Download - C2 Prob Extra
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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
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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)
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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
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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
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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
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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
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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
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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
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2.122
B)'(A a)
A
CB
B)'(A b)
A
CB
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B)(A c) C
A
CB
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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
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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
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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)
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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
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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)
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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
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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
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P(N)
N)P(CN)|P(C
0.8905
0.88350.007
)C'P(NC)P(NP(N)
00786.08905.0
007.0
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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
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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
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P(P)
)P(PP)|P(A
A
0.01188
0.009980.0019
)'P(PA)P(PP(P)
A
1599.00118.0
0019.0
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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
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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).
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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
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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
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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
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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
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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
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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)
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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
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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
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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
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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