn truyn hnh, B l bin c nm c thng tin qua i pht thanh. Gi C
l bin c chn ngu nhin 1 khch hng th ngi nm c thng tin v sn phm ca
cng ty) Theo bi ra ta c: P(A) = 0.25; P(B) = 0.34; P(AB) = 0.1 P(C)
= P(A+B) = P(A) + P(B) - P(AB) = 0.25 + 0.34 - 0.1 = 0.49 Bi 1.52
P(A)=1/2 , P(B)= 11/36, P(AB)= 1/6 P(A+B)= P(A)+ P(B)- P(AB)=23/36
P(AB)= 1- P(AB)= 5/6 1.53. Gi A l bin c mch khng c in do bng hng.
gi A l bin c bng n th nht hng; A l bin c bng n th 2 hng a. nu 2 bng
n mc ni tip: mch khng c in th phi c t nht 1 bng hng. khi l bin c
khng c bng n no hng ta c: = . => P() = P( . ) = 0.8 . 0.9 = 0.72
P(A) = 1 - P () = 0.28 b. nu 2 bng n mc song song: mch khng c in th
c 2 bng n u hng. khi ta c: A = A A P(A) = P(A A) = 0.1 . 0.2 = 0.02
1.54. Gi A l bin c ly c chnh phm t l i th l bin c ly c ph phm t l
i, i = 1,2 a.gi A l bin c ly c 1 chnh phm theo bi ra ta c P(A) = ;
P(A) = ; P() = ; P() = vy P(A) = P(A) P() + P() P(A) = . + . = 0.26
c. gi B l bin c ly c t nht mt chnh phm th l bin c khng ly c chnh
phm no
P() = P()P() = . = 0.02 Vy P(B) = 1 - P() = 0.98 1.55. Gi A l
bin c ln th i ly ra 3 sn phm mi kim tra. (i = ). Gi A l bin c sau 3
ln kim tra tt c cc sn phm u c kim tra. A = A A A V cc bin c l ph
thuc nn: P(A) = P( A )P(A /A )P(A /A A ) = 1 . . = 1.56. Gi A l bin
c vin n th i trng ch, i = Gi A l bin c bn n vin n c th hi vng rng
khng c vin no trt Ta c: P(A) = P(A) = = P(A) = 0.8 P(A) =
P(A)P(A)P(A) = 0.8 Theo gi thit xc sut ca bin c A nh hn 0.4 nn ta c
bt phng trnh sau: 0.8 < 0.4 => lg (0.8) < lg 0.4 =>
n.lg 0.8 < lg 0.4 => n > => n > 4.106 Vy n 5 1.57.
Gi A l bin c tung ln th i c mt 6 chm, i = v A l bin c trong n ln
tung th c t nht 1 ln c mt 6 chm.Vy A = A . Cc bin c A l khng xung
khc v c lp ton phn vi nhau, do : P(A) = 1 - P() V P(A) = P(A) =
P(A) == P(A) = (xc sut mi ln tung c mt 6 chm) do : P() = P() =.=
P() = Nn ta c: P(A) = 1 - ( ) Theo gi thit xc sut ca bin c A ln hn
0.5 do ta thu c bpt sau: 1 - ( ) > 0.5 => ( ) < 0.5 =>
n > => n > 3.802 1.58. Coi vic bn hng mi ni ca ngi l 1 php
th th ta c 10 php th c lp. trong mi php th ch c 2 kh nng i lp: hoc
bn c hng, hoc khng bn c hng. Xc sut bn c hng mi ni u bng 0.2. Vy bi
ton tha mn lc Bernoulli. a. Theo cng thc Bernoulli P (2) = C . 0.2
. 0.8 = 0.302 b. P (1) + P (2) + +P (10) = 1 - P (0) = 1- C . 0.2 .
0.8 = 0.8926 1.59.
Coi vic sn xut ra sn phm ca my l 1 php th th ta c 12 php th c
lp. trong mi php th ch c 2 kh nng i lp: hoc sn xut ra ph phm hoc sn
xut ra chnh phm. Xc sut my sn xut ra ph phm l 0.05. Vy bi ton tha
mn lc Bernoulli. a. Theo cng thc Bernoulli ta c: P (2) = C . 0.05 .
0.95 = 0.109 b. Gi A l bin c my sn xut ra c khng qu 2 ph phm, ta c:
P(A) =P (0) + P (1) + P (2) = C . 0.05 . 0.95 + C . 0.05 . 0.95 + C
. 0.05 . 0.95 = 0.9804 1.60. Coi vic tr li cu hi l 1 php th th ta c
10 php th c lp. Trong mi php th ch c 2 kh nng i lp: hoc tr li ng
hoc tr li sai. Xc sut tr li ng 1 cu hi l ( v mi cu c 5 cch tr li,
trong ch c 1 cch tr li ng). Vy bi ton tha mn lc Bernoulli Gi A l
bin c ngi thi , ta c: P(A) = P (8) + P (9) + P (10) = C . ( ) . ( )
+ C . ( ) . ( ) + C . ( ) . ( ) = 0.000078 Bi 1. 61: Gi A l bin c c
chung ku khi chy th no ku khi chy. Gi
A
l bin c khng c chung
H
i
(i= 1,4 ) ln lt l bin c chung th i khng ku khi c chyi1,4
Ta c P( Ta cA
Bn bin c ny ging nh 4 php th c lp. Trong mi php th u hoc xy ra
bin c chung ku, hoc chung khng ku. Xc sut mi ln chung khng ku l 0,
05. Vy n tha mn lc Bernoulli. Xc sut c 4 chung u khng ku l P( A
)=P(
H )=1-0, 95=0, 05 (i= =H H H H1 2 3 4
)
HH H H1 2 3
. . 4 )= C 4 0,054
4
(1 ,0 ) 0 5
0
= 6, 25.
10
6
Vy P(A)=1-P( A )=0, 999994 Bi 1. 62 Gi A l bin c sn phm ly ra l
tt. Gi H1 l bin c sn phm ly ra l ca my I. Gi H2 l bin c sn phm ly
ra l ca my II. Bin c A c th xy ra vi 1 trong hai bin c H1, H2 to
thnh 1 nhm bin c y . Do theo cng thc xc sut y P(A)=P(H1).
P(A/H1)+P(H2). P(A/H2)=2/3. 0, 97+1/3. 0, 98=0, 9733
Bi 1. 63: a. Gi A l bin c vin n trng ch Gi H1 l bin c x th c ly
l x th loi I th P(H1)=
C C
1
2 1 10
=1/5
H2 l bin c x th c ly l x th loi II th P(H2)=4/5 Ta c A xy ra ng
thi vi 2 bin c I, II l 2 bin c lp thnh mt nhm bin c y nn theo cng
thc xc sut y ta c: P(A)= P(H1). P(A/H1)+P(H2). P(A/H2)=1/5. 0,
9+4/5. 0, 8=0, 82 b. Gi G1 l bin c ngi th nht bn trng G2 l bin c
ngi th 2 bn trng B l bin c c 2 ngi bn trng Ta c P(G1)=P(G2)=0, 82
theo cu a Ta c B=G1. G2 m 2 bin c ny c lp nn P(B)=P(G1). P(G2)=0,
82. 0, 82=0, 6724 Bi 1. 64: Gi Hi l bin c ly dc l i (i= 1,2 )
P(H1)=P(H2)=1/2 Gi A l bin c ly c chnh phm: P(A/H1)=1; P(A/H2)=4/5
Bin c A xy ra ng thi vi 2 bin c H1, H2 lp thnh 1 nhm bin c y nn
theo cng thc xc sut y : P(A)=P(H1). P(A/H1)+P(H2). P(A/H2)=0, 9
Theo cng thc Bayes P(H1/A)=P ( H 1). P ( A / H 1) P ( H 1) P ( A /
H 1) + P ( H 2) P ( A / H 2)
=5/9
Tng t P(H2/A)=4/9 Gi B l bin c Sn phm ly ra ln th 2 l chnh phm B
vn c th xy ra vi mt trong 2 bin c H1, H2. Do theo cng thc xc sut y
P(B)=P(H1/A)P(B/H1A)+P(H2/A)P(B/H2A)=4/45 Bi 1. 65: Gi A l bin c my
bay ri. Hi l bin c bn trng i pht (i= 1,3 ) Gi l bin c bn trng pht
th i(i= 1,3 ) Ta c H1=G1. G 2 . G 3 + G1 . G2. G 3 + G1 . G 2 . G3
nn P(H1)=0, 33 Tng t, P(H2)=0, 41; P(H3)=0, 14 A xy ra ng thi vi 3
bin c H1, H2, H3 l 1 nhm bin c y nn theo cng thc xc sut y ta c :
P(A)=P(H1). P(A/H1)+P(H2). P(A/H2)+P(H3). P(A/H3)=0, 36. 0, 2+0,
41. 0, 6+0, 14. 1=0, 458
Bi 1. 66 Gi A l bin c ly ra c t nht 1 chnh phm th A l bin c ly c
ton ph phm. Gi H1 l bin c ly c 2 sn phm ly ra u thuc l 1 H2 l bin c
ly c 2 sn phm ly ra thuc l 2 H3 l bin c ly c 2 sn phm th 1 sp thuc
l 1, 1 sp thuc l 2. Ta c P(H1)=A
C C
2 2 2 5
=1/10, P(H2)=
C C
2 3 2 5
=3/10, P(H3)=
C 2 .C 3 C2 5
1
1
=3/5
xy ra ng thi vi 3 bin c trn v 3 bin c ny lp thnh 1 nhm bin c y
.
C Ta c P( A /H1)= C
2
3 2
10
C =3/45, P( A /H2)= C
2
2 2
=1/45, P( A /H3)= C . C3
1
1 2
10
10
10
=0, 6 Theo cng thc xc sut y ta c P( A )=P(H1). P( A /H1)+P(H2).
P( A /H2+P(H3). P( A /H3)= 37/750 Vy P(A)=1-P( A )=1-37/750 0, 951
Bi 1. 67: Gi A l bin c ly c chnh phm. H1 l bin c thnh phn ca l th
nht khng thay i H2 l bin c l th nht mt ph phm c thay th bng mt chnh
phm. H3 l bin c l th nht mt chnh phm c thay th bng mt ph phm. Ta c
: Vi H1, do thnh phn ca l 1 khng i tc l t l th nht b sang l th 2 l
sn phm nh th no th sn phm b tr li s nh th. Do P(H1)=
C a . C c +1 + C b . C d +1 C11
1
1
1
1
. a + b C c + d +1
1
Vi H2, ta c l 1 lc u b ph phm, chuyn 1 ph phm sang l 2 th l 2 c
c+d+1 sn phm, sau t l 2 li chuyn 1 chnh phm sang l 1 P(H2)=
C .Ca +b
C b .C c1 1 c +d +1
1
Vi H3, ta c lc u l 1 c a chnh phm, chuyn sang l 2 mt ph phm th l
2 c c+d+1 sn phm, sau t l 2 li chuyn sang l 1 mt ph phm
P(H3)=
C
C a .C d1 1
1
1
. a + b C c + d +1a
Ta c: P(A/H1)= a + b ; P(A/H2)= a + b ; P(A/H3)= a + b Bin c A
xy ra ng thi vi 3 bin c H1, H2, H3 l mt nhm bin c y . Theo cng thc
xc sut y ta c: P(A)=P(H1). P(A/H1)+P(H2). P(A/H2)+P(H3). P(A/H3)= a
+ b +bc ad2
a +1
a 1
a
(a +b)
(c + d +1)
Bi 1. 68: a. Gi A l bin c tm c mt ngi vim hng H1 l bin c tm c
ngi nghin thuc H2 l bin c tm c ngi khng nghin thuc Bin c A xy ra ng
thi vi 2 bin c H1, H2 l 1 nhm bin c y nn theo cng thc xc sut y ta
c: P(A)=P(H1). P(A/H1)+P(H2). P(A/H2)=0, 3. 0, 6+0, 7. 0, 4=0, 46
Khi theo cng thc Bayes: P(H1/A)=P ( H 1). P ( A / H 1) P ( A)
=0, 3913
b. Gi B l bin c tm c ngi khng b vim hng th P(B)=1-P(A)=10, 46=0,
54 B vn xy ra vi 2 bin c H1, H2 nn theo cng thc Bayes ta c:
P(H2/B)=P ( H 1). P ( B / H 1) P( B)
=0, 222
Bi 1. 69 Gi A l bin c c 2 b phn hng H1 l bin c ch c b phn 1 v 2
b hng H2 l bin c ch c b phn 1 v 3 b hng H3 l bin c ch c b phn 2 v 3
b hng Gi (i= 1,3 ) l bin c b phn th i b hng Ta c H1=G1. G2. G 3 nn
P(H1)=0, 056 Tng t P(H2)=0, 036; P(H3)=0, 096 Ta c A=H1+H2+H3 m 3
bin c ny xung khc tng i mt nn P(A)=P(H1)+P(H2)+P(H3)=0, 188 Vy xc
sut 2 b phn hng l 1 v 2 l Bi 1. 70 Gi G l bin c tm c bnh nhn l k s
H1 l bin c bnh nhn tnh A.P ( H 1) P ( A)
=0, 298
H2 l bin c bnh nhn tnh B H3 l bin c bnh nhn tnh C Ta c G xy ra
ng thi vi 3 bin c H1, H2, H3 l 1 nhm bin c y . Theo cng thc xc sut
y ta c: P(G)= P(H1). P(G/H1)+P(H2). P(G/H2)+P(H3). P(G/H3) =0, 25.
0, 02+0, 35. 0, 03+0, 4. 0, 035=0, 0295 1.71, A= ch cu c mt con c
trong 3 ln cu Hi= ngi ny ngi ch th i, i = 1,2,3 A c th xy ra ng thi
vi 1 trong 3 bin c v to nn mt nhm cc bin c y . Theo cng thc Bayes
ta c:P ( H 1 / A) = P ( H1 ).P ( A / H1 )
P( H ).P( A / H )i =1 i i
3
Trong : P(H1) = P(H2) = P(H3) = 1/3 P(A/H1) = 0,6.0,42 = 0,096
P(A/H2) = 0,7.0,32 = 0,063 P(A/H3) = 0,8.0,22 = 0,032 Do , ta c:P (
H 1 / A) = 0,096 96 = 0,5026 0,096 + 0,063 + 0,032 191
p n: 0,5026 1.72, Xc sut ca bin c A l 0,7. iu c ngha l t s gia
kt cc thun li cho A v tng s cc kt cc duy nht ng kh nng c th xy ra
khi thc hin php th l 0,7 1.73, A = c mt hp no c 1 ph phm Gi s 3 hp
ny phn bit vi nhau S kt cc duy nht ng kh nng c th xy ra khi xp 30
sn phm vo 10 10 10 3 chic hp sao cho mi hp c 10 sn phm l: C30 .C 20
.C10 Xt bin c A = c 3 ph phm u nm trong mt hp + Nu c 3 ph phm u nm
trong hp 1. 7 10 10 S kt cc thun li l C 27 .C 20 .C10 + Vi trng hp
c 3 ph phm cng nm trong hp 2, hoc 3 u chung mt kt qu nh trn 7 10 10
Vy, s kt cc thun li cho bin c A l: 3.C27 .C20 .C10 Do ,P( A ) =7
3.C 27 10 C 30
Suy ra,P( A) = 1 P ( A ) = 1 7 3.C 27 0,91133 10 C 30
1.74, a, A= Mi ngi khch x ng i giy ca mnh S kt cc thun li ng kh
nng l : (N!)2 (Do c N i giy, v s cch i N i giy y cho chn tri hay
chn phi ca N ngi l N!) S kt cc thun li l: 1 Do :P ( A) = 1 ( N !)
2
b, B= Mi ngi khch x ng hai chic giy ca mt i giy no S kt cc thun
li ng kh nng l : (N!)2 S kt cc thun li l N! Do :P( B) = N! 1 = 2 (
N !) N!
1.75, A= Sn phm sn xut ra l chnh phm B= Sn phm sn xut ra l ph
phm H1= Sn phm c thit b kt lun l chnh phm H2= Sn phm c thit b kt
lun l ph phm Ta c: P(A) = 0,95 P(B) = 0,05 P(H1/A) = 0,04 P(H2/B) =
0,01 a, X = Sn phm c kt lun l chnh phm nhng thc ra l ph phm Ta c: X
= BH1 P(X) = P(B). P(H1/B) = 0,05.0,01 = 0,0005 Vy: T l sn phm b kt
lun l chnh phm nhng thc t l ph phm l 0,05% b, Y= Sn phm c kt lun l
ph phm nhng thc ra l chnh phm Ta c: Y =AH2 P(Y) = P(A).P(H2/A) =
0,95.0,04 = 0,038 Vy: T l sn phm c kt lun l ph phm nhng thc ra l
chnh phm l 3,8% c, T l sn phm b thit b kim tra kt lun nhm l:
3,85%
1.76, a, A= Ly c nam sinh vinP ( A) = 400 + 800 = 0,6 2000
b, B= Ly c sinh vin hc kinh tP( B) = 400 + 500 = 0,45 2000
c, C= Ly c hoc nam sinh vin, hoc hc kinh tP (C ) = P ( A + B ) =
P ( A) + P ( B ) P ( AB ) = P ( A) + P ( B ) P ( A). P ( B / A) =
0,6 + 0,45 0,6. 400 = 0,85 400 + 800
d, D= Ly c nam sinh vin v hc kinh t D = ABP ( D) = P ( AB ) = P
( A). P ( B / A) = 0,6. 400 = 0,2 400 + 800
e, E= Ly c l sinh vin kinh t khi ngi l nam sinh vinP ( E ) = P (
B / A) =
1.77,P ( A) =
400 1 = 400 + 800 3
a, Trc khi m kin hng, xc sut kin hng l ca x nghip l:60 = 0.6
100
b, X= Sn phm ly ra l ph phm H1 = Sn phm ca x nghip A H2 = Sn phm
ca x nghip B Ta c:P( H 1 / X ) = P ( X / H 1 ). P( H 1 ) P( X / H 1
). P ( H 1 ) + P ( X / H 2 ). P ( H 2 )
Trong : P(H1) = 0,6; P(H2) = 0,4 P(X/H1) = 0,3; P(X/H2) = 0,1 Do
:P( H 1 / X ) = 0,6.0,3 9 = 0,81 0,6.0,3 + 0,4.0,1 11
c, Y = C 2 sn phm u ra ph phmP( H 1 / Y ) = P (Y / H 1 ). P ( H
1 ) P (Y / H 1 ).P ( H 1 ) + P (Y / H 2 ).P ( H 2 )
Gi s: L hng kim tra c 10N sn phm2 C 32N CN P (Y / H 1 ) = 2 ;
P(Y / H 2 ) = 2 C10 N C10 N
Do :
P( H 1 / Y ) =
C 32N 2 C 32N + C N
1.78, X= v thng xem chng trnh th thao Y= chng thng xem chng trnh
th thao P(X) = 0,3; P(Y) = 0,5; P(Y/X) = 0,6 a, A = C hai cng xem
chng trnh th thao A = XY P(A) = P(X).P(Y/X) = 0,3.0,6 = 0,18 b, B =
C t nht mt ngi thng xem B=X+Y P(B) = P(X+Y) = P(X) + P(Y) P(XY) =
0,3 + 0,5 0,18 = 0,62 c, C = Khng c ai thng xem P(C) = 1-P(B) =
0,38 d, D = Nu chng xem th v xem cngP ( D) = P( X / Y ) = P(Y / X
). P ( X ) 0,6.0,3 = = 0,36 P (Y ) 0,5
e, E = Nu chng khng xem th v vn xemP( E ) = P( X / Y ) =
P (Y / X ). P( X ) [1 P(Y / X ) ].P ( X ) (1 0,6). 0,3 = = =
0,24 P (Y ) 1 P (Y ) 1 0,5
1.79, Hi = Bn c hng ln th i Ta c: P(H1) = 0,8 P(H2/H1) = 0,9;
P(H3/H2) = 0,9P ( H 2 / H 1 ) = 0,4; P ( H 3/ H 2 ) = 0,4
a, A= C 3 ln u bn c hng P(A) = P(H1.H2.H3) = P(H1).P(H2/
H1).P(H3/ H2. H1) = = 0,8.0,9.0,9 = 0,648 b, B = C ng 2 ln bn c
hngP( B) = P( H 1 H 2 H 3 + H 1 H 2 H 3 + H 1 H 2 H 3 ) = P ( H 1
). P ( H 2 / H 1 ). P ( H 3 / H 2 H 1 ) + P ( H 1 ). P ( H 2 / H 1
). P ( H 3 / H 2 H 1 ) + P ( H 1 ). P ( H 2 / H 1 ). P ( H 3 =
0,2.0,4.0,9 + 0,8.0,1.0,4 + 0,8.0,9.0,1 = 0,176
1.80, E1 = Cp sinh i ng trng E2 = Cp sinh i khc trng P(E1) = P;
P(E2) = 1- P A = Cp sinh i c cng gii tnh
P ( E1 / A) =
P ( A / E1 ).P ( E1 ) 1 .P 2P = = 1 P ( A / E1 ).P ( E1 ) + P (
A / E2 ).P ( E2 ) 1.P + .(1 P ) P + 1 Vy: 2
Nu cp tr sinh i c cng gii tnh th xc sut chng l cp sinh i cng gii
tnh l:2P P +1
1.81.
= C9 C6 C3 a. Gi A = Mt phn gm 3 cu P( A) = 1
3
3
3
CCC9 6
3
3
3 3
=
1 1680
b. Gi B = Mi phn c 1 cu Nh vy 2 cu cn li mi phn l cu xanh
P( B) =1.82.
C3C6 + C3 C 4 + C3C 2 CCC9 6 3 3 3 3
1
2
2
2
1
2
=
58 1680
a. A= H thng phun nc b hng P( A) = 0,1 B= H thng bo ng b hng P(
B) = 0,2 C= C t nht 1 h thng hot ng bnh thngC = ABP (C ) = P ( A )
= 0,04 B
P (C ) =1 P (C ) = 0,96
P( D ) = 1 0,26 = 0,74
b. D= C 2 h thng hot ng bnh thng D = A + B , A,B khng xung khc P
( D ) = P ( A + B ) = P ( A) + P ( B ) P( AB ) = 0,26
1.83. H= Chai ru ly ra thuc loi A A= 3 ngi kt lun chai ru thuc
loi A B= 1 ngi kt lun chai ru thuc loi B
P( A / H ) = C 4 .0,8 3.0,2 = 0,4096P( B / H ) = C 4 .0,2.0,8 3
= 0,4096P ( A / H ) = C 4 .0,2 3.0,8 = 0,0256 P ( B / H ) = C 4
.0,8.0,2 3 = 0.0256 1 (0,4096) 2 P ( H ) P ( AB / H ) 2 P ( H / AB)
= = = 0,996 1 1 2 2 P ( H ) P ( AB / H ) + P ( H )P ( AB / H )
(0,4096) + (0,0256) 2 21 3
3
1
1.84. a. A= 2 mu hng cng loi A1= 2 mu hng loi A A2= 2 mu hng loi
B
6 3 = 0,18 10 10 4 7 P ( A2 ) = = 0,28 10 10 A = A1 + A2 , A1 ,
A2 P ( A1 ) =
A1, A2 xung khc 1.
P ( A) = P ( A)1 + P ( A2 ) = 0,18 + 0,28 = 0,46
2.
b. H1= Mu ly ra hp 1 H2= Mu ly ra hp 2 B= Mu ly ra thuc loi B
P(H1)=0,45 P(H2)=0,55
P(B)=P(H1)P(B/H1)+P(H2)P(B/H2)=0,45.0,4+0,55.0,7=0,565 C= Mu ly ra
thuc loi A
P ( H 1 / A) =
P ( H 1) P ( A / H 1) 0,45 .0,6 = = 0,62 P ( H 1) P ( A / H 1) +
P ( H 2) P ( A / H 2) 0,435 P ( H 2) P ( A / H 2) 0,55 .03 P ( H 2
/ A) = = = 0,38 P ( H 2) P ( A / H 2) + P ( H 1) P ( A / H 1)
0,435
Vy mu nhiu kh nng thuc H1 hn 1.85. ok H1= Chic 1 ly c l ca i
khng chic no hng H2= Chic 1 ly c l ca i c mt chic hng H3= Chic mt
ly c l ca i c hai chic hng P(H1)=.9, P(H2)=0.08,P(H3)=0.02 A= Chic
mt ly ra b hng B= Chic hai ly ra b hng
P(A)=P(H1)P(A/H1)+P(H2)(A/H2)+P(H3)P(A/H3)
=0,9.0+0,08.0,5+0,02.1=0,06 P(B)=P(H3/A)=P ( H 3) P ( A / H 3) 0,02
1 = = P ( A) 0,06 3
Bi 1.86:ok a. Gi X l bin c rt phi a b li bin c X c th xy ra vi 1
trong 2 gi thuyt : H1 a rt ra ca ca hng A H2 a rt ra ca ca hng B
P(X)=P(H1).P(X/H1)+ P(H2).P(X/H2)=0,6.0,1+0,4.0,2=0.014
b.P(H1/X)=P(H1).P(X/H1)/P(X)=0,6.0,1/0,014=0,4285 Bi 1.87: a. Gi A
l bin c 2 sn phm ly ra u xu l bin c 2 sn phm ly ra c t nht 1 sn phm
tt B l bin c sn phm 1 ly ra l xu Bin c B c th xy ra vi 1 trong 2 gi
thuyt: H1: sn phm l ca my 1 H2: sn phm l ca my 2
P(B)=P(H1).P(B/H1)+ P(H2).P(B/H2)=0,01.0,4+0,02.0,6=0,016
P(H1/B)=P(H1).P(B/H1)/P(B)=o,01.0,4/0.016=0,25
P(H2/B)=P(H2).P(B/H2)/P(B)=0,75 P(A)=P(H1/B).P(A/H1B)+
P(H2/B).P(A/H2B)=0,25.0,01+0,75.0,02=0,0175 P()=1-P(A)=0,9825
b.P(B)=1-P(B)=0,984 P(H1/B)=P(H1).P(B/H1)/
P(B)=0,4.0,99/0,984=0,402 P(H2/ B)= P(H2).P(B/H2)/ P(B)=0,598 Gi C
l bin c sn phm th 2 l tt P(C )= P(H1/B).P(C/ H1B)+ P(H2/B).P(C/
H2B)=0,402.0,99+0,598.0,98=0,98402 P(H1 B/C)= P(H1/B).P(C/
H1B)/P(C)=0,404 P(H2B/C)= P(H2B).P(C/ H2 B)/P(C)=0,596 Gi D l bin c
sn phm th 3 l sn phm tt P(D)= P(H1 B/C).P(D/ H1 BC)+ P(H2B/C).P(D/
H2BC) =0,404.0,99+0,596.0,98=0,98404 P(H1 BC/D)= P(H1 B/C).P(D/ H1
BC)/P(D)=0,406 P(H2 BC/D)= P(H2 B/C).P(D/ H2 BC)/P(D)=0,594 Gi E l
bin c sn phm th 4 cng l sn phm tt P(E)= P(H1 BC/D).P(E/ H1 BCD) +
P(H2 BC/D).P(E/ H2 BCD) =0,406.0,99+0,594.0,98= 0,98406
Bi 1.88: ok Gi A l bin c anh ta v nh sau 6 gi Bin c A c th xy ra
vi 1 trong 2 gi thuyt: H1 : i theo ng ngm H2 : i qua cu
P(A)=P(H1).P(A/H1)+P(H2).P(A/H2)=1/3.0,25+2/3.0,3=17/60
P(H2/A)=P(H2).P(A/H2)/P(A)=0.706 Bi 89: Gi A l bin c trong 8 sp u
tin c 2 ph phm Hi (i=1,2,3) l bin c 8sp do ngi th I lm ra P(Hi)=1/3
P(A/H1)= P(A/H2)= C68.0,96.0,12 P(A/H3)= C68.0,86.0,22
P(A)=P(H1).P(A/H1)+ P(H2).P(A/H2)+ P(H3).P(A/H3)=0,2 P(H1/A)=
P(H1).P(A/H1)/P(A)= 0,25 P(H2/A)= P(H2).P(A/H2)/P(A)=0,25 P(H3/A)=
P(H3).P(A/H3)/P(A)=0,5 Gi B l bin c trong 8 sp sau c 6 chnh phm
P(B)= P(H1/A).P(B/ H1A)+ P(H2/A).P(B/ H2A)+ P(H3/A).P(B/
H3A)=0,23
Bi 90: Gi A l bin c ly c 3 chnh phm v 1 ph phm B l bin c ly c 2
ph phm v mt chnh phm Hi l bin c trong 8 sp c i chnh phm P(A)=
P(H3).P(A/H3) + P(H4).P(A/H4)+ P(H5).P(A/H5)+ P(H6).P(A/H6 )+
P(H7).P(A/H7) =1/9.(1.5/C48+ C34.4/C48+ C35.3/ C48 + C36.2/ C48+
C37.1/ C48 ) =0,2 B xy ra khi i=4,5 P(H4/A)=
P(H4).P(A/H4)/P(A)=0,127 P(H5/A)= P(H5).P(A/H5)/P(A)=0.238 P(B)=
P(H4/A).P(B/ H4A)+ P(H5/A).P(B/ H5A) =0,127.C23/C34 + 0,238.C12/
C34=0,214 Bi 1.91 Gi A l bin c ly c sn phm tt Hi l bin c lc ban u
hp c i sn phm tt i = 0, nP ( Hi ) = i n i +1 n +1
P ( A \ Hi ) =
V b vo hp c n sn phm 1 sn phm ri sau ly ngu nhin ra 1 sn phm nn
p dng cng thc xc sut y , ta c:P ( A) = P ( Hi ) P ( A \ Hi )i=0
n
Ta c P( A) = 2(n + 1) p s: 2(n + 1) Bi 1.92 Gi A l bin c k sn
phm ly ra u l chnh phm. Hi ln lt l bin c trong hp c i chnh phm, vi
i=1,2,6. Theo cng thc Bayes th:n+2
n+2
P(Hi/A) = V ta ly ngu nhin ln lt k sn phm theo phng thc c hon li
nn tng s kt cc ng kh nng xy ra ng bng s chnh hp lp chp k ca n, tc l
bng nk. Mt khc, nu hp cha i chnh phm, th c C cch xy ra ( c C cch
chn i sn phm trong n sn phm), vi mi cch nh th, s cch ly c k chnh
phm trong s i chnh phm theo phng thc c hon li l ik. Vy
P(Hi).P(A/Hi) = P(Hn/A) = =1. P(A) = P(H1).P(A/H1) + P(H2).P(A/H2)
++P(Hn).P(A/Hn) = ++.+
=( n + 2k +
3k ++ nk) / nk
Vy, P(Hn/A) =
Bi 1.93 TH1) H1 = A thng chung cuc trong 2 vn P(H1) = C 1 .
p(1-p) + C 2 .p2 = 2p(1-p) + p2 2 2 TH2) H2 = A thng chung cuc
trong 4 vn P(H2) = C 2 .p2(1-p)2 + C 3 .p3(1-p) + C 4 .p4 4 4 4 2 2
2 4 = 6p (1-p) + 4p (1-p) +p A thng chung cuc trong 2 vn d hn trong
4 vn P(H1) > P(H2) 2p(1-p) + p2 > 6p2(1-p)2 + 4p3(1-p) + p4 (
0 6p(1-p)2 +4p2(1-p) + p3 (p-1)2(p- 3 ) < 0 p 2 p>0 32
Vy A thng chung cuc trong 2 vn d hn trong 4 vn th Bi 1.94 Gi A l
bin c sn phm th nht l ph phm B l bin c sn phm th 2 l ph phm.
2 > p>0 3
Vy bin c trong 2 sn phm mua c 1 sn phm ph phm l bin c tng A. +
.B y l 2 nhm bin c xung khc nn xc sut tng bng tng cc xc sut Gi H1,
H2 l bin c sn phm mua thuc day chuyn s 1, s 2. Ta tnh P(A. )trc. P
(H1) = P(H2) = 0,5 P(A) = 0,5.0,2 +0,5.0,03 = 0,025 Sau khi A xy ra
th cc xc sut ca H1, H2 thay i theo cng thc Bayes nh sau: P(H1/A) =
= P(H2/A) = = P( /A ) = .0,98 + .0,97 = 0,974 Vy P(A, ) =
0,025.0,0974 = 0,02435.
Tnh P( , B)
Tng t, P( ) = 0,5.0,98 + 0,5.0,97 = 0,975 P(H1/ ) = = P(H2/ ) =
= P(B/ ) = 0,02. + 0,03. = Vy P( , B) = 0,975. = 0,02435 Vy P =
0,02435 + 0,02435 =0,0487 1.95 Co cac truong hop xay ra la: Goi x
la so lan nem trung ro cua nguoi 1, y la so lan nem trung ro cua
nguoi 2. a. co cac truong hop thoa man la: (2,1);(2,0);(1,0)( cac
truong hop nay la xung khac), xac suat nem trung cua nguoi 1 va
nguoi 2 la hoan toan doc lap. Theo becnuli va quy tac nhan: 2 1 1
P(a)= c 2 (0.6)^2* c 2 0.7*0.3+ c 2
0.6*0.4*0,3^2+0,6^2*0,3^2=0,2268 b. tuong tu co cac cap thoa man
la` (2,2),(1,1),(0,0) 1 1 P(B)=0,6^2.0,7^2+ c 2 .0,6.0,4. c 2
.0,7.0,3+0,4^2.0,3^2=0,3924 Bi 1.96 gi s ln rt th i ngi 1 rt c qu
trng. ln 1: ln 2: ln 3: P= a/(a+b) P= b/(a+b). b/(a+b). a/(a+b) = [
b/(a+b)]^2 .a/(a+b) P= [ b/(a+b)^4 . a/(a+b)
p l xc xut ngi th nht rt c qu trng trc
P= a/(a+b) + [b/(a+b)] ^2 . a/(a+b) a/(a+b) +............ P=
a/(a+b) . { [ 1 + [b/(a+b)]^2 t [b/(a+b)]^2 = X +
+ [ b/(a+b)] ^4 . [b /(a+b)]^4...........}
ta c ( [ 1 + [b/(a+b)]^2
+
[b /(a+b)]^4...........} )
= 1 + X + X^2 + X^3 +............. = 1/(1-X) (dy cp s nhn li v
hn: co so la X < 1) = P= a/(a+b) . 1/ (1 - X) = (a+b) / (a+2b)
Bi 1.97: Trong rp c n ch c nh s , n ngi c v vo ngi mt cch ngu nhin.
Tm xc sut c m ngi ngi ng ch. Yu cu bi ton l tm xc sut m ngi ngi ng
ch v (n- m) ngi cn li ngi sai ch. A= m ngi ngi ng ch. S trng hp ng
kh nng c th xy ra khi xp m ngi vo n v tr l S trng hp thun li cho
bin c A l mA= Vy P(A) =
A
m n
.
C
m n
B= n- m ngi cn li ngi sai ch. C= C t nht 1 ngi trong n- m ngi cn
li ngi ng ch n- m v tr cn li. Ta c P(B)= P(C) Ai= Ngi th i ngi ng
ch. Vi i= 1, n m . P(Ai)=1 n m
C A
m
n m n
=
1 m!
v C=
n m i= 1
Ai< j
i
V cc bin c Ai khng xung khc vi nhau nn P i - P ( Ai A j ) + ...
+ (-1)n-1P(A1A2...An-m) P(C)=
(A)i
C n-m ch nn
1 P( A ) =(n-m). n m =1i i
V Ai, Aj l cc bin c c iu kin nn P(AiAj)= P(Ai)P(Aj/Ai)= n m n m
11 1
P( A A ) = Ci< j i j
2 n m
.
1 1 1 . = n m n m 1 2!1n m
Tng t P(A1A2...An-m)= (n m)!
1 P(C)= 1- 2! + 3! +... + ( )
1
1
.
n m n m 1 1 = ) . ( 1 ( n m)! i =1 ( n m)!
P(B)=1-P(C) = 1-
n m i= 1
( 1 )1 m!n m h= 0
n m
n m 1 . = ( n m)! h= 0
( ) 1 h!
h
Theo nh l nhn xc sut, v A, B c lp P(AB)= P(A)P(B)=
( ) 1 h!
h
Bi 1.98: Mt h thng k thut gm n b phn vi xc sut hot ng tt ca mi b
phn l p. H thng s ngng hot ng khi c t nht mt b phn b hng. nng cao
tin cy ca h thng ngi ta d tr thm n b phn na theo 2 phng thc( hnh v
SBT/T37) Hi phng thc d tr no em li tin cy cao hn cho h thng? tin cy
ca h thng d tr theo phng thc a: Pa = [1- (1-p)2]n = pn(2-p)n (V:
(1-p)2 l xc sut c 2 b phn cng mt nhm b hng, khi h thng b hng) tin
cy ca h thng d tr theo phng thc b: Pb = 1- (1-pn)2 = pn(2-pn) (V
(1-pn)2 l xc sut m trong c hai nhnh, mi nhnh u c 1 b phn b hng) Ta
cn chng minh: Pa>Pb hay (2-p)n > (2-pn) (2-p)n + pn > 2 t
q = 1-p ta c: (1+q)n + (1-q)n >2 (ng theo khai trin nh thc
Newton) Vy: phng thc d tr a mang li tin cy cao hn cho h thng. Bi
1.99 Hai nguoi nay choi toi da la m+n-1 van thi moi biet duoc minh
thang hay thua. Xet cac truong hop thang cho nguoi thu nhat: Ta
phan thanh cac truong hop nguoi do phai danh lan luot la : m, m+1,
m+2,.,m+n-1 tran moi biet minh thang( so thu tu cau van thang cuoi
cung lan luot la m, m+1, m+2,.,m+n-1). Cac truong hop nay lan luot
xung khac. Ap dung becnuli P(m)=(1/2)^m m 1 1 P(m+1)= c m
.(1/2)^(m-1).1/2.1/2=1/2. c m .(1/2)^m P(m+2)= c m+1
.(1/2)^(m-1).(1/2)^2.1/2=1/2^2. c m+1 .(1/2)^mm 1 2
. m 1 n P(m+n-1)= c m+n 2 .(1/2)^(m-1).(1/2)^(n-1).1/2=(1/2)^n.
c m+n 2 .(1/2)^(n-1) Nen xac suat nguoi thu nhat: P(1)=
(1/2)^m(1+1/2.
c
1 m
+1/2^2. c m+1 +.+ (1/2)^(n-1). c m+n 2 )n
2
Hoan toan tuong tu cho nguoi thu 2. ti le chia tien la
p(1)/p(2)
Bi 1.100: Mt ngi b 2 bao dim vo ti, mi bao c n que dim. Mi khi
ht thuc ngi rt ngu nhin mt bao v nh mt que. Tm xc sut khi ngi pht
hin mt bao ht dim th bao kia cn li ng r que dim. Ngi pht hin mt bao
ht v bao cn li r que dim nn ngi ly dim 2n-r+1 ln ( v 1 ln l pht hin
1 bao ht nn ko ly c dim!) S cc trng hp ng kh nng l : 22n-r+1 (v c 2
bao nn mi ln rt dim c 2 kh nng) c 1 bao cn li r que dim nn trong
2n-r+1 ln ly dim phi c r ln khng r ly phi bao . S trng hp r ln ly
ko vo bao : C 2 n r +1 Vy xc sut khi ngi pht hin mt bao ht dim v
bao kia cn li ng r que dim l :r C 2 n r +1 P = 2 n r +1 2
