2.115 A large number industrial firm uses 3 local motels to……

Plumbing faulty20 % - Ramada Inn ~ 5%30 % - Lakeview Motor Lodge ~ 8%50 % - Sheraton ~ 4%

Let R ~ Ramada Inn L ~ Lakeview Motor Lodge S ~ Sheraton P ~ Plumbing faulty

P(R) = 0.2 P(P | R) = 0.05P(L) = 0.3 P(P | L) = 0.08 P(S) = 0.5 P(P | S) = 0.04

9

4

054.0

0.08(0.3)

P(P)

L)P(L)|P(P

P(P)

L)P(PP)|P(L b)

0.054

0.04(0.5)0.08(0.3)0.05(0.2)

S)P(S)|P(PL)P(L)|P(PR)P(R)|P(P

S)P(PL)P(PR)P(PP(P) a)

2.116 4 men, 5 women, choose 3 committees

ways84 nrestrictio no if way of No. ) a 39 C

ways40 women2 and man 1 way withof No. b) 25

14 CC

ways15

in must man certain a if woman1 and men 2 way withof No. c)

15

13

11 CCC

2.117 Let Rn ~ n patient recover from a delicate heart operation

Given P(R1) = 0.8

a) Exactly 2 of next 3 patients P1 P2 P3

b) All the 3 next patients

384.03)2.0((0.8))P(R 122

0.512(0.8))P(R 33

2.118 3

2)25(P

5

3)Male(P

8

5)Female 25(P

5

2

5

31)Female(P

120

13

8

5

3

1

5

2

Female)25P(P(Female)25)P(Female) 25P(

3

1

3

21)25(P

2.119 4 Red 5 Green 6 Yellow

Choose 9 apples if 3 of each color are to be selected

Number of ways =

ways80036

35

34 CCC

2.120 6 black balls 4 green balls3 balls are drawn with replacement

Let B ~ black balls G ~ green balls

a) All 3 are the same color

28.010

4

10

6 P(3G) P(3B)

33

b) Each color is represented

0.72

0.28-1

] P(3G) P(3B) [-1

d)represente is color P(each

2.121 12 TV 3 defective

Let D ~ defective TV set

Hotel receive at least 2 defective sets from 5 sets

ways504378126

defective 1defective zero defective 2least at

49

13

59

03

CCCC

2.122

B)'(A a)

A

CB

B)'(A b)

A

CB

B)(A c) C

A

CB

2.123

Let A ~ consulting firm A B ~ consulting firm B C ~ consulting firm C O ~ cost overrun

P(A) = 0.40 P(O|A) = 0.05P(B) = 0.35 P(O|B) = 0.03P(C) = 0.25 P(O|C) = 0.15

P(O’ | A) = 0.95

P(O | B) = 0.03

P(O | A) = 0.05

P(O’ | B) = 0.97

P(A) = 0.4

P(C) = 0.25

P(B) = 0.35

P(O | C) = 0.15

P(O’ | C) = 0.85

0.068

(0.25) 0.15(0.35) 0.03(0.40) 0.05

P(C)C)|P(OP(B)B)|P(OP(A)A)|P(OP(O)

5515.0 068.0

(0.25) 0.15

P(O)

P(C) C)|P(O

P(O)

O)P(CO)|P(C a)

2941.0 068.0

(0.40) 0.05

P(O)

P(A) A)|P(O

P(O)

O)P(AO)|P(A b)

2.124 3 temperatures 4 cooking times 3 oils

studied be to nscombinatio 36a) 13

14

13 CCC

oil of type each for

used be willnscombinatio 12b) 13

14

11 CCC

c) Because the arrangement among temperature, cooking time and oil are not important here

2.125 Suppose that the manufacturer can try only two combinations in a day

18

1

36

2

day) a in nscombinatio twoonly P(try a)

b) Let H ~ Highest temperature is used in either of these two combinations Total number of way,

12 )n(H 13

14

11 CCC

3

1

36

12P(H)

2.126 Let C ~ a woman over 60 has the disease P ~ positive result after the test N ~ negative result after the test

P(C) = 0.07 P(C’) = 0.93

Given that 10% incorrectly gives a negative resultP( N | C ) = 0.10

5% incorrectly gives a positive resultP( P | C’) = 0.05

P(P | C) = 0.90

P(N | C’) = 0.95

P(C) = 0.07

P(C’) = 0.93

P(N | C) =

0.10

P(P |C’) = 0.05

0.0070.070.1P(C) A)|P(NC)P(N

0.04650.930.05)P(C' )C'|P(P)C'P(P

0.88350.930.95)P(C' )C'|P(N)C'P(N

0.0630.070.9P(C) C)|P(PC)P(P

P(N)

N)P(CN)|P(C

0.8905

0.88350.007

)C'P(NC)P(NP(N)

00786.08905.0

007.0

2.128 Let A ~ affected P ~ positive result

N ~ negative result

P(A) = 0.002 P(A’) = 0.998P( P | A ) = 0.95 P( P | A’ ) = 0.01P( N | A ) = 0.05 P( N | A’ ) = 0.99

P(N | A) = 0.05

P(P | A’) = 0.01

P(A) = 0.002

P(C’) = 0.998

P(P | A) =

0.95

P(N | A’) = 0.99

0.00190.0020.95P(A) A)|P(P)P(P A

0.988020.9980.99)P(A' )A'|P(N)'P(N A

0.009980.9980.01)P(A' )A'|P(P)A'P(P

0.00010.0020.05P(A) A)|P(N)P(N A

P(P)

)P(PP)|P(A

A

0.01188

0.009980.0019

)'P(PA)P(PP(P)

A

1599.00118.0

0019.0

2.129 Let 1 ~ Engineer 1 2 ~ Engineer 2

E ~ error

P(E’ | 1) = 0.98

P(E | 2) = 0.04

P(1) = 0.7

P(2) = 0.3

P(E | 1) =

0.02

P(E’ | 2) = 0.96

0.0140.70.02P(1) 1)|P(E)1P(E

0.0120.30.04P(2) 2)|P(E)2P(E

5385.0026.0

014.0

P(E)

1)P(EE)|P(1

0.026

0.0120.014

)2P(E)1P(EP(E)

4615.0026.0

012.0

P(E)

2)P(EE)|P(2

Engineer 1 has the higher probability because if there is an error occur, probability for Engineer 1 did the work is 0.5385 which is higher than Engineer 2 (0.4615).

2.130 Let D ~ defective

P(D) = 0.20

a) If three items arrive off the process line in succession

b) If four items arrive off the process line in succession

008.0(0.20)defective) are 3 P(all 3

0256.0)80.0((0.20) 4defective) are P(3 3

2.131 Let A ~ admitted to hospital B ~ back on the job the next day

P(A) = 0.10P(B) = 0.15

0.02B)P(A

0.23

0.02 -0.15 0.10

B)P(A-P(B)P(A)B)P(A

2.132 Let T ~ attend the training course M ~ able to meet their production quotas

Given P(M | T) = 0.90P(M | T’) = 0.65P(T) = 0.5

5806.0775.0

0.45

)5.0(65.0)5.0(90.0

0.90(0.5)

))P(T'T'|P(MT)P(T)|P(M

T)P(T)|P(M

P(M)

M)P(TM)|P(T

2.133 Let D ~ dissatisfied A ~ purchase from vendor A

Given P(D) = 0.10P(A | D) = 0.50P(A) = 0.20

0.25 0.20

)10.0( 50.0

P(A)

P(D) D)|P(AA)|P(D

2.134

Union Nonunion TOTAL

Same Company

40 15 55

New Company (same field)

13 10 23

New field 4 11 15

Unemployed 2 5 7

TOTAL 59 41 100

Let U ~ union member NS ~ new company (same field) UE ~ unemployed

23

13

10023

10013

P(NS)

NS)P(UNS)|P(U a)

59

2

10059

1002

P(U)

U)P(UEU)|P(UE b)

2.135 Let C ~ the queen is a carrier H ~ the prince has hemophilia

P( H | C) = 0.5 P( H’ | C ) = 0.5P( H | C’) = 0 P( H’ | C’) = 1

We want to find for )H'H'P(H'

)H'H'H'P(C)H'H'H'|P(C

H

H’

H

H’

0.5

0.5

0.5

0.50

1

0

1

0

1

0

1

H

H’H

H’H

H’

H

H’

C

C’

0.5

0.5

H

H’

H

H’

0.5

0.5

0

1

H

H’

H

H’

0

1

0

1

H

H

H’

H

H’H

H’

H’

0.5

0.5

0.5

0.5

0.5

0.5

0.5

0.5

16

9

2

1

16

1

)H'H'H'P(C')H'H'H'P(C)H'H'P(H'

2

11

2

1)H'H'H'P(C'

16

1

2

1)H'H'H'P(C

3

4

9

1

169

161

)H'H'P(H'

)H'H'H'P(C)H'H'H'|P(C

2.136

P( no two students will have the same birth date in a size of 60 class )

60(365) ! 305

! 364

365

306

365

363

365

364

365

1