Chuong 2-Dai Luong Ngau Nhien Va Cac Phan Phoi Xac Suat

1

BAI GIAI XAC SUAT THONG KE

(GV: Tran Ngoc Hoi 2009)

CHNG 2

AI LNG NGAU NHIEN VA PHAN PHOI XAC SUAT

Bai 2.1: Nc giai khat c ch t Sai Gon i Vung Tau. Moi xe ch 1000 chai bia Sai Gon, 2000 chai coca va 800 chai nc trai cay. Xac suat e 1 chai moi loai b be tren ng i tng ng la 0,2%; 0,11% va 0,3%. Neu khong qua 1 chai b be th lai xe c thng. a) Tnh xac suat co t nhat 1 chai bia Sai Gon b be. b) Tnh xac suat e lai xe c thng. c) Lai xe phai ch t mat may chuyen e xac suat co t nhat mot chuyen c thng khong nho hn 0,9?

Li giai Tom tat:

Loai Bia Sai Gon

Coca Nc trai cay

So lng/chuyen 1000 2000 800 Xac suat 1 chai be

0,2% 0,11% 0,3%

- Goi X1 la LNN ch so chai bia SG b be trong mot chuyen. Khi o,

X1 co phan phoi nh thc X1 B(n1,p1) vi n1 = 1000 va p1 = 0,2% = 0,002. V n1 kha ln va p1 kha be nen ta co the xem X1 co phan phan phoi Poisson:

X1 P(a1) vi a1 = n1p1 = 1000.0,002 = 2, ngha la X1 P(2).

- Tng t, goi X2 , X3 lan lt la cac LNN ch so chai bia coca, chai nc trai cay b be trong mot chuyen. Khi o, X2 , X3 co phan phoi Poisson:

X2 P(2000.0,0011) = P(2,2); X3 P(800.0,003) = P(2,4).

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a) Xac suat co t nhat 1 chai bia Sai Gon b be la

2 02

1 1e 2P(X 1) 1 P(X 0) 1 1 e 0, 8647.0!

= = = = =

b) Tnh xac suat e lai xe c thng. Theo gia thiet, lai xe c thng khi co khong qua 1 chai b be, ngha la

X1 + X2 + X3 1.

V X1 P(2);X2 P(2,2); X3 P(2,4) nen X1 + X2 + X3 P(2+2,2 + 2,4) = P(6,6) Suy ra xac suat lai xe c thng la:

P(X1 + X2 + X3 1) = P[(X1 + X2 + X3 =0) + P(X1 + X2 + X3 = 1)]=

6 ,6 0 6 , 6 1e (6, 6 ) e (6, 6 )0 ! 1 !

+ = 0,0103.

c) Lai xe phai ch t mat may chuyen e xac suat co t nhat mot chuyen c thng khong nho hn 0,9? Goi n la so chuyen xe can thc hien va A la bien co co t nhat 1 chuyen c thng. Yeu cau bai toan la xac nh n nho nhat sao cho P(A) 0,9. Bien co oi lap cua A la: A khong co chuyen nao c thng. Theo cau b), xac suat e lai xe c thng trong mot chuyen la p = 0,0103. Do o theo cong thc Bernoulli ta co:

n n

n

P(A) 1 P(A) 1 q 1 (1 0, 0103)1 (0, 9897) .

= = = =

Suy ra

n

n

P(A) 0, 9 1 (0, 9897) 0, 9(0, 9897) 0,1n ln(0, 9897) ln 0,1

ln 0,1n 222, 3987ln(0, 9897)

n 223.

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Vay lai xe phai ch t nhat la 223 chuyen. Bai 2.2: Mot may tnh gom 1000 linh kien A, 800 linh kien B va 2000 linh kien C. Xacsuat hong cua ba linh kien o lan lt la 0,02%; 0,0125% va 0,005%. May tnh ngng hoat ong khi so linh kien hong nhieu hn 1. Cac linh kien hong oc lap vi nhau. a) Tnh xacsuat e co t nhat 1 linh kien B b hong. b) Tnh xac suat e may tnh ngng hoat ong. c) Gia s trong may a co 1 linh kien hong. Tnh xac suat e may tnh ngng hoat ong.

Li giai Tom tat:

Loai linh kien A B C So lng/1may 1000 800 2000 Xac suat 1linh kien hong 0,02% 0,0125% 0,005%

- Goi X1 la LNN ch so linh kien A b hong trong mot may tnh. Khi

o, X1 co phan phoi nh thc X1 B(n1,p1) vi n1 = 1000 va p1 = 0,02% = 0,0002. V n1 kha ln va p1 kha be nen ta co the xem X1 co phan phan phoi Poisson:

X1 P(a1) vi a1 = n1p1 = 1000.0,0002 =0,2, ngha la

X1 P(0,2).

- Tng t, goi X2, X3 lan lt la cac LNN ch so linh kien B, C b hong trong mot may tnh. Khi o, X2 , X3 co phan phoi Poisson nh sau:

X2 P(800.0,0125%) = P(0,1);

X3 P(2000.0,005%) = P(0,1).

a) Xac suat co t nhat 1 linh linh kien B b hong la:

0,1 0

0,12 2

e (0,1)P(X 1) 1 P(X 0) 1 1 e 0, 0952.0!

= = = = =

b) Tnh xac suat e may tnh ngng hoat ong.

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Theo gia thiet, may tnh ngng hoat ong khi so linh kien hong nhieu hn 1, ngha la khi

X1 + X2 + X3 > 1.

V X1 P(0,2);X2 P(0,1); X3 P(0,1) nen X1 + X2 + X3 P(0,2+0,1 + 0,1) = P(0,4) Suy ra xac suat e may tnh ngng hoat ong la:

P(X1 + X2 + X3 > 1) = 1 - P(X1 + X2 + X3 1) = 1- [P(X1 + X2 + X3 = 0) + P(X1 + X2 + X3 = 1)] =

0,4 0 0,4 1e (0, 4) e (0, 4)10! 1!

= 1-1,4.e-0,4 = 0,0615 = 6,15%. c) Gia s trong may a co 1 linh kien hong. Khi o may tnh ngng

hoat ong khi co them t nhat 1 linh kien hong na, ngha la khi

X1 + X2 + X3 1.

Suy ra xac suat e may tnh ngng hoat ong trong trng hp nay la:

P(X1 + X2 + X3 1) = 1 - P(X1 + X2 + X3 < 1) = 1- P(X1 + X2 + X3 = 0) =

0,4 0e (0, 4)10!

= 1-e-0,4 = 0,3297 = 32,97%.

Bai 2.3: Trong lng cua mot loai san pham c quan sat la mot ai lng ngau nhien co phan phoi chuan vi trung bnh 50kg va phng sai 100kg2 . Nhng san pham co trong lng t 45kg en 70kg c xep vao loai A. Chon ngau nhien 100 san pham (trong rat nhieu san pham). Tnh xac suat e a) co ung 70 san pham loai A. b) co khong qua 60 san pham loai A. c) co t nhat 65 san pham loai A.

Li giai

Trc het ta tm xac suat e mot san pham thuoc loai A.


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Goi X0 la trong lng cua loai san pham a cho. T gia thiet ta suy ra X0 co phan phoi chuan X0 N(0, 02) vi 0 = 50, 02 = 100 (0 = 10). V mot san pham c xep vao loai A khi co trong lng t 45kg en 70kg nen xac suat e mot san pham thuoc loai A la P(45 X0 70). Ta co

0 0

00 0

70 45 70 50 45 50P(45 X 70) ( ) ( ) ( ) ( )10 10

(2) ( 0,5) (2) (0,5) 0, 4772 0,1915 0, 6687.

= = = = + = + =

(Tra bang gia tr ham Laplace ta c (2) = 0,4772; (0,5) = 0,1915).

Vay xac suat e mot san pham thuoc loai A la p =0,6687. Bay gi, kiem tra 100 san pham. Goi X la so san pham loai A co trong 100 san pham c kiem tra, th X co phan phoi nh thc X B(n,p) vi n = 100, p = 0,6687. V n = 100 kha ln va p = 0,6687 khong qua gan 0 cung khong qua gan 1 nen ta co the xem X co phan phoi chuan nh sau:

X N(, 2) vi = np = 100.0,6687 = 66,87;

npq 100.0, 6687.(1 0, 6687) 4,7068. = = = a) Xac suat e co 70 san pham loai A la:

1 70 1 70 66, 87P (X 70) f ( ) f ( )4,7068 4,7068

1 0, 3209f (0, 66) 0, 0681 6, 81%.4,7068 4,7068

= = = = = = =

(Tra bang gia tr ham Gauss ta c f(0,66) = 0,3209). b) Xac suat e co khong qua 60 san pham loai A la:

60 0 60 66,87 0 66,87P (0 X 60) ( ) ( ) ( ) ( )4,7068 4,7068

( 1,46) ( 14,21) (1,46) (14,21) (1,46) (5)0,4279 0,5 0,0721 7,21%.

= = = = + = + = + = =

(Tra bang gia tr ham Laplace ta c (14,21) = (5) = 0,5; (1,46) = 0,4279).

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c) Xac suat e co t nhat 65 san pham loai A la:

100 65 100 66,87 65 66,87P (65 X 100) ( ) ( ) ( ) ( )4,7068 4,7068

(7,0388) ( 0,40) (5) (0,4) 0,5 0,1554 0,6554 65,54%.

= = = = + = + = =

(Tra bang gia tr ham Laplace ta c (7,7068) (5) = 0,5; (0,4) = 0,1554). Bai 2.4: San pham trong mot nha may c ong thanh tng kien, moi kien gom 14 san pham trong o co 8 san pham loai A va 6 san pham loai B. Khach hang chon cach kiem tra nh sau: t moi kien lay ra 4 san pham; neu thay so san pham thuoc loai A nhieu hn so san pham thuoc loai B th mi nhan kien o; ngc lai th loai kien o. Kiem tra 100 kien (trong rat nhieu kien). Tnh xac suat e a) co 42 kien c nhan. b) co t 40 en 45 kien c nhan. c) co t nhat 42 kien c nhan.

Li giai

Trc het ta tm xac suat e mot kien c nhan. Theo gia thiet, moi kien cha 14 san pham gom 8A va 6B. T moi kien lay ra 4 san pham; neu thay so san pham A nhieu hn so san pham B, ngha la c 3A,1B hoac 4A, th mi nhan kien o. Do o xac suat e mot kien c nhan la:

3 1 4 08 6 8 6

4 4 4 4 414 14

C C C CP (3 k 4) P (3) P (4) 0, 4056C C

= + = + = Vay xac suat e mot kien c nhan la p = 0,4056. Bay gi, kiem tra 100 kien. Goi X la so kien c nhan trong 100 kien c kiem tra, th X co phan phoi nh thc X B(n,p) vi n = 100, p = 0,4056. V n = 100 kha ln va p = 0,4056 khong qua gan 0 cung khong qua gan 1 nen ta co the xem X co phan phoi chuan nh sau:

X N(, 2) vi = np = 100.0,4056 = 40,56;

npq 100.0, 4056.(1 0, 4056) 4, 9101. = = =

a) Xac suat e co 42 kien c nhan la:


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1 42 1 42 40,56 1P (X 42) f ( ) f ( ) f (0, 29)4, 9101 4, 9101 4, 9101

0, 3825 0, 0779 7,79%.4, 9101

= = = = = = =

(Tra bang gia tr ham Gauss ta c f(0,29) = 0,3825).

b) Xac suat e co t 40 en 45 kien c nhan la 45 40 45 40,56 40 40,56P (40 X 45) ( ) ( ) ( ) ( )

4,9101 4,9101(0,90) ( 0,11) (0,90) (0,11) 0,3159 0,0438 0,3597 35,97%.

= = = = + = + = =

(Tra bang gia tr ham Laplace ta c (0,9) = 0,3519; (0,11) = 0,0438).

c) Xac suat e co t nhat 42 kien c nhan la

100 42 100 40,56 42 40,56P (42 X 100) ( ) ( ) ( ) ( )4,9101 4,9101

(12) (0,29) 0,50 0,1141 0,3859 38,59%.

= = = = = =

(Tra bang gia tr ham Laplace ta c (12) = (5) = 0,5; (0,29) = 0,1141). Bai 2.5: San pham trong mot nha may c ong thanh tng kien, moi kien gom 10 san pham So san pham loai A trong cac hop la X co phan phoi nh sau:

X 6 8

P 0,9 0,1 Khach hang chon cach kiem tra nh sau: t moi kien lay ra 2 san pham; neu thay ca 2 san pham eu loai A th mi nhan kien o; ngc lai th loai kien o. Kiem tra 144 kien (trong rat nhieu kien). a) Tnh xac suat e co 53 kien c nhan. b) Tnh xac suat e co t 52 en 56 kien c nhan. c) Phai kiem tra t nhat bao nhieu kien e xac suat co t nhat 1 kien c nhan khong nho hn 95%?

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Li giai

Trc het ta tm xac suat p e mot kien c nhan. Goi C la bien co kien hang c nhan. Ta can tm p = P(C). T gia thiet ta suy ra co hai loai kien hang: Loai I: gom 6A, 4B chiem 0,9 = 90%. Loai II: gom 8A, 2B chiem 0,1 = 10%. Goi A1, A2 lan lt la cac bien co kien hang thuoc loai I, II. Khi o A1, A2 la mot he ay u, xung khac tng oi va ta co

P(A1) = 0,9; P(A2) = 0,1. Theo cong thc xac suat ay u ta co:

P(C) = P(A1) P(C/A1) + P(A2) P(C/A2). Theo gia thiet, t moi kien lay ra 2 san pham; neu ca 2 san pham thuoc loai A th mi nhan kien o. Do o:

2 06 4

1 2 210

C C 1P(C / A ) P (2) ;C 3

= = =

2 08 2

2 2 210

C C 28P(C / A ) P (2) .C 45

= = = Suy ra P(C) = 0,9. (1/3) + 0,1.(28/45) = 0,3622. Vay xac suat e mot kien c nhan la p = 0,3622. Bay gi, kiem tra 144 kien. Goi X la so kien c nhan trong 144 kien c kiem tra, th X co phan phoi nh thc X B(n,p) vi n = 144, p = 0,3622. V n = 144 kha ln va p = 0,3622 khong qua gan 0 cung khong qua gan 1 nen ta co the xem X co phan phoi chuan nh sau:

X N(, 2) vi = np = 144.0,3622 = 52,1568;

npq 144.0,3622.(1 0, 3622) 5,7676. = = = a) Xac suat e co 53 kien c nhan la P(X=53) = 6,84% (Tng t Bai 21). b) Xac suat e co t 52 en 56 kien c nhan la P(52 X 56) = 26,05% (Tng t Bai 21). c) Phai kiem tra t nhat bao nhieu kien e xac suat co t nhat 1 kien c nhan khong nho hn 95%? Goi n la so kien can kiem tra va D la bien co co t nhat 1 kien c nhan. Yeu cau bai toan la xac nh n nho nhat sao cho P(D) 0,95.


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Bien co oi lap cua D la D: khong co kien nao c nhan. Theo chng minh tren, xac suat e mot kien c nhan la p = 0,3622. Do o Theo cong thc Bernoulli ta co:

n n nP(D) 1 P(D) 1 q 1 (1 0, 3622) 1 (0, 6378) .= = = =

Suy ra n

n

P(D) 0, 95 1 (0, 6378) 0, 95(0, 6378) 0, 05n ln(0, 6378) ln 0, 05

ln 0, 05n 6, 6612ln(0, 6378)

n 7.

Vay phai kiem tra t nhat 7 kien. Bai 2.6: Mot may san xuat san pham vi t le san pham at tieu chuan la 80% va mot may khac cung san xuat loai san pham nay vi t le san pham at tieu chuan la 60%. Chon ngau nhien mot may va cho san xuat 100 san pham. Tnh xac suat e

a) co 70 san pham at tieu chuan. b) co t 70 en 90 san pham at tieu chuan. c) co khong t hn 70 san pham at tieu chuan.

Li giai

Goi X la LNN ch so san pham at tieu chuan trong 100 san pham. A1, A2 lan lt la cac bien co chon c may 1, may 2. Khi o A1, A2 la mot he ay u, xung khac tng oi va ta co:

P(A1) = P(A2) = 0,5. Theo cong thc xac xuat ay u, vi moi 0 k 100, ta co:

1 1 2 2

1 2

P(X = k) = P(A )P(X=k/A ) + P(A )P(X= k/A )1 1= P(X=k/A )+ P(X=k/A )2 2

(1)

Nh vay, goi X1, X2 lan lt la cac LNN ch so san pham at tieu chuan trong trng hp chon c may 1, may 2. Khi o:

(1) cho ta 1 21 1P(X = k) = P(X =k)+ P(X =k)2 2

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X1 co phan phoi nh thc X1 B(n1,p1) vi n1 = 100, p1 = 80% = 0,8. V n1 = 100 kha ln va p1 = 0,8 khong qua gan 0 cung khong qua gan 1 nen ta co the xem X1 co phan phoi chuan nh sau:

X1 N(1, 12) vi 1 = n1p1 = 100.0,8 = 80;

1 1 1 1n p q 100.0, 8.0, 2 4. = = = X2 co phan phoi nh thc X2 B(n2,p2) vi n2 = 100, p2 = 60% =

0,60. V n2 = 100 kha ln va p2 = 0,60 khong qua gan 0 cung khong qua gan 1 nen ta co the xem X2 co phan phoi chuan nh sau:

X2 N(2, 22) vi 2 = n2p2 = 100.0,60 = 60;

2 2 2 2n p q 100.0, 60.0, 40 4, 8990. = = = a) Xac suat e co 70 san pham at tieu chuan la:

1 21 2

1 1 2 2

70 701 1 1 1 1 1P(X = 80) = P(X =70)+ P(X =70) = f ( ) f ( )2 2 2 2

1 1 70 80 1 1 70 60 1 1 1 1= . f ( ) . f ( )= . f ( 2,5) . f (2,04)2 4 4 2 4,8990 4,8990 2 4 2 4,89901 1 1 1= . 0,0175 . 0,0498 0,0007272 4 2 4,8990

+ + +

+ =

b) Xac suat e co t 70 en 90 san pham at tieu chuan la:

1 2

1 1 2 2

1 1 2 2

1 1P(70 X 90) = P(70 X 90)+ P(70 X 90)2 2

90 70 90 701 1= [ ( ) ( )] [ ( ) ( )]2 21 90 80 70 80 1 90 60 70 60= [ ( ) ( )] [ ( ) ( )]2 4 4 2 4,899 4,8991= [ (2,5) ( 2,5) (6,12) (2,04)]21= (0,49379 0,2

+ +

+

+ 49379 0,5 0,47932)0,50413

+ =

c) Xac suat co khong t hn 70 san pham at tieu chuan la P(70 X 100) =0,5072

(Tng t cau b) Bai 2.7: Mot may san xuat san pham vi t le phe pham la 1% va mot may khac cung san xuat loai san pham nay vi t le phe pham la 2%.


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Chon ngau nhien mot may va cho san xuat 1000 san pham. Tnh xac suat e a) co 14 phe pham. b) co t 14 en 20 phe pham.

Li giai

Goi X la LNN ch so phe pham trong 1000 san pham. A1, A2 lan lt la cac bien co chon c may 1, may 2. Khi o A1, A2 la mot he ay u, xung khac tng oi va ta co:


1 1 2 2

1 2


(1)

Nh vay, goi X1, X2 lan lt la cac LNN ch so phe pham trong trng hp chon c may 1, may 2. Khi o:

(1) cho ta 1 21 1P(X = k) = P(X =k)+ P(X =k)2 2 X1 co phan phoi nh thc X1 B(n1,p1) vi n1 = 1000 va p1 = 1% =

0,001. V n1 kha ln va p1 kha be nen ta co the xem X1 co phan phan phoi Poisson:


X2 co phan phoi nh thc X2 B(n2,p2) vi n2 = 1000 va p2 = 2% = 0,002. V n2 kha ln va p2 kha be nen ta co the xem X2 co phan phan phoi Poisson:


a) Xac suat e co 14 phe pham la: 10 14 20 14

1 21 1 1 e 10 1 e 20P(X = 14) = P(X =14)+ P(X =14) = 0,04542 2 2 14! 2 14!

+ =

b) Xac suat e co t 14 en 20 phe pham la:

1 2

20 2010 k 20 k

k 14 k 14

1 1P(14 X 20) = P(14 X 20)+ P(14 X 20)2 2

1 e 10 1 e 20= 31,35%2 k! 2 k !

= =

+ =

12

Bai 2.8: Mot x nghiep co hai may I va II. Trong ngay hoi thi, moi cong nhan d thi c phan mot may va vi may o se san xuat 100 san pham. Neu so san pham loai A khong t hn 70 th cong nhan o se c thng. Gia s oi vi cong nhan X, xac suat san xuat c 1 san pham loai A vi cac may I va II lan lt la 0,6 va 0,7. a) Tnh xac suat e cong nhan X c thng. b) Gia s cong nhan X d thi 50 lan. So lan c thng tin chac nhat la bao nhieu?

Li giai

Goi Y la LNN ch so san pham loai A co trong 100 san pham c san xuat. A1, A2 lan lt la cac bien co chon c may I, may II. Khi o A1, A2 la mot he ay u, xung khac tng oi va ta co:


1 1 2 2

1 2

P(Y = k) = P(A )P(Y=k/A ) + P(A )P(Y= k/A )1 1= P(Y=k/A )+ P(Y=k/A )2 2

(1)

Nh vay, goi X1, X2 lan lt la cac LNN ch so san pham loai A co trong 100 san pham c san xuat trong trng hp chon c may I, may II. Khi o:

(1) cho ta 1 21 1P(Y = k) = P(X =k)+ P(X =k)2 2 X1 co phan phoi nh thc X1 B(n1,p1) vi n1 = 100, p1 = 0,6. V

n1 = 100 kha ln va p1 = 0,6 khong qua gan 0 cung khong qua gan 1 nen ta co the xem X1 co phan phoi chuan nh sau:

X1 N(1, 12) vi 1 = n1p1 = 100.0,6 = 60;

1 1 1 1n p q 100.0, 6.0, 4 4, 8990. = = = X2 co phan phoi nh thc X2 B(n2,p2) vi n2 = 100, p2 = 0,7. V n2

= 100 kha ln va p2 = 0,7 khong qua gan 0 cung khong qua gan 1 nen ta co the xem X2 co phan phoi chuan nh sau:

X2 N(2, 22) vi 1 = n2p2 = 100.0,7 = 70;

2 2 2 2n p q 100.0,7.0, 3 4,5826. = = = a) Xac suat e cong nhan X c thng la:


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1 2

1 1 2 2

1 1 2 2

1 1P(70 Y 100) = P(70 X 100)+ P(70 X 100)2 2

100 70 100 701 1= [ ( ) ( )] [ ( ) ( )]2 21 100 60 70 60 1 100 70 70 70= [ ( ) ( )] [ ( ) ( )]2 4,899 4,899 2 4,5826 4,58261= [ (8,16) (2,04) (6,55) (0)2

+ +

+ 1]= (0,5 0,47932 0,5) 0,26032

+ =

b) Gia s cong nhan X d thi 50 lan. So lan c thng tin chac nhat la bao nhieu?

Goi Z la LNN ch so lan cong nhan X c thng. Khi o Z co phan phoi nh thc Z B(n,p) vi n = 50, p = 0,2603. So lan c thng tin chac nhat chnh la Mod(Z). Ta co: Mod(Z) k np q k np q 1 50.0,2603 0,7397 k 50.0,2603 0,7397 1 12,2753 k 13,2753 k 13

= + + =

Vay so lan c thng tin chac nhat cua cong nhan X la 13 lan. Bai 2.9: Trong ngay hoi thi, moi chien s se chon ngau nhien mot trong hai loai sung va vi khau sung chon c se ban 100vien an. Neu co t 65 vien tr len trung bia th c thng. Gia s oi vi chien s A, xac suat ban 1 vien trung bia bang khau sung loai I la 60% va bang khau sung loai II la 50%. a) Tnh xac suat e chien s A c thng. b) Gia s chien s A d thi 10 lan. Hoi so lan c thng tin chac nhat la bao nhieu? c) Chien s A phai tham gia hoi thi t nhat bao nhieu lan e xac suat co t nhat mot lan c thng khong nho hn 98%?

Li giai

Goi X la LNN ch so vien trung trong 100 vien c ban ra. Goi A1, A2 lan lt la cac bien co chon c khau sung loai I, II. Khi o A1, A2 la mot he ay u, xung khac tng oi va ta co:


14

1 1 2 2

1 2


(1)

Nh vay, goi X1, X2 lan lt la cac LNN ch so vien trung trong 100 vien c ban ra trong trng hp chon c khau loai I, II. Khi o:

(1) cho ta 1 21 1P(X = k) = P(X =k)+ P(X =k)2 2 X1 co phan phoi nh thc X1 B(n1,p1) vi n1 = 100, p1 = 0,6. V n1


X1 N(1, 12) vi 1 = n1p1 = 100.0,6 = 60;

1 1 1 1n p q 100.0, 6.0, 4 4, 8990. = = = X2 co phan phoi nh thc X2 B(n2,p2) vi n2 = 100, p2 = 0,5. V n2


X2 N(2, 22) vi 1 = n2p2 = 100.0,5 = 50;

2 2 2 2n p q 100.0,5.0,5 5. = = = a) Xac suat e chien s A c thng la:

1 2

1 1 2 2

1 1 2 2

1 1P(65 X 100) = P(65 X 100)+ P(65 X 100)2 2

100 65 100 651 1= [ ( ) ( )] [ ( ) ( )]2 21 100 60 65 60 1 100 50 65 50= [ ( ) ( )] [ ( ) ( )]2 4,899 4,899 2 5 51 1= [ (8,16) (1,02) (10) (3)]= (0,5 0,32 2

+ +

+ 4614 0,5 0,49865) 0,0776.+ =

b) Gia s chien s A d thi 10 lan. So lan c thng tin chac nhat la bao nhieu? Goi Y la LNN ch so lan chien s A c thng. Khi o Y co phan phoi nh thc Y B(n,p) vi n = 10, p = 0,0776. So lan c thng tin chac nhat chnh la mod(Y). Ta co: mod(Y) k np q k np q 1 10.0,0776 0,9224 k 10.0,0776 0,9224 1 0,1464 k 0,8536 k 0

= + + =


15

Vay so lan c thng tin chac nhat cua chien s A la 0 lan, noi cach khac, thng la chien s A khong c thng lan nao trong 10 lan tham gia. c) Chien s A phai tham gia hoi thi t nhat bao nhieu lan e xac suat co t nhat mot lan c thng khong nho hn 98%? Goi n la so lan tham gia hoi thi va D la bien co co t nhat 1 lan c thng. Yeu cau bai toan la xac nh n nho nhat sao cho P(D) 0,98. Bien co oi lap cua D la D: khong co lan nao c thng. Theo chng minh tren, xac suat e mot lan c thng la p = 0,0776. Do o Theo cong thc Bernoulli ta co:

n n nP(D) 1 P(D) 1 q 1 (1 0, 0776) 1 (0, 9224) .= = = =

Suy ra n

n

P(D) 0, 98 1 (0, 9224) 0, 98(0, 9224) 0, 02n ln 0, 9224 ln 0, 02

ln 0, 02n 48, 43ln 0, 9224

n 49.

Vay chien s A phai tham gia hoi thi t nhat la 49 lan. Bai 2.10: Mot ngi th san ban 4 vien an. Biet xac suat trung ch cua moi vien an ban ra la 0,8. Goi X la ai lng ngau nhien ch so vien an trung ch. a) Tm luat phan phoi cua X. b) Tm ky vong va phng sai cua X.

Li giai

a) Ta thay X co phan phoi nh thc X B(n,p) vi n = 4, p = 0,8. X la LNN ri rac nhan 5 gia tr: 0, 1, 2, 3 , 4. Luat phan phoi cua X co dang:

X 0 1 2 3 4 P p0 p1 p2 p3 p4

16

Theo cong thc Bernoulli ta co: 0 0 441 1 342 2 243 3 144 4 04

P(X 0) (0, 8) (0, 2) 0, 0016;

P(X 1) (0, 8) (0, 2) 0, 0256;

P(X 2) (0, 8) (0, 2) 0,1536;

P(X 3) (0, 8) (0, 2) 0, 4096;

P(X 4) (0, 8) (0, 2) 0, 4096.

CCCCC

= = == = == = == = == = =

Vay luat phan phoi cua X la:

X 0 1 2 3 4 P 0,0016 0,0256 0,1536 0,4096 0,4096

b) Tm ky vong va phng sai cua X. - Ky vong: M(X) = np = 3,2. - Phng sai: D(X) = npq = 0,64. Bai 2.11: Co hai lo hang I va II, moi lo cha rat nhieu san pham. T le san pham loai A co trong hai lo I va II lan lt la 70% va 80%. Lay ngau nhien t moi lo 2 san pham. a) Tnh xac suat e so san pham loai A lay t lo I ln hn so san pham loai A lay t lo II. b) Goi X la so san pham loai A co trong 4 san pham c lay ra. Tm ky vong va phng sai cua X.

Li giai Goi X1, X2 lan lt la cac LNN ch so sp loai A co trong 2 sp c

chon ra t lo I, II. Khi o X1 co phan phoi nh thc X1 B(n1, p1); n1 = 2; p1 = 70% = 0,7

vi cac xac suat nh bi: k k 2 k

1 2P(X k) (0,7) (0, 3)C = = Cu the

X1 0 1 2 P 0,09 0,42 0,49

X2 co phan phoi nh thc X2 B(n2, p2); n2 = 2; p2 = 80% = 0,8 vi cac xac suat nh bi:

k k 2 k2 2P(X k) (0, 8) (0, 2)C = =

Cu the X2 0 1 2 P 0,04 0,32 0,64


17

a) Xac suat e so san pham loai A lay t lo I ln hn so san pham loai A lay t lo II la: P(X1 X2) = P[(X1 =2)(X2 =0)+ (X1 =2)(X2 =1)+ (X1 =1)(X2 =0)] = P(X1 =2)P(X2 =0)+ P(X1 =2)P(X2 =1)+ P(X1 =1)P(X2 =0) = 0,1932.

b) Goi X la so sp loai A co trong 4 sp chon ra . Khi o

X = X1 + X2 V X1 , X2 oc lap nen ta co: - Ky vong cua X la M(X) = M(X1) + M(X2) = n1p1 + n2p2 = 3 - Phng sai cua X la D(X) = D(X1) + D(X2) = n1p1q1 + n2p2q2 = 0,74.

Bai 2.12: Cho hai hop I va II, moi hop co 10 bi; trong o hop I gom 6 bi o, 4 bi trang va hop II gom 7 bi o, 3 bi trang. Rut ngau nhien t moi hop hai bi. a) Tnh xac suat e c hai bi o va hai bi trang. b) Goi X la ai lng ngau nhien ch so bi o co trong 4 bi c rut ra. Tm luat phan phoi cua X.

Li giai Goi X1, X2 lan lt la cac LNN ch so bi o co trong 2 bi c chon ra t hop I, hop II. Khi o

- X1 co phan phoi sieu boi X1 H(N1, N1A, n1); N1 = 10; N1A= 6; n1 = 2 vi cac xac suat nh bi:

k 2 k

6 41 2

10

P(X k) .C CC

= =

Cu the

X1 0 1 2 P 6/45 24/45 15/45

- X2 co phan phoi sieu boi X2 H(N2, N2A, n2); N2 = 10; N2A = 7; n2

= 2 vi cac xac suat nh bi:

k 2 k

7 32 2

10

P(X k) .C CC

= =

Cu the

18

X2 0 1 2 P 3/45 21/45 21/45

Goi X la ai lng ngau nhien ch so bi o co trong 4 bi c rut ra. Khi o

X = X1 + X2 Bang gia tr cua X da vao X1, X2 nh sau:

X X2 X1

0 1 2

0 0 1 2 1 1 2 3 2 2 3 4

a) Xac suat e c 2 bi o va 2 bi trang la:

P(X = 2) = P[(X1=0) (X2=2)+ (X1=1) (X2=1)+ (X1=2) (X2=0)] = P(X1=0) P(X2=2)+ P(X1=1)P(X2=1)+ P(X1=2)P(X2=0)] = (6/45)(21/45) + (24/45)(21/45) + (15/45)(3/45) = 1/3. b) Luat phan phoi cua X co dang:

X 0 1 2 3 4 P p0 p1 p2 p3 p4

trong o: p0 = P(X = 0)= P(X1 =0) P(X2 = 0) = 2/225; p1 = P(X = 1)= P(X1 =0) P(X2 = 1) + P(X1 =1) P(X2 = 0)= 22/225; p2 = P(X = 2) = 1/3; p3 = P(X = 3)= P(X1 =1) P(X2 = 2) + P(X1 =2) P(X2 = 1)= 91/225; p4 = P(X = 4)= P(X1 =2) P(X2 = 2) = 7/45. Vay luat phan phoi cua X la :

X 0 1 2 3 4 P 2/225 22/225 1/3 91/225 7/45


19

Bai 2.13: Mot may san xuat san pham vi t le phe pham 10%. Mot lo hang gom 10 san pham vi t le phe pham 30%. Cho may san xuat 3 san pham va t lo hang lay ra 3 san pham. Goi X la so san pham tot co trong 6 san pham nay. a) Tm luat phan phoi cua X. b) Khong dung luat phan phoi cua X, hay tnh M(X), D(X).

Li giai

Goi X1, X2 lan lt la cac LNN ch so sp tot co trong 3 san pham do may san xuat; do lay t lo hang. Khi o X1, X2 oc lap va ta co:

- X1 co phan phoi nh thc X1 B(n1, p1); n1 = 3; p1 = 0,9. Cu the ta co:

0 0 2 31 3

1 1 2 21 3

2 2 1 21 3

3 3 0 31 3

P(X 0) p q (0,1) 0, 001;

P(X 1) p q 3(0, 9)(0,1) 0, 027;

P(X 2) p q 3(0, 9) (0,1) 0, 243;

P(X 3) p q (0, 9) 0,729.

CCCC

= = = == = = == = = == = = =

- X2 co phan phoi sieu boi X2 H(N2, N2A, n2); N2 = 10; N2A = 7; n2

= 3 (v lo hang gom 10 san pham vi t le phe pham la 30%, ngha la lo hang gom 7 san pham tot va 3 san pham xau). Cu the ta co:

0 3

7 32 3

101 2

7 32 3

102 1

7 32 3

103 0

7 32 3

10

1P(X 0) ;120

21P(X 1) ;120

63P(X 2) ;120

35P(X 3) .120

C CC

C CCC CC

C CC

= = =

= = =

= = =

= = =

a) Ta co X = X1 + X2. Luat phan phoi cua X co dang:

X 0 1 2 3 4 5 6 P p0 p1 p2 p3 p4 p5 p6

20

trong o: p0 = P(X = 0)= P(X1 = 0)P(X2 = 0) = 1/120000; p1 = P(X = 1)= P(X1 = 0)P(X2 = 1) + P(X1 = 1)P(X2 = 0) = 1/2500; p2 = P(X = 2) = P(X1 = 0)P(X2 = 2) + P(X1 = 1)P(X2 = 1) + P(X1 = 2)P(X2 =0) = 291/40000 p3 = P(X = 3) = P(X1 = 0)P(X2 = 3) + P(X1 = 1)P(X2 = 2) + P(X1 = 2)P(X2 =1) + P(X1 = 3)P(X2=0) = 473/7500 p4 = P(X = 4) = P(X1 = 1)P(X2 = 3) + P(X1 = 2)P(X2 = 2) + P(X1 = 3)P(X2 = 1)

= 10521/40000 p5 = P(X = 5) = P(X1 = 2) P(X2 = 3) + P(X1 = 3)P(X2 = 2) = 567/1250 p6 = P(X = 6) = P(X1 = 3)P(X2 = 3) = 1701/8000. Vay luat phan phoi cua X la: X 0 1 2 3 4 5 6 P 1/120000 1/2500 291/40000 473/7500 10521/40000 576/1250 1701/8000

b) V X = X1 + X2 va X1 , X2 oc lap nen ta co:

- Ky vong cua X la M(X) = M(X1) + M(X2) = n1p1 + n2 p2 = 4,8 (vi p2 = N2A/N2)

- Phng sai cua X la D(X) = D(X1) + D(X2) = n1p1q1 + n2 p2q2(N2-n2)/(N2-1)= 0,76.

Bai 2.14: Cho hai hop I va II, moi hop co 10 bi; trong o hop I gom 8 bi o, 2 bi trang va hop II gom 6 bi o, 4 bi trang. Rut ngau nhien t hop I hai bi bo sang hop II, sau o rut ngau nhien t hop II ba bi. a) Tnh xac suat e c ca 3 bi trang. b) Goi X la ai lng ngau nhien ch so bi trang co trong ba bi c rut ra t hop II. Tm luat phan phoi cua X. Xac nh ky vong va phng sai cua X.

Li giai

Goi X la LNN ch so bi trang co trong 3 bi rut ra t hop II. Ai (i = 0, 1, 2) la bien co co i bi trang va (2-i) bi o co trong 2 bi lay ra t hop I. Khi o A0, A1, A2 la he bien co ay u, xung khac tng oi va ta co:


21

0 2

2 80 2

101 1

2 81 2

102 0

2 82 2

10

28P(A ) ;45

16P(A ) ;45

1P(A ) .45

C CC

C CC

C CC

= =

= =

= =

Vi moi k = 0, 1, 2, 3 theo cong thc xac suat ay u, ta co

P(X = k) = P(A0)P(X = k/A0) + P(A1)P(X = k/A1) + P(A2)P(X = k/A2)

a) Xac suat e c ca ba bi trang la: P(X = 3) = P(A0)P(X = 3/A0) + P(A1)P(X = 3/A1) + P(A2)P(X = 3/A2)

Ma

3 0

4 80 3

123 0

5 71 3

123 0

6 62 3

12

4P(X 3 / A ) ;220

10P(X 3 / A ) ;220

20P(X 3 / A ) .220

C CC

C CC

C CC

= = =

= = =

= = =

nen P(X= 3) = 73/2475. b) Luat phan phoi cua X co dang:

X 0 1 2 3 P p0 p1 p2 p3

trong o, tng t nh tren ta co:

22

0 3 0 3 0 3

4 8 5 7 6 60 3 3 3

12 12 121 2 1 2 1 2

4 8 5 7 6 61 3 3 3

12 12 122 1 2 1 2 1

4 8 5 7 6 62 3 3 3

12 12 12

28 16 1p P(X 0) . . . 179 / 825;45 45 45

28 16 1p P(X 1) . . . 223 / 450;45 45 45

28 16 1p P(X 2) . . . 1277 / 4950;45 45 45

C C C C C CC C C

C C C C C CC C CC C C C C CC C C

= = = + + =

= = = + + =

= = = + + =

p3 = P(X= 3) = 73/2475. Suy ra luat phan phoi cua X la:

X 0 1 2 3 P 179/825 223/450 1277/4950 73/2475

T o suy ra ky vong cua X la M(X) = 1,1 va phng sai cua X la D(X) = 0,5829. Bai 2.15: Co ba lo san pham, moi lo co 20 san pham. Lo th i co i+4 san pham loai A (i = 1, 2, 3). a) Chon ngau nhien mot lo roi t lo o lay ra 3 san pham. Tnh xac suat e trong 3 san pham c lay ra co ung 1 san pham loai A. b) T moi lo lay ra 1 san pham. Goi X la tong so san pham loai A co trong 3 san pham c lay ra. Tm luat phan phoi cua X va tnh Mod(X), M(X), D(X).

Li giai

a) Goi C la bien co trong 3 san pham c lay ra co ung 1 san pham loai A. Goi A1, A2, A3 lan lt la cac bien co chon c lo I, II, III. Khi o A1, A2, A3 la mot he ay u, xung khac tng oi va P(A1) = P(A2) = P(A3) = 1/3. Theo cong thc xac suat ay u, ta co:

P(C) = P(A1)P(C/A1) + P(A2)P(C/ A2)+ P(A3)P(C/A3)

Theo Cong thc xac suat la chon:


23

1 25 15

1 3

201 2

6 142 3

201 2

7 133 3

20

525P(C / A ) ;1140

546P(C / A ) ;1140

546P(C / A ) .1140

C CC

C CC

C CC

= =

= =

= =

Suy ra P(C)= 0,4728. b) Luat phan phoi cua X co dang:

X 0 1 2 3 P p0 p1 p2 p3

Goi Bj (j = 1, 2, 3) la bien co lay c sp loai A t lo th j. Khi o B1, B2, B3 oc lap va

1 1

2 2

3 3

5 15P(B ) ; P(B ) ;20 206 14P(B ) ;P(B ) ;20 207 13P(B ) ;P(B ) .20 20

= =

= =

= =

Ta co 1 2 3 1 2 3

1 2 3 1 2 3 1 2 3

1 2 3 1 2 3 1 2 3

1 2 3 1 2 3 1 2 3

1 2 3 1 2 3

" X 0" B B B P(X 0) P(B )P(B )p(B ) 273 / 800" X 1" B B B B B B B B B

P(X 1) P(B )P(B )P(B ) P(B )P(B )P(B ) P(B )P(B )P(B ) 71 / 160" X 2" B B B B B B B B B

P(X 2) P(B )P(B )P(B ) P(B )P(B )P(B )

= = = = = = = + +

= = + + = = = + +

= = + + 1 2 31 2 3 1 2 3

P(B )P(B )P(B ) 151 / 800" X 3" B B B P(X 3) P(B )P(B )P(B ) 21 / 800

= = = = = =

Vay luat phan phoi cua X la

X 0 1 2 3 P 273/800 71/160 151/800 21/800

T luat phanphoi cua X ta suy ra mode, ky vong va phng sai cua X :

- Mode: Mod(X) = 1. - Ky vong: M(X) = 0,9. - Phng sai: D(X) = 0,625.

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2.16: Mot ngi co 5 cha khoa be ngoai rat giong nhau, trong o ch co 2 cha m c ca. Ngi o tm cach m ca bang cach th tng cha mot cho en khi m c ca th thoi (tat nhien, cha nao khong m c th loai ra). Goi X la so cha khoa ngi o s dung. Tm luat phan phoi cua X. Hoi ngi o thng phai th bao nhieu cha mi m c ca? Trung bnh ngi o phai th bao nhieu cha mi m c ca?

Li giai

Ta thay X la LNN ri rac nhan 4 gia tr: 1, 2, 3, 4. Luat phan phoi cua X co dang:

X 1 2 3 4 P p1 p2 p3 p4

Goi Aj (j = 1,2, 3, 4) la bien co cha khoa chon lan th j m c ca. Khi o: P(X=1) = P(A1) = 2/5.

1 2 1 2 1

1 2 3 1 2 1 3 1 2

1 2 3 4 1 2 1 3 1 2 4 1 2 3

P(X 2) P(A A ) P(A )P(A / A ) (3 / 5)(2 / 4) 3 / 10;

P(X 3) P(A A A ) P(A )P(A / A )P(A / A A ) (3 / 5)(2 / 4)(2 / 3) 1 / 5

P(X 4) P(A A A A ) P(A )P(A / A )P(A / A A )P(A / A A A ) (3 / 5)(2 / 4)(1 / 3)(2 / 2) 1 / 10

= = = = == = = = == = =

= = Vay luat phan phoi cua X la:

X 1 2 3 4 P 2/5 3/10 1/5 1/10

T luat phan phoi tren ta suy ra:

- Mode cua X la Mod(X) = 1.

- Ky vong cua X la i iM(X) x p 2= = . Vay ngi o thng phai th 1 chia th m c ca. Trung bnh ngi o phai th 2 cha mi m c ca. Bai 2.17: Mot ngi th san co 5 vien an. Ngi o i san vi nguyen tac: neu ban trung muc tieu th ve ngay, khong i san na. Biet xac suat


25

trung ch cua moi vien an ban ra la 0,8. Goi X la ai lng ngau nhien ch so vien an ngi ay s dung trong cuoc san. a) Tm luat phan phoi cua X. b) Tm ky vong va phng sai cua X.

Li giai

a) Ta thay X la LNN ri rac nhan 5 gia tr: 1, 2,..., 5. Luat phan phoi cua X co dang:

X 1 2 3 4 5 P p1 p2 p3 p4 p5

Goi Aj (j = 1,2,..., 5) la bien co vien an th j trung ch. Khi o:

j jP(A ) 0,8;P(A ) 0,2= = Ta co: P(X=1) = P(A1) = 0,8.

1 2 1 2

1 2 3 1 2 3

1 2 3 4 1 2 3 4

1 2 3 4 1 2 3 4

P(X 2) P(A A ) P(A )P(A ) 0,2.0,8 0,16;

P(X 3) P(A A A ) P(A )P(A )P(A ) 0,2.0,2.0,8 0,032;

P(X 4) P(A A A A ) P(A )P(A )P(A )P(A ) 0,2.0,2.0,2.0,8 0,0064;

P(X 5) P(A A A A ) P(A )P(A )P(A )P(A ) 0,2.0,2

= = = = == = = = == = = = == = = = .0,2.0,2 0,0016.=


X 1 2 3 4 5 P 0,8 0,16 0,032 0,0064 0,0016

b) T luat phan phoi cua X ta suy ra: - Ky vong cua X la M(X) = 1,2496. - Phng sai cua X la D(X) = 0,3089.

Bai 2.18: Mot ngi th san co 4 vien an. Ngi o i san vi nguyen tac: neu ban 2 vien trung muc tieu th ve ngay, khong i san na. Biet xac suat trung ch cua moi vien an ban ra la 0,8. Goi X la ai lng ngau nhien ch so vien an ngi ay s dung trong cuoc san. a) Tm luat phan phoi cua X. b) Tm ky vong va phng sai cua X.

Li giai

a) Ta thay X la LNN ri rac nhan 3 gia tr: 2, 3, 4. Luat phan phoi cua X co dang:

26

X 2 3 4 P p2 p3 p4

Goi Aj (j = 1,2, 3, 4) la bien co vien an th j trung ch. Khi o:

j jP(A ) 0,8;P(A ) 0,2= = Ta co:

1 2 1 2

1 2 3 1 2 3 1 2 3 1 2 3

1 2 3 1 2 3

1 2 3 1 2 3 1 2 3 1 2 3

1 2 3

P(X 2) P(A A ) P(A )P(A ) 0,8.0,8 0,64;

P(X 3) P(A A A A A A ) P(A A A ) P(A A A )

= P(A )P(A )P(A ) P(A )P(A )P(A ) 0,2.0,8.0,8 0,8.0,2.0,8 0,256

P(X 4) P(A A A A A A A A A A A A )

P(A )P(A )P(A ) P

= = = = =

= = + = ++ = + =

= = + + += + 1 2 3 1 2 3 1 2 3(A )P(A )P(A ) P(A )P(A )P(A ) P(A )P(A )P(A )

0,2.0,2.0,2 0,8.0,2.0,2 0,2.0,8.0,2 0,2.0,2.0,8 0,104+ +

= + + + =


X 2 3 4 P 0,64 0,256 0,104

b) T luat phan phoi cua X ta suy ra:

- Ky vong cua X la M(X) = 2,464. - Phng sai cua X la D(X) = 0,456704.

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Chuong 2-Dai Luong Ngau Nhien Va Cac Phan Phoi Xac Suat

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