Phương pháp giải toán tổ hợp xác suất

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CNU
TO HQP VA XAC SUAT

A - CHUAN KIEN TH LfC KI NING
1. Kién thi?c
- Blt: ' 'l•y lac c0ng vé guy tac nhdn; hoiin vi, chinh h pp, to hpp ch8p k ciia n phdn
lii; cfing thiic nhi thiic Niu-tern (a + b)‘.
- Biet dupc phép thii ngdu nhién; khfing gran mau; bién co lien ‹juan den phép thii
ng8u nhién; d|nh nghia ct dicn, dinh nghia thing ké x:ic sudt cua bien ct.

- Biet cJc tinh chilt: P(Ol) = 0; P(f2) = 1 ; 0 y P(A) < I .
— Bi t (khGng chfirig minh) dinh li c0ng x:ic sudl va dinh li nhdn x:ic sudt.

- Dici khii i ni in x:ic sudt co dieu ki n.

- Bucic dfiu v8n dung duoc quy lac cfing vé quy tac nhdn.

- Tinh dupc sfi c:ic ho:in v], chinh hpp, lfi hpp ch8p k cua n phdn tit.

- Bi6t khai trim nhi thiic Niu-ton véi mGt sG
- Tim d ii‹ic hc .so ciia x‘ trong khai trien nh] thuc Niuton thanh da thuc.
- X:ic d)nh dupc: phép thii ngdu nhi6n; khiing gian mdu; bien co lien quan den
phép thii ngdu nhien.

- Bill v n dung guy tac cong xiic sual, guy tac nhan xac sudt trong c:ie bai flip don
yr an.
- Sir dung dvpc x:ie sudt co diéu ki(n dé tinh to:in trong cac tinh huong don gién.

35

- Biet sit dung in:iy tinh bfi t6i hfi trp tinh xiic sudt.
3. Céc vi d g
Vi du 1. M0t doi thi ddu béng ban g8m 8 v8n dong vien nam va 7 v8n dGng vien
nii. Hfii co bao nhieu c:ich cii v8n dong vien thi ddu:
a) Don nam, don nii ?
b) D0i nam-rift ?

Léi giiii
a) 56 c:ich cii v8n d ng vién thi dltu don nam, don nii voi doi ban la:
+ Don nam cé: C}

(c:ich);

+ Don rift co: C'= 7 (cc::iicchh);
b) Sfi c:ich cii v n dGng vien thi dttu d6i nam-rift véi dfii ban la:
Véi moi vBn dong vien nam ta déu co the ghép voi mot trong 7 v n dGng vien nii de
cii ra dupc mot dGi nam-nii thi d6u, nhu v8y ta co 7 ciich;

hoan vi cua 5 chit

1,; 22;; 3; 4; 5. Do
do

nhién nay ciing chinh la s6

ho:in vi ciia 5 phlin tir dGi mot kh:ic nhau: PQ = 5! = 120.

Vi du 3. Hfii co bao nhifi u c:ich chia mot lép 40 hoc sinh thanh c:ic nhfiin hoc t p
ma m8i nhom co 8 hpc sinh ?
Liii gin:li

Ddy chinh la so tfi hpp ch p 8 ciia tfip hpp co 40 phdn tit:

40! = 76904685.
= 8!32!

36

= 1024x" + 5l20x’ + l 1520x' + 15360x’ + 13440x6 + b064x’ + 3360x4 +
+ 960x' + 180x' + 20x + 1 ;
b) H( sG ciia x’ trong khai tri6n la 8064.
Vt du S.
a) Chiing minh rang Cn ‘ C + C2 +

+ C) '2‘

b) Chiing minh rang:

C2n ‘ C + C +
nn

+ C2 _
2n «2. + c„ + c2. +

+ c"‘

-1

a) Ap dung khai trien nh] thiic Niu-ton:

b) Ap dung khai trién nh] thiic Niu-ton:

?n—I
2n

Léi gift
56 khé nang chon chit sfi hang don vi theo yéu cdu cua bai to:in la sfi khé uang
chon m6t trong 5 chit sG lé 1; 3; 5; 7; 9 la: C5

37

56 khé nang chon chit so hang van sau khi da chon chit so hang den vi la so khé
nang chon in(or trong 9 chit sfi kh:ic 0 va kh:ic chit so hang don vi la: C[ = 8 ;
Tiep luc chpn 1fin lupt c:ic chfi st› hang chuc, hang tram, hang nghin (kh:ic vcii ciic
chit sfi da chon) ta co sG khii nang chon c:ic cliii so hang chuc, hang tram, hang

nghin l6n luot la 8, 7, 6.

Theo quy iac nhdn, la co the 18p dunc 5 x 5 x 8 x 7 × 6 = 13440 so tq nhién Ie
ma
moi sG gfim 5 chit so d6i in t kh:ie nhau.
Vi du 7. Cé 30 cau hfii kh:ic nhau cho mot mtin hoc, trong do cé 5 chu hfii kho, 10
cdu hfii irung binh va 15 cilu hoi de. Tit c:ic cdu hfii do co the l0p dune bao nhicu
de kiém ma, moi de gom 5 cau hfii khac nhau sao cho trong moi de phéi cé du ba
loai cllu hfii (kho, trung bi nh, de) va sfi cdu de khong ii hon 2 ?
Liii giai

Theo yf:u c6u ta co:
- Néu mii de co 3 cdu de thi chac chan phai co them 1 chu trung binh va 1 cilu khfi;
- Néu mGi dé co 2 cdu de thi kém theo phéi la 2 cdu trung binh va 1 cdu kho hoac
I cdu trung binh va 2 cau kho.

Vi d;i 8. Gieo mOt con siic sac dong chfit.
a) Hay m6 té kh8ng gian mdu;
b) Hay xéc dinh bien co “xudt hi(n mat co so ch6m la so Ie”.
a) Mfit con sdc sac co s:iu mat la: 1 chhm; 2 chdm; 3 chdm; 4 chdm; 5 chdm;
6 chain. Do v8y, kh8ng gian mdu la f2 =|l ; 2 ; 3 ; 4 ; 5 ; 6[ ;
b) Bien cfi A: “xudt hi(n ma;t co so chain la so lé” la tdp h pp bao g6m c:ic
phép thii gieo con siic sac xuat hi(n mat co so chdm la so lé, nhu v)y x:ic dJnh
bien co A chinh la x:ic dinh t8p con A = ]1 ; 3 ; 5) ciia khong
gian m6u f2 = ]l ; 2 ; 3 ; 4 ; 5 ; 6}.

Vi dii 9. Gieo hat con silc sac dong chdt. Tinh x:ic suat ciia bien co “tong so chain
trcn m a't xu I hi(n ciia hai con siic sac bang 8”.

Lcii gicii
Khong gian in.a u co 6 x ft = 36 bien co x:iy ra. Bien co “tong so chain trén mat
xudt hien cua hat con siic sae bang 8” x:iy ra voi 5 khé nang sau:
2

3

6

4

4

3

2

36
Vt du 10. Chon nyau nhi n 5 so In nhien tit 1 den 20tJ. Tf nh gdn diing x:ic suat de
5 so nay deu nhfi hon 51).
Léi giiii
So phép chon 5 so In nhién lii 1 den 2011 la C/„ ;
So phép chpn 5 so tu nhién tit I dcn 49 Ta C„ .

V8y x:ic sufil ‹Ie 5 so dvpc chon deu nho hon 50 la
Vt d¿ 11. Mfit hop dimg 4 vicn bi xanh, 3 vién bi dfi va 2 vién bi vang.
a) Chon ngd u nhién liar vien bi tir hop bi do. "Finh x:ic suat de chon duoc hat
vien bi ciing in u va xac sudt de chon dv‹c hat vicn bi kh:ic mau.
.b) Chon ngau nhién ba vién bi lii hop bi do. Tinh x:ie suat dc chon dii‹ic ba
vicn bi he n loin kh:ic in u.
Léi giai
a) Khfing gian riiau cé C/, phép chon;
56 phép chon de duoc hat vien ciing mau xanh la C² ;
56 phép chon de dvoc hat vién ciing mau dfi lé C; ;

39

Sti phép chpn d dune hat vi8n ciing mau vang lé C,.

Vtly x:ic sudt dé chpn dupc hai vien bi ciing mau It P -

5

c;

xéc suit dé chpn dunc hat vifin bi kh:ic mau la I — P = I — — = —
b) KhGng gian méu cé C, phép chon, s6 phép chpn theo yeu cdu cua bai la

C4 .C].C[. V8y x:ic sudt dé chpn ducic ba vi8n bi hoan toan kh:ic milu lé
p

CC 44 '’CC Q .C,

,

22



’7

(véi mii can tit: s6 ’I dé“n sc›“ 87 duét’ déy, c’é nhiéu phitong an In:a chon, hiy
khoanh tron véo c’hii: c’éi dii:rig din phu:on$ én ma em cho let diing, trit céc’ can
sci“ 4, 41, 55)
Céu 1. to hai tsp hpp htm han A va B, ki hiéu n(X) la sG ph8n tit cua mot t8p hpp
X. Khi dé:
a) n(A

B) = n(A)

n(B);

b) n(A in B) = n(A) — n(B);
c) n(A in B) = n(A) + n(B);
d) n(A in B) = n(A) + n(B) — n(A in B).
Can 2. to hai t8p hpp hint han A va B kh0ng cé phdn tit chung, ki hieu n(X) la sfi
phdn tit ciia mGt t8p hpp X. Khi dé:
a) n(A

B) = n(A)

n(B);

b) n(A w B) = n(A) in n(B);

c) n(A

B) = n(A) + n(B);

d) n(A in B) = n(A) — n(B).
Cd u 3. to hai tdp hpp htm han A va B, ki hi(u n(X) la sfi phdn tit cfia mGt t8p
hop
X. Khi do:
a) n(A X B) = n(A) — n(B);
b) n(A X B) = n(A) — n(B) + n(A in B);
c) n(A X B) = n(A) — n(B) — n(A in B);
d) n(A X B) = n(A) — n(A in B).

Céu 4. to brit khang d)nh nko sau ddy la sai:
a) Néu A vd B la hat tttp kh6ng giao nhau thi n(A

B) - n(A) + n(B);

b) Gié sit m6t cGng vi(c co thé du‹ic thuc hi(n theo mGt trong hai phuong :in
A hoac B. Co n c:ich thuc hi(n phuong :in A va in c:ich thuc hi(n phuong :in B.
Khi do c0ng vi(c co thé dvoc thuc hi(n béi in + n c:ich;
c) Gié sit phéi thqc hi(n hai c0ng vi(c A ho0c B. Cé n c:ich thvc hi(n cfing vi(c A
va in c:ich thirc hi(n cGng vi(c B. Khi dé co the dupc thuc hien hai cGng vi(c b6i in +
n c:ich;

d) Gié sit phéi thuc hi(n hai cong viec A hoa) c B d6c 18p véi nhau. Co n
c:ich thvc hi(n cfing vi(c A va in c:ich thuc hi6n cfing vi(c B. Khi do co the thqc
hicn hai cong vi(c b6i in + n c:ich.
Ciiu 5. Mht ban co 20 quyen s:ich, 30 quyf:n ver. Khi dé, tong sG s:ich ver ciia ban
:iy la:
a) 20;

b) 30;

c) 50;

d) 10.

Ciiu 6. M0t khung gfi co hinh ngfi gi:ic Hi ABCDE (c:ic dinh ldy theo thu tu do)
va co in t thanh go riot du rig chéo AD. Mot con kien di tit A den D infit c:ich
ngau nhién. Khi do st c:ich kh:ic nhau ma con kién co the di la:
a) 1;

b) 2;

c) 3;

d) 4.

Clin 7. Mot tru rig Trung hpc phO thong co 150 hpc sinh khoi 10; 250 hpc sinh
khoi 11; 180 hpc sinh khoi 12. Khi do, iong so hoc sinh cua triiéing la:
a) 150;

b) 250;

c) 180;

d) 580.

Ciiu 8. Moi h6p co 10 vien bi trang; 20 vién bi xanh; 30 vién bi dfi. So c:ich
chon ng6u nhién m6t trong sG c:ic vién bi thufic h6p do la:
41

a) l tl;

b) 20;

c) 30;

d) 60.

Ciu 9. MW i d‹)i the" thao cd 10 v4n d ›ng vién nam va IS vAn d6ng vi£n nii tham
gia ihi dim b‹›ng ban. Khi dé, so c$ch khlic nhau cé the cii ngdu nhién mt›t v8n
dting vi6n ra ihi ‹lilu la:

a) 10;

b) 15:

c) 25;

d) 5.

Cir 10. Mhi Ill ct 411 hpc sinh, trong dfi céi 15 ban hpc giéi m6n Van; 20 ban
hpc gioi m6n To:in; ltl ban hpc gioi c$ hat mon V6n va Toén. Khi d6, sfi ban
khdng hpc gioi mon nao (ir‹›ng sG hat mdn Van va Toén ) cfia lép d6 la:
a) 5;

b) IS;

c) 20;

d) 25.

Céu 11. M6t edu lac b6 c‹› 6() ngitiii dang ki hpc hat m6n co vua vé bfing dé.
Bi6t rang trong dé cé 511 ngiiiii dang ki hpc cd vua, 30 nguiii dang ki h9c béng
d:i. Khi d‹i. sfi nj;i ‹i ding Lf line cii li.ii m6n cd vaa va bdng d:i la:
a) Ill;

bJ 29:

c) 30:

‹il 9.

a) 5;

b) 6;

cI I I;

d) 30.

Cir J‹f. C1i‹› tip hcp A g6m in philn itt. tip l«q› b g‹›in n phan iii. Khi dé. so c ch
chon ng$u nhic³ii mlii ciip J x; y). trong d‹i x lliu‹›c I:ip hop A cor y thu6c t p hpp B

Céu i5. Cli‹› e:ie tap hpp A; II; C lfin liter ‹ ‹› tit. ii: p phiiii iii.

a) in;

li) in + n ‹ p;

‹: i init + rip + pm;

minh. Klli ‹l‹i, s‹› ‹.it'Il tit‹) ID! tiiC kli‹i:i I I \.i‹ nli:iti I.i:

d) m.n.p.

a) 27;

b) 311;

c) 729;

d) I()0t1.

Ciiu 17. Co 8 0 vufing difoc xep thanh mot hang dpc. Co hat loai bra hinh vufing
dupe t6 mau dfi hoa' c xanh. Moi fi vufing dvpc gan ngau nhien mol mieng bra
hinh vuong va mii c:ich gan nhv the got la mob tin hieu. Khi do, so tin hieu kh:ic
nhau dv‹ic tao thanh ino) t c:ich ngdu nhien theo cach trcn la:
a) 16;

b) 64;

c) 128;

d) 256.

Céu 18. Mfit truéing Trung hoc pho thGng co 100 hoc sinh khoi 10; I 5tl hoc sinh
khoi I I va 200 hoc sinh khoi 12. Ngvcii ta muon cu ra ba ngifcii, moi ngticii thuoc
mol kh‹ii dé thay mat hoc sinh nha truériy di dii irai hé. Khi do,
“ Sch cii ba hoc
sinh cua truéing di du trai hé la:

Ciiu 19. Dfiu xudn, bon ban A, B, C. D muon ru nhau di chi i nhurig chua bict
kh‹› i hanh nhv the nao cho tieii. Vi vay, he guy voc ncu at xuat phat dan lien st
din nha ban lhii hat, sau dé cé hai ban ciing dcn nhii ban thii ba, ... cho den khi
ga' p mat ca 4 ban. Khi do, sG c:ich cé the x iiy ra mol ciich ngdu nhien la:
a) 1 ;

b) 4;

c) 16;

d) ?º4.

a) 5;

b) 25;

c) 120;

‹I) 3125.

a) l ;

b) 36;

c) 7211;

d J 46656.

a) 1 ;

b) 36;

c) 72tJ;

d) 144().

a) 1;

b) 100;

c) 3628800;

d) I 00tJ00tltJtl00.

a) 1;

b) 100;

c) 3628800;

d) 10000000000.

Cm 25. Co 10 ban nam va 10 ban nii xép ngfiu nhién thanh hang dpc nhimg xen
ké in t nam, mot rift. Khi dé, so tGi da c:ic kha nEng co thé xay ra la:

a) 10;

b) 100;

c) 10!;

d) 2 x 10!.

Cdu 26. to tttp hpp A gfim n phdn tu va k la in t s6 tq nhién thuGc [1; n]. Mfii
c:ich lily ra k phhn tir

a) ph6n bi(t ciia A dupc gpi la ino) t chinh hpp ch8p k ciia n phdn tit da cho;
b) doi in(ot kh:ic nhau cua A dupe gpi la infit chinh hpp chdp k ciia n phdn tir
da cho;

c) co phdn bi t thii tu ciia A dupc gpi la mtit chinh hpp ch8p k ciia n phdn tit
da cho;
d) kh0ng phdn bi(t thii tq ciia A dupc gpi la in t chinh hpp chsp k ciia n phdn tit
da cho.
Céu 27. MGt giai thé thao chi co ba giéi la nh6t, nhi va ba. Trong so 20 v8n dong
vi6n di thi, sfi khé nang ma ba nguiii cé thé dupc ban to chiic trao gi:ii nhttt, nhi va
ba mGt c:ich ngéu nhi6n la:

a) I ;

b) 3;

c) 6;

d) 6840.

Ciiu 28. to cac chit sfi 1; 2; 3; 4; 5; 6. Khi dfi, so ciic so tu nhién co 3 chit s8,
doi
in t kh:ic nhau, dupc thanh lsp tit c:ie chit sG da cho lé:
a)6;

b) 18;

c) 120;

d)729.

Ciiu 29. M(ot lép co 40 hoc sinh. Khi d6, sG c:ich kh:ic nhau de co the cii mot
céch
ng5u nhit:ii 10 hoc sinh
a I di truc
a) 4;

c) Ph = 30!;

b) P, =
10!;
d) C40 '

847660582.

Clin 30. Trén dudng tron cho n di6m (ph6n bi(I). So céc tain giiic cé dinh l$y trong
c:ic diem da cho la:

a) n;

Clin 31. Mot hop co 10 vien bi trang, 20 vién bi xanh va 3tJ vién bi dfi. Sd ciich
chon ngdu nhién 5 trong so c:ic vien bi thufic hop do de duoc 8 vién bi trang la:
d) C60

Céu 32. M t hGp cé 10 vién bi trang, 20 vién bi xanh va 30 vien bi dfi. 56 c:ich
chpn ng5u nhien 8 trong sG c:ic vién bi thu0c http dé de dupc 8 vién bi ciing mau
lé:

d) C60

Céu 33. Trén mat phang P cé hai during thang cat nhau d va d’. Trén P co in dvéng
thang phdn bi(t va song song v6i during thang d; dong théi cé n duéng thang phdn
bi(t va song song voi duéfng thang d’. Khi do so c:ie hinh binh hénh dupc tao thanh
iii c:ic dvéng thang song song riot trén (trii d va d') la:
a) in.n;

d)

b) C +n '

2

Ciiu 34. to tain gi:ic ABC, tren ’i canh AB, BC, CA lan luot l4y in, n, p diem
phdn bi(t (va khong trung vdi dinh ciia tain gi:ic). Khi dé, so tain gi:ic co dinh
trong st c:ic diem da cho la:
a) in.n.p;
c)

3

Ciiu 35. Cho c:ic chfi sfi 0; 1; 2; 3; 4; 5; 6. Khi
ciic so tu
do,
dfii mGt kh:ic nhau diipc thanh 18p tit céc chfr so da cho la:
a) A 6 — 360
;

b) A4 — 840 ;

c) C7‘ = 35 ;

d) 720.

Ciiu 36. M0t hop co 10 vién bi trang, 211 vién bi xanh va 30 vién bi dfi. 56 ciich
chon ngdu nhién trong sfi ciic vién bi thu0c hGp do de duoc 8 vién bi ma khfing
co vién bi xanh nao Id:

C‹iu 37. M0t h6p co 10 vién bi trang, 20 vién bi xanh vé 30 vién bi dfi. 56 c:ich
chon ngdu nhién trong sG c:ic vién bi thufic hop dfi de duoc 8 vien bi, trong do
co diing m6t vién bi xanh la:
45

a)

Ciiu 38. Moi hGp co 10 vicn bi trang, 20 vien bi xanh va 30 vien bi dfi. 56 c:ich
chon ngdu nhien trong so c:ic vien bi thufic hop dfi de duoc 5 vi?n bi, trong do co
it nhal m6t vién bi xanh la:
1

7

1

2

3

4

5

6

7

7

Clin 39. Mfit hop co 10 vien bi trang, 20 vién bi xanh va 30 vién bi dfi. So c:ich
chon ngdu nhien trong sit c:ic vién bi thuoc h p dfi de dvoc 8 vi‹:n bi, trong d‹i co
dung m6t vien bi xanh va co diing 2 vien bi dfi la:

Goi S = C’n —

n -3

b) S '

C n —I ³

d) S = 3Ck.
Cbu 41. Dang thfic nao sau d6y la sai:
7

c) C,007

Clin 42. Ta co:

6

n—I
2n .

n+I

+

+ c,n

Clin 43. Khai trién P(x) = (x + y

)6

thanh da thiic, ta co:

c) P(x) = x + 6x'y
d) P(x) = x‘ + 6x’y
Ciiu 44. Khai trién P(x) = (x — 2y)' thanh da thiic, ta co:
b) P(x) — 6

6x’ 2y + 15x’ 2y2 20x’ 2y' + l5x' 2y4

6x 2y5 + 2y6

c) P(x)
12x y + 60x’y'— 160x'y' + 240x'y4 192xy5 + 64y6
32 +
2g 3
4+ 5
Ciiu 45. G9i S = 2’ + 5.24g 3 + I 0.2'.
+ 5.2.
ta c‹›:
10.2
d) P(x)

6

a) S = 625;

b) S = 3125;

c) S = 1875fl;
d) S = I .
4
Ciiu 46. Gpi S — 7’— 5.7‘.3 + I 0.7'.3'— 10.7'.3' + 5.7.
, ta co:
a) S = 100000;
b) S = 1024;
c) S = 1025;
y)4
6
Clin 47. Got S = x
6x’3y + l 5x (3y)'— 20x'(3y)’ + l5x’(3

Can 48. Gpi S = 32x'— 80x
l0x

d) S = (x -- 3y)6
4 + S0x'— 40x' +

a) S = (1 — 2x)";

b) S = (I + 2x) ;

c) S = (2x — l) ;

d) S = (x — I ) .

Ciiu 49. Ta co:
a) 1 + 2 + 3 + ... + n = C',
45

l, ta

d) S = I .
6x(3y)’ + (3y)‘, la

47

Clin 50. Ta co:
2n

’ ‘’ +

2n
2n—1

2n—I

2n

n

2

a) S = 0;

Ciiu 52. Got P(x) = (3x

b) S = n;

c) S = 2";

d) S = gnp

n

Ciiii 55. Ucii n, k, p la c:ic so tu nhie“ n va k, p ciing thufic [1; n] thi dang thiic nao
sau dtly la sai:

2

C) C n'CC nn
--44

k

” 4C n-4

Ciiu 56. Xét phép thii gieo hat ding tien ciing mot liic, hai ldn (kh6ng tinh truéng
hpp hai ding tién xep de lén nhau) ta co khfing gian m6u la
a) f2 = { SS; SN; NS; NN };
b) f2 = SS; SN;
c) II — |(SS; SS); (SS; SN); (SS; NN); (SN; NN); (SN; SS); (NN; SS); (NN;
d) f2 = ] (SS; SS); (SS; SN); (SS; NN); (SN; SS); (SN; SN); (SN; NN); (NN;
SS; SN; SN); (NN; NN) }.

Clin 57. Xét phép thii giGo hai dong tién ciing mGt fire, hat 16n (khGng tinh
truéng hpp hat dGng tien xép dé len nhau). Gpi A la bien co “ket qué cua hai lhn
gieo la nhv nhau” thi
a) ºA = { SS; NN};
b) ×A 'I (SS; SS); (NN; NN); (SN; SN) ] ;
c) ×A = I (SS; SS; (SS; NN); (NN; SS), (NN; NN) ] ;
= ] (SS; SS); (SS; SN); (SS; NN); (SN; SS); (SN; SN); (SN; NN);
(NN,SS); (SN; SN); (NN; NN) ] .

Ciiu 58. X ét phép ihir gico mot con siic sac hai lfin. Gpi N la bien co “13n da'u
xu3t hien mat 5 chdm” thi

a) ºN = { 5; 5 J ;
b) ×N 'I (6; l ); {6; 2); 56; 3); (6; 4); (6; 5) ;

c) ³²N = ] (5; I ); (5; 2); (5; 3); (5; 4); (5; 5); (5; 6) ] ;
d)

N 'I (

); (I ; 2); (1 ; 3); (I ; 4); (1 ; 5); (1; 6) J .

Céu 59. Xét phép thu gico mot con siic sac hat tan. Got C la bien co “tong so
chilm trén mat xudt hien cua hat lan gieo bang 9” ihi

50

c) D, —

(9; II; (8; I ); (7; 2); (6,3); (5; 4); (4,5); (3; 6); (2; 7);
(1; 8); (II; 9) J ;

‹1) I2 = ] (6; 3); (5; 4); (4; 5); (3; 6) |.

Ciiu 60. Xé1 phép t hii gieo mGt con siic sac hat ldn. Goi A la bien cfi “tong s6
chhm tren mat xu llt hien ciia hat lan gieo lé mot sG chan”; B la bien co “tong sG
chain trc:n ma' t xuat hi(n ciia hai lan gieo bang 7” thi

Ciiu 61. Xét phép th ii gieo mot con site sac hat ldn. Goi A la bien co “tong so
chdm men mat xudi hien cua hat ldn gieo lé m6t s6 chan”; B la bien co “tong so
chdm tren mat xudt hien ciia hat 16n gieo la mot sfi ie” thi A in B
“ doi ciia A;
Ciiu 62. Xéi phép Thu gieo mot con siic sac hai ldn. Gpi N la bien co “lan ddu
xudt hien mat 5 chdm”; M la bien co “ldn thii hat xudt hi(n mat 5 chain” thi
a) M in N =|5; 5 } ;

b) M in N — ] (5; 1); (5, 2); (5; 3); (5; 4); (5; 5); (5; 6) |;
— { (1; 5; (2; 5); (3; 5); (4; 5); (5; 5); (6; 5) } ;
d) M in N = |(5;
l ); (›; 2); (5; 3); (5; 4); (5; 5); QS; 6); (1; 5);
(2; 5); (3; 5); (4; 5); (6; 5) ] .

Can 63. Xét phép thii gieo mot con sue sac hai 1$n. Gpi N la bien co “ldn ddu
xudi hicn mai 5 chhm”, got M Ta bien cfi “lan hat xult hi(n mat 5 chhm” thi
u) M in N = |5; 5 ] ;
b) M in N = (5; 1); (5, 2); (5; 3); (5; 4); (5; 5); (5; 6) [ ;

c) M in N = ] (l ; 5); (2; 5); (3; 5); (4; 5); (5; 5); (6; 5) ];
d) M in N — ( (5; 1); (5; 2); (5; 3); (5; 4); (5; 5); (5; 6); (I ; 5);
(2; 5); (3; 5); (4; 5); (6; 5) ] .

Ciiu 64. Mot hop chiia 15 vien bi trang, 20 vien bi xanh va 25 vien bi dfi. L6y
ngdu nhien lii hop ra mGt vien bi, khi dfi x:ic sultt dé ldy dupc mot vién bi do la: