C2 Prob Extra

2.115 A large number industrial firm uses 3 local motels to…… Plumbing faulty 20 % - Ramada Inn ~ 5% 30 % - Lakeview Motor Lodge ~ 8% 50 % - Sheraton ~ 4% Let R ~ Ramada Inn S ~ Sheraton P(R) = 0.2 P(L) = 0.3 P(S) = 0.5

L ~ Lakeview Motor Lodge P ~ Plumbing faulty P(P | R) = 0.05 P(P | L) = 0.08 P(P | S) = 0.04

a) P(P) = P(P ∩ R) + P(P ∩ L) + P(P ∩ S) = P(P | R)P(R) + P(P | L)P(L) + P(P | S)P(S) = 0.05(0.2) + 0.08(0.3) + 0.04(0.5) = 0.054

P(P ∩ L) P(P) P(P | L)P(L) = P(P) 0.08(0.3) = 0.054 4 = 9

b) P(L | P) =

2.116

4 men,

5 women,

choose 3 committees

a ) No. of way if no restriction = 9C3 = 84 ways

b) No. of way with 1 man and 2 women = 4C1×5 C2 = 40 ways

c) No. of way with 2 men and 1 woman if a certain man must in

(

)

= 1C1 ×3 C1 ×5 C1 = 15 ways

2.117 Let Rn ~ n patient recover from a delicate heart operation Given P(R1) = 0.8 a) Exactly 2 of next 3 patients P(R 2 ) = (0.8)2 (0.2)1 × 3 = 0.384

b) All the 3 next patients P(R 3 ) = (0.8)3 = 0.512

P1 P2 P3

2.118

2 P(< 25 ) = 3

5 8 3 2 P(Female ) = 1 − = 5 5

3 P(Male) = 5

P( > 25 ∪ Female ) =

P( > 25) = 1 −

2 1 = 3 3

P( > 25 ∩ Female) = P( > 25) + P(Female) − P( > 25 ∪ Female) 2 1 5 = + − 5 3 8 13 = 120

2.119

4 Red

5 Green

6 Yellow

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

4 5 6 Number of ways = C3 × C3 × C3 = 800 ways

2.120

6 black balls 4 green balls 3 balls are drawn with replacement

Let B ~ black balls

G ~ green balls

a) All 3 are the same color 3 3 6 4     P(3B) + P(3G) =   +   = 0.28  10   10  b) Each color is represented P(each color is represente d) = 1 - [ P(3B) + P(3G) ] = 1 - 0.28 = 0.72

2.121

12 TV

3 defective

Let D ~ defective TV set Hotel receive at least 2 defective sets from 5 sets

at least 2 defective = zero defective + 1 defective = 3C0 ×9 C5 + 3C1×9 C4 = 126 + 378 = 504 ways

2.122 a) (A ∩ B)'

b) (A ∪ B)' A

B

A

C

B

C

c) (A ∩ C ) ∪ B A

B

C

2.123 Let A ~ consulting firm A B ~ consulting firm B C ~ consulting firm C O ~ cost overrun P(A) = 0.40 P(B) = 0.35 P(C) = 0.25

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

P(O | A) = 0.05

A) ( P

.4 0 =

P(O’ | A) = 0.95 P(O | B) = 0.03

= 0. ) B ( P

P(C )

=0

.25

35

P(O’ | B) = 0.97 P(O | C) = 0.15 P(O’ | C) = 0.85

P(O) = P(O | A) • P(A) + P(O | B) • P(B) + P(O | C) • P(C) = 0.05 (0.40) + 0.03 (0.35) + 0.15 (0.25) = 0.068 P(C ∩ O) a) P(C | O) = P(O) P(O | C) P(C) = P(O) 0.15 (0.25) = 0.068 = 0.5515

P(A ∩ O) P(O) P(O | A) P(A) = P(O) 0.05 (0.40) = 0.068 = 0.2941

b) P(A | O) =

2.124

3 temperatures

4 cooking times

3 oils

a) 3C1×4 C1×3 C1 = 36 combinatio ns to be studied b) 1C1×4 C1×3 C1 = 12 combinatio ns will be used for each type of oil

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

a) P(try only two combinatio ns in a day) 2 1 = = 36 18 b) Let H ~ Highest temperature is used in either of these two combinations Total number of way, n(H) = 1C1×4 C1×3 C1 = 12 12 1 P(H) = = 36 3

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 result P( N | C ) = 0.10 5% incorrectly gives a positive result P( P | C’) = 0.05

)= C P(

7 0 . 0

P(C ’) =

P (N

) |C

.1 0 =

0

P(N ∩ C) = P(N | A) P(C) = 0.1× 0.07 = 0.007

P(P | C) = 0.9 0

P(P ∩ C) = P(P | C) P(C) = 0.9 × 0.07 = 0.063

0.95 P(N | C’) = P(N ∩ C' ) = P(N | C' ) P(C' ) = 0.95 × 0.93 0.9 3

P(P |C’) =

= 0.8835

P(P ∩ C' ) = P(P | C' ) P(C' ) = 0.05 × 0.93 = 0.0465 0. 0

5

0.007 P(C ∩ N) = = 0.00786 P(C | N) = 0.8905 P(N) P(N) = P(N ∩ C) + P(N ∩ C' ) = 0.007 + 0.8835 = 0.8905

2.128 Let A ~ affected P ~ positive result N ~ negative result P(A) = 0.002 P( P | A ) = 0.95 P( N | A ) = 0.05

P(A’) = 0.998 P( P | A’ ) = 0.01 P( N | A’ ) = 0.99

P (A

)=

0 0.0

P(C ’) =

P (P

2

) |A

.95 0 =

P(N | A) = 0.0 5

P(P ∩ A) = P(P | A) P(A) = 0.95 × 0.002 = 0.0019 P(N ∩ A) = P(N | A) P(A) = 0.05 × 0.002 = 0.0001

0.01 P(P | A’) = 0. 9 98

P(P ∩ A' ) = P(P | A' ) P(A' ) = 0.01× 0.998 = 0.00998

P(N

| A’ )=

P(N ∩ A' ) = P(N | A' ) P(A' ) = 0.99 × 0.998 = 0.98802

0.9 9

0.0019 P(P ∩ A) = = 0.1599 P(A | P) = 0.0118 P(P) P(P) = P(P ∩ A) + P(P ∩ A' ) = 0.0019 + 0.00998 = 0.01188

2.129 Let 1 ~ Engineer 1 2 ~ Engineer 2 E ~ error

P (E ) 1 ( P

.7 0 =

P(2 )=

) |1

=

2 0 . 0

P(E ∩ 1) = P(E | 1) P(1) = 0.02 × 0.7 = 0.014

P(E’ | 1) = 0.9 8

.04 P(E | 2) = 0 0. 3 P(E ’|2

)=

0.9 6

P(E ∩ 2) = P(E | 2) P(2) = 0.04 × 0.3 = 0.012

P(E) = P(E ∩ 1) + P(E ∩ 2) = 0.014 + 0.012 = 0.026

P(E ∩ 1) 0.014 P(1 | E) = = = 0.5385 P(E) 0.026 P(2 | E) =

P(E ∩ 2) 0.012 = = 0.4615 P(E) 0.026

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 P(all 3 are defective) = (0.20)3 = 0.008

b) If four items arrive off the process line in succession P(3 are defective) = 4 (0.20)3 (0.80 ) = 0.0256

2.131 Let A ~ admitted to hospital B ~ back on the job the next day P(A) = 0.10 P(B) = 0.15 P(A ∩ B) = 0.02

P(A ∪ B) = P(A) + P(B) - P(A ∩ B) = 0.10 + 0.15 - 0.02 = 0.23

2.132 Let T ~ attend the training course M ~ able to meet their production quotas Given P(M | T) = 0.90 P(M | T’) = 0.65 P(T) = 0.5

P(T ∩ M) P(T | M) = P(M) P(M | T)P(T) = P(M | T)P(T) + P(M | T' )P(T' ) 0.90(0.5) = 0.90(0.5) + 0.65(0.5) 0.45 = = 0.5806 0.775

2.133 Let D ~ dissatisfied A ~ purchase from vendor A Given P(D) = 0.10 P(A | D) = 0.50 P(A) = 0.20

P(A | D) P(D) P(D | A) = P(A) 0.50 (0.10 ) = 0.20 = 0.25

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

P(U ∩ NS) a) P(U | NS) = P(NS) 13 = 100 23 100 13 = 23

P(UE ∩ U) b) P(UE | U) = P(U) 2 = 100 59 100 2 = 59

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

P( H | C) = 0.5 P( H | C’) = 0

P( H’ | C ) = 0.5 P( H’ | C’) = 1

P(C ∩ H' H' H' ) We want to find for P(C | H' H' H' ) = P(H' H' H' )

H 0.5

0.5 0.5

0.5

C

0.5

C’

0

H 0.5

H’

H

H’ 0.5 H 0.5

H’

H’ 0.5

0.5

H

0.5

H’

0

H

1

H’

H

H

0.5

H’

0.5

0

1

H

H’ H 0 H’ 1 H

1

H’

0.5

H’

0.5 0.5

0 1

H

H

H’

H’ 1 H

0

H’

1

0

4

1 1 P(C ∩ H' H' H' ) =   =  2  16 1 3 1 P(C'∩H' H' H' ) =  (1) = 2 2 P(H' H' H' ) = P(C ∩ H' H' H' ) + P(C'∩H' H' H' ) 1 1 9 = + = 16 2 16

1 P(C ∩ H' H' H' ) 16 1 P(C | H' H' H' ) = = = 9 9 P(H' H' H' ) 16

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

 1  364  363   306  =       365  365  365   365  364 ! = 305 ! (365) 60