CIVIL ENGINEERING STATISTICS
The Defi ni ti on of Ba si c Term s in Probability (i)
A random experiment ~ a process leadin g to at l east t wo p oss ibl e outc certainty a s to whic ome h thesywit willh un occur.
(ii)
Basic outc ome s ~ the possible out com es of a rando m experiment
(iii)
Sample Space ~ th e set of all b asi c out com es ( de not ed by S )
Example Tos si ng a fa ir co in
Random Experiment
The re su lt wi ll be eit her a “ head ” or a “ tail ” Basic Outcomes
Sample Spac e S = { H ,T } H ~ head , T ~ tail
(iv) Event ~ is a sub set of ba sic out com es f ro m the sa mpl e spa ce. Example Roll ing a die the ba sic ou tco mes are th e numbers 1, 2, 3, 4, 5 and 6. Thus S = { 1, 2, 3, 4, 5 ,6 } If A = { 1, 3, 5 }
2 4
1 3 5
6
Then A is an eve nt of gett ing odd num bers
Example Li st a sample space wh en t wo di ce are tos sed or a die is t os sed tw ic e . S = { (1,1) , (1,2) , (1,3) , (1,4) , (1,5) , (1,6) (2,1) , (2,2) , (2,3) , (2,4) , (2,5) , (2,6) (3,1) , (3,2) , (3,3) , (3,4) , (3,5) , (3,6) (4,1) , (4,2) , (4,3) , (4,4) , (4,5) , (4,6) (5,1) , (5,2) , (5,3) , (5,4) , (5,5) , (5,6) (6,1) , (6,2) , (6,3) , (6,4) , (6,5) , (6,6) } Thu s n(S) = 36
Let A be an event of gettin g the sum of two numb ers is 6 Thus A = { (1,5) , ( 2,4) , (3,3) , (4,2) , (5,1) } n(A) = 5 Let B be an event of ge tt ing th e su m of two numb ers is a mult iple of 5 Thus B = { (1,4) , ( 4,1) , (2,3) , (3,2) , (5,5) , (4,6) , (6,4) } n(B) = 7
Mut ually E xc lu si ve Events
Two events A and B are called mutually exclusive if t here is no i ntersectio n betwe en the two events.
A A
B= B
Example Rolli ng a die Let A : gett ing even nu mbers B : getting odd nu mbe rs A={2,4,6} B={1,3,5}
Therefo re,
A
B 2
4 6
A
B=
1
3 5
Exh aus ti ve Events
Two events A 1, A 2, A 3,…., An are said t o b e exhaustive events if nA A2 ( 1A
A1
A ... 3
A2
) (n)
nS
A3
A4
A5
A6
A7
A8
Example
Let If
S={1,2,3,4,5,6,7,8,9,10}. A={1,2,3,4,5,6} and B ={5,6,7,8,9,10}, then n (A B) =n (S).
Therefore, A & B are exhaustive events.
Example Let
S={1,2,3,4,5,6}.
If
A={2,4,6} and B={1,3}, th en n (A
B) ≠ n (S).
Therefore, A and B are not exhaustive events.
The Pro babil it y Of An Eve nt Definition The pr obabili ty of an event A occ ur ri ng is deno ted by P(A) ,where P(A) = =
number of pos sibl e outco mes in A numbe r of all possible outcom es in S
n(A) n(S)
P(A) sa tisf ies t he follow ing cond itio ns 0 ≤ P(A) ≤ 1
If P(A) = 1 , => the event A is a su re event If P(A) = 0 , => the event A is an i mpo ssi ble e vent
The com pl ement o f an e vent A is denot ed as A with
P(A)
1
P(A)
where P( A ) or P(A’) means the probability that A doesn’t happen
Example Two dic e are tos sed, find the pro babil it y (a) the sum of the two numbe rs is 8 (b) the sum of t he two numbers is a prime number Solut ion : (a) Let A be th e event th e su m of th e tw o numb ers is 8 A = { (2,6) , (6,2) , (3,5), (5,3) , (4,4) } n(A) = 5 , n(S) = 36 Thu s P(A) =
5 36
(b)
Let B be th e event of getti ng th e sum of t he two numb ers i s a prime number.
S = { (1,1) , (1,2) , (1,3) , (1,4) , (1,5) , (1,6) (2,1) , (2,2) , (2,3) , (2,4) , (2,5) , (2,6) (3,1) , (3,2) , (3,3) , (3,4) , (3,5) , (3,6) (4,1) , (4,2) , (4,3) , (4,4) , (4,5) , (4,6) (5,1) , (5,2) , (5,3) , (5,4) , (5,5) , (5,6) (6,1) , (6,2) , (6,3) , (6,4) , (6,5) , (6,6) }
(b)
Let B be th e event of getti ng th e sum of t he two numb ers i s a prime number. B = { (1,1) , (1,2) , (1,4) , (1,6) , (2,1), (2,3) , (2,5) , (3,2) , (3,4) , (4,1) ,(4,3) , (5,2) , (5,6), (6,1) , (6,5) } n(B) = 15 , n(S) = 36 Thus P(B) =
15
5
36
12
Example Two fair c oi ns are tos sed sim ul tane ou sl y. Find t he pro babili ty of gettin g (a) exact ly tw o h eads (b) at least one h ead
Solut ion : Start
After 1 st toss
Aft er 2 nd toss H
H
T H
T T
(a)
S={HH,HT,TH,TT}
n(S) = 4
Let A be the e vent o f getti ng 2 heads A = {HH} Thu s P(A) = (b)
1
4
Let B be th e event of getti ng at least on e head B = { HH,HT,TH} Thus P(B) =
3 4
Venn Diagr am A Venn Diagram can al so b e used to solve p robability problems
Example There a re 100 FKA AS fi rs t y ear st ud ent s , of wh om 20 are stu dyi ng mathematics , 15 are studyi ng sta tist ics a nd 8 are studyi ng both mathe matics and st atis ti cs. W e can ill ust rate th is in a Venn di agr am.
There a re 100 FKA AS fir st year st ud ent s , of w ho m 20 are stu dyi ng mathema ti cs , 15 are stu dyi ng st atis ti cs and 8 are st udyi ng b oth m athema tic s and st ati st ic s. We can il lu st rate th is in a Venn d iagram.
TRY THIS In a jun ior scho ol class of 28 pup il s, 7 are in bot h a spo rt s t eam and t he scho ol band. T here are 16 pup il s i nvo lved in spo rts teams and 10 in the schoo l band. Find t he pro babil it y that a pup il cho sen a t random. (a) is only in the school band (b)
is in eit her a sp or ts team o r t he sc ho ol band
(c)
is in neit her a sp or ts team n or t he sc ho ol band
DEFINITION Probability of tw o events A or B occu rri ng can be calculate d usi ng addi tio nal probabili ty rul es s uch as b elow, P(A or B o r bo th ) = P(A) + P(B) – P(A and B) , can be deno ted as P(A P(A
B),
B) = P(A) + P(B) – P(A
B)
An integer is selected randomly from a set of integer s {1, 2,3, 4, 5, 6, 7, 8, 9, 10, 11,12 }. Fi nd the probabilit y t hat t he int eger is (a) an eve n nu mb er o r i s d iv is ib le by 3 (b) an eve n nu mb er and i s no t d iv is ibl e by 3 (c) not an eve n n umb er and is not div isi ble by 3.
An integer is selected randomly from a set of integer s {1, 2,3, 4, 5, 6, 7, 8, 9, 10, 11,12 }. Find the probability that the inte ger is (c) a n neve nu mber i ssin div isi ble3by 3 (b) an aorndisai snd di vi bl3. e by (a) not an eve eve n nu nunmber mber dinviot blot e si by
SOLUTION : Let A = e vent th at even num ber is c ho sen 2 , 4 , 6 , 8 , 10 , 12
Let B = event that th e cho sen n umb er i s divisible by 3 3 , 6 , 9 ,12 a ) (P A ) B( ) P () A(
P )B
P A
6
4
2
2
12
12
12
3
B
b) P ( A
')B
(P) A(
P) A
6
2
12
12
B
1 3
( A c) P
)'B1 (
P) A 1
2 3
1 3
B
Example Reco rd s s ho wed that 8 0% of all d ri vers w ho are sum mon ed for v ari ous t raffic of fence s are man dr iv ers . 17% are dri vers b elo w 30 years ol d, wit h 13% are man d ri vers wh o are belo w 30 years . If a driv er w ho i s s ummon ed is randoml y selected, wha t is t he probabili ty that the dri ver is a man or below 30 years old or both?
SOLUTION : Let, L = event t o g et a ma n d ri ver T = event to get a dri ver aged belo w 30 years old
According to the information given, P(L) = 0.80, P(T) = 0.17 and P(L T) = 0.13 We want to fi nd P(L T). Usi ng th e add it io nal rul e, we ge t P(L
T) = P(L) + P(T) – P(L = 0.80 + 0.17 – 0.13 = 0.84
T)
pro babili ty t hat t he dri ver i s a ma n o r belo w 30 years ol d i s 0.84
NOTE :
Sometimes infor matio n is a lso gi ven in t abl e fo rm . The value of pr obabili ty is d efi ned based on tha t info rma tio n.
A,itt B,en C in and areefour events that can be wr th eDtabl below C
D
A
P(A
C)
P(A
D)
P(A)
B
P(B
C)
P(B
D)
P(B)
P(C)
P(D)
Example A surve y is conduc ted on a group of w orke rs comprising of produc tion ope rators , admi nis trative off icers and secur ity guards . The survey is to determine the tota l wo rking hours in a week. Production operator hrs 40< 50 – 70 hrs hrs 70 >
63
Administrative Security officer guard 21
46
4
14
87
10
8 196
88
17 43
70 112 31
Production operator hrs 40< 50 – 70 hrs hrs 70 >
63
Administrative Security officer guard 21
46
4
14
87
10
8 196
88
17 43
70 112 31
One of t he wor kers i n t he sur vey is rando mly sele ct ed. Base d on th e in fo rmatio n pr ov id ed, calcu late the probabilit y o f c) d) work wo er er being aa secur admi nis ity een gu trative working b) a) wor work rkker erbeing who beingwan ork pro betw duct ionard ope 50offic –rator 70erhrs. .and less working t han mo 4 0 re hothan ur s . 7 0 ho ur s .
SOLUTION : Let P = th e wo rker is a pr odu ct ion ope rato r A = the worker is an administrative officer S = the wo rk er is a secur it y gu ard
a) P(P)
196 196
43
31
196 270 98 135
b) P working between 50 70 88
70
7 27
112
70hrs
c) P(A
wo rking more than7 0hrs)
8 270
4 135
d) P(S 4 270 2 135
wo rking less than 40hrs)
MUTUALLY EXCLUSIVE EVENTS • Two events A and B are ca ll ed mutually exclusive if the y c annot occu r at the same ti me • Neit her eve nt A n or event B c an o ccu r simul taneousl y thus th e prob abili ty of A and B occuring at the same time is zero. P( A
B)
0
Defin iti on 1 If A and B are mu tu all y excl us iv e event s , th en
P (A
B
)
P (A )
(P)B
sin ce P( A B)=0 On a Venn diagr am A
B
s
The con cept o f mu tually e xcl usi ve events can be e xt end ed t o m or e th an tw o events . Defi nit ion 2 If n events X1 , X 2 excl usi ve then, PX (
1X 2
, X n are mu tu all y
X P)X (PX) .... 1n 2
X ( P).....
( ).
n
Example C and D are tw o events wher e P(C) = 0.1, P(D)= 0.2 and P(C D) = 0.3. (a) Determi ne wh eth er C and D are two mutually exclusive events. (b) Fin d PC ( 'and ) P(C
D'
)
'
Solution (a) C and D a re mut ually excl usi ve event s if P(C D) = P(C) + P(D) Given P(C
D) = 0.3
P(C) = 0.1 + y0.2 = lu 0.3 Thu s C and+DP(D) are mu tu all exc si ve events . '
(b) P(C ) = 1 – P(C) = 1 – 0.1 = 0.9
'
'
) can b e determ ined by u sin g th e Venn di agr am b elo w
P(C
D
C
( PC
'
D
' )1 ( PC )D = 1 - 0.3 = 0.7
D
S
Example The event s A, B , C a nd D are mu tu all y exc lu siv e with P(A) = P(B) = 0.3 and P(C) = P(D) = 0.1 If E and F are event s defi ned by E = A and F = B C , fi nd
a) P(E
F)
b) P( E
F)
D
Solution
a) P(E
F)
P( (A
D)
(B
C))
P( A ) P ( B ) P(C) P( D )
= 0.3 + 0.3 + 0.1 + 0.1 = 0.8
b) P(E
F)
P( (A
D)
(B
=0
si nc e A, B, C a nd D are mu tu all y exc lu si ve events
C))
Example A bag c on tains 4 re d m arb les, 2 wh it e marbl es and 8 bla ck marb les. What is the pro babi lit y that a ma rb le pick ed f ro m t he ba g at random is either re d or wh it e ?
Solut ion : Let
R : event th at red marbl e is pi ck ed W :eve nt t hat whi te ma rb le is picked
P( R U W )
= P(R) + P(W) 4
2
14
14
6 14
3 7
Example The resul t o f t he fin al exam in a FKA AS are as follows: 600 st ud ent s passed t he Hydr aul ic pape r. 300 st ud ent s passed t he Geot ech ni cal p aper. 175 st ud ent s p ass ed bo th p apers 50 st udents faile d bo th pape rs . From the inf ormation above , calcu late : (a) howprobabilit many of the stu dents he no Geote chni (b) the (c) prob ability yof ofcollege sstu tudents dents cho chos sen entook who whot did passed t both pass papers. incal Geand otechnic al. Hydra ulic pape r?
From the inf or mati on abo ve, calcu late : (a) how many of t he colle ge stude nts to ok the G eotechni cal and Hydraulic paper? (b) the pro babili ty of stud ents chosen wh o passe d bo th p apers. (c) the prob abili ty o f stud ents chos en wh o did n ot pass in G eot echn ical.
Solution Let
H : event that stu dents passe d Hydraulic G : eve nt t hat stu dents passed G eot ech ni c F : event that stu dents f ail ed both pa pers .
Therefore, n(H) = 600, n(G) = 300, n(H
G) = 175 and n(F) = 50
(a)The total numb er o f s tu dents who too k both papers , n(S) = n(F (H G))
Sin ce the e vents of passed a nd failed the cou rses a re mutu ally excl usi ve, n(S) = n(F) + n(H G) To find n(H n(H
G),
G) = n(H) + n(G) – n(H = 600 + 300 – 175 = 725 st ud ent s.
So, n (S) = n(F) + n(H G) = 50 + 725 = 775 st ud ent s.
G)
(b) Prob abil it y of st ud ent wh o pa ssed both courses
n(A P(A
M)
M)
n(S)
175 775 7 31
(c) P( G’
) = 1 - P (G ) 1
300 775
475
775 19
31
Con dit ion al Events When a n eve nt oc cu rs with t he condit ion that anot her e vent has occ ur red, th en t he event is a con dit ional eve nt.
Condi tion al Probability
- is the proba bili ty th at an event will occur gi ven that a no th er eve nt has a lr eady oc cu rr ed 53
Condi tion al Probability For event s A and B i n a sampl e sp ace S, the condi tio nal pro babili ty of A give n B is defi ned b y
P ( A| B )
P( A
B)
P(B)
P( B )
0
P (A|B) is read as“the probability of A givenB”
Example A and B are two events such that P(A) = 1/3, P(B) =1/4 and P(A U B) = 1/2 . Find a) P(A ∩ B) c) P(B|A)
b) P(A|B)
Solution:
a) P(A U B) = P(A) + P(B) - P(A ∩ B)
1 2 P(A ∩ B)
1
1
3
4
1
1
P(A
3
1 12
4
1
2
B)
b ) P( A | B )
P( A
B)
P(B) 1 12 1 4 1 3
c) P(B|A)
P( B
A)
P(A) 1 12
1
1
3
4
59
Example 30 professors out of 100 who are examined were found to be overweight (W). 10 of them had high blood pressure (H). Only 4 of the professors who were not overweight had high blood pressure. Find the probability that a professor (a) is overweight if heblood had high bloodifpressure, (b) will not have high pressure he is overweight.
Solution:
W
W'
Total
H
10
4
14
H'
20
66
86
Total
30
70
100
(a) P(W | H)
=
P(W Ç H) P(H) 10
=
100 14
=
2 7
100
(b ) P(H ' | W)
=
P( H' Ç W ) P(W) 20
=
100 30 100
2 =
3
Example Harry travels to work by either route A or route B. The probability that he chooses route A is 1 . 4
The probability that he is late for work if he chooses route A is
2 3
and the probability that he is late for work
if he chooses route B is
1
.
3
(a) (b)
What is the probability that he is late for work on a particular day ? Given that he is not late for work, what is the probability that he chooses route B ?
Solution: 2
L
3 1 4
A
L’
1 3
1 3
3 4
B 2 3
64
L L’
(a)
P( L)
12
3 1
4 3 4 3
5
(b)
12
P( B | L' )
P ( B L ') P(L') 3 4
x
2 3
7 12
6 7
The probability rule for conditional events,
P( A | B)
P( A
B)
P( B)
Then, we have
P( A
B )
( )P B (|)
P A B 66
Definition If A and B are independent events, it means that the outcome of one event does not affect the outcome of the other, then P ( A| )B
( )P
A
and P ( B| )A
( )P
B
Thus, P( A
B )
(P | ) A( )B
P B
P( A )
(P) B
if A and B are two 67
independent events
Remark If A and B are independent events, then (A and B’), (A’ and B), and (A’ and B’) are
also independent events.
P( A
B')
(P)
A(
')P
P ( A'
)B ( P ')
A(
)
P ( A'
B ')
B
P B
( P') A ( ')P
B
Example
Suppose two events A and B are independent. Given P(A) = 0.4 and P(B) = 0.25. Find a )(P A ) B
)(
b) P A
B
69
Solution: a)P( A
B)
P( A )
(P) B
A and B are independent
(0.4)(0.25)
A and B are
0.1 b) P ( A
B)
P( A )
independent ( )P (B P ) A B
) P( A
(P)
0.4
0.25
0.55
B(
)
) P( A
P B
0.1 70
Example
A, B and C are three events such that A and B are independent, A and C are mutually exclusive . Given P(A) = 0.4 , P(B) = 0.2 , P(C) = 0.3 and P(B ∩ C ) = 0.1 . Find (a) P(A U B) (b) P( C | B )
( c) P( C | A’)
71
A and B are independent
Solution: () a) P(A U B) = P
A( ) P( B
P( A )
(P)
0.4
0.2
B(
)P )
A
P( A )
B P B
(0.4)(0.2)
0.52
b) P( C | B )
P (C
B)
P(B)
0.1 0.2
0.5 72
c) P( C | A’)
P (C
A ')
P ( A ') P (C )
1
P ( A)
A
B
C
0.3 1
0.4
0.5 C
A'
C
73
Example
A mathematics puzzle is given to three students Aziz, Bong and Samy. From the past experience, known that the probabilities Aziz, Bong and Samy will get the correct solutions are 0.65, 0.6 and 0.55 respectively. If three of them attempt to solve the puzzle without consulting each other, find the probability that: a) the puzzle will be solved correctly by all of them. 74
b) only one of them will get the correct solution.
Solution:
Let A= the event that Aziz answers correctly B= the event that Bong answers correctly C= the event that Samy answers correctly
P(A) = 0.65, P(B) = 0.60 and P(C)=0.55
75
a)The event that the puzzle will be solved correctl by all of them is the event P( A
B
( )A P
C)
A
B
C
( )P B ( ) P C
(0.65)(0.60)(0.55) 0.21
b) The events that only one of them will get the correct solution will occur if one of the events
(A
B '
')C ,( ' A
B')
C
or (P' A'
)B
occurs and all these events are mutually exclusive.
C 76
Thus, P( A
' B
P A
P A'
')C P B' P B'
(P ' A P C'
B
') C
P A'
(P' A '
) B
P B
P C'
C
P C
(0.65)(0.4)(0.45)
(0.35)(0.6)(0.45)
(0.35)(0.40)(0.55) 0.29 77
Example
The probability that Roy is late for college on any day is 0.15 and is independent of whether he was late on the previous day. Find the probability that he a) is late on Monday and Tuesday b) arrives on time on one of these days
78
Solution:
0.15 0.85
0.15 LATE
0.85
LATE ON TIME
0.15
LATE
0.85
ON TIME
ON TIME
Monday
Tuesday
79
a) P( late on Monday and Tuesday ) = (0.15)(0.15) =
0.0225
b) P( arrives on time on one of these days ) = (0.15) (0.85) + (0.15)(0.85) = 0.1275 + 0.1275 = 0.2550 80
TRY THIS There are 60 students in a certain college, 27 of them are taking Mathematics, 20 are taking Biology and 22 are taking neither Mathematics nor Biology. a)Find the probability that a randomly selected student takes (i) both Mathematics and Biology. (ii) Mathematics only. b) A student is selected at random. Determine whether the event ‘ taking Mathematics’ is statistically independent 81
of the event ‘ taking Biology’ .
Solution:
M B 20-x 27-x x 22
Let B= the event of taking Biology M= the event of taking Mathematics x = the number of students taking both Biology and Mathematics
82
Total number of students ,n(S) =60 a)
27 - x + x + 20 - x + 22 = 60 x=9
(i)
(ii)
P(B
P( M
M)
B ')
9
3
60
20
18
3
60
10 83
b)
P( M
3
B)
20
27 9 , () 60 20
) P( M
P( M )
(P)
P( M
B
B )
P B
20 1 60 3
9 20
( )P M ()
1 3
3 20
P B 84
Hence, the two events are independent.
Let A be an event of gettin g the sum of two numb ers is 6 Thus A = { (1,5) , ( 2,4) , (3,3) , (4,2) , (5,1) } n(A) = 5 Let B be an event of ge tt ing th e su m of two numb ers is a mult iple of 5 Thus B = { (1,4) , ( 4,1) , (2,3) , (3,2) , (5,5) , (4,6) , (6,4) } n(B) = 7
RECALL the conditional probability of A given B is written as
P(A | B)
The event whose probability is to be
P (A B ) P (B ) The event that has already occurred 86
determined
If you are given How do you find ?
P(B | A) = 0.2 P(A | B) = ? When the condition is reversed , Bayes’ Theorem is used to solve
such problems. 87
TOTAL PROBABILITY OF EVENT B = P(B)
P( B | A ) P( A)
B
A
B’
P( B | A’ ) P( A’)
B
A’ B’
P( B ) = P(A) x P( B | A ) +
P(A’) x P( B | A’ ) 88
THE TOTAL PROBABILITY THEOREM TOTAL PROBABILITY OF EVENT B = P(B)
In general, if events A1 ,A2, .……., An are mutually exclusive and exhaustive events, then the probability of event B is given by : P( B ) = P(A 1) x P( B | A 1 ) + P(A 2) x P( B | A2 ) + P(A3) x P( B | A 3 ) +…..+ P(An) x P( B | A n ) 89
P (A|B) : “the probability of A given B”
P ( A P ( A |B ) P () P ( A |B ) ) ( P (
P (B|A) : “the probability of B given A”
P ( B P ( B |A ) P () P ( B |A ) ) ( P ( 90
P ( A P ( A |B ) P ()
(1)
P ( B |A ) ) ( P ( (2) S i n c e P A B ( Substitute (2) into (1) , we get : BAYES’
P ( B |A ) ( P ( A |B ) P ( B )
THEOREM 91
BAYES’ THEOREM
P ( B |A ) (i i P ( A |B ) i P ( B ) where A1 , A2 , ….. , An are n mutually exclusive and exhaustive events so that A1 A2 ……. An = S , the possibility space, and B is an arbitrary event of S ( i = 1,2,3,…..,n ) .
P(B) is the total probability of event B. Bayes’ Theorem is useful when we have 92
to ‘ reverse the conditions ’ in a problem.
Example
There are 12 red balls and 8 green balls in a bucket. Two balls are taken out in sequence without replacement. By using a tree diagram , find the probability that (a) the first ball is red (b) the second one is red if the first is red (c) the second one is red if the first is green (d) the second one is red (e) the first one is red if the second is red 93
Solution:
1 1
R2
1 9
12
R1
20
R ~ red ball G ~ green ball
G2
8 1 9 1 2
1 9 8 20
G1 7 1 9
1st draw
R2 G2
2nd draw
94
(a) P( first ball is red) = P(R1)
(b) P( R2 | R1 )
12
3
20
5
11 19
Or using the formula of conditional probability P( R2 | R1 )
P (R R ) 1 2
12
11
20
19 12
P (R ) 1 11
20 95
19
(c) P( R2 | G1 )
Direct from the tree diagram
(d) P( R2) = P( R1 ∩ R2) + P(G1 ∩ R2)
96
(e) P(R1 | R2 ) =
‘Reverse condition’ use Bayes’
Theorem
P (R )P ( | R 2 1
1
P ( R ) 2 11
12
19
20 3 5
11 19 97
Example Harry travels to work by route A or route B. 1 The probability that he chooses route A is
.
4
The probability that he is late for work if he goes to 2
work by route A is
and the corresponding 3
probability if he goes to work by route B is (a) (b)
1 3
.
What is the probability that he is late for work on Monday ? Given that he is late for work, what is the probability that he went to work by route B ? 98
Solution: 2
L (late) P(A) x P(L|A)
3 1
A
4
L’ (not late)
1 3
1 3
3
B
4
2 3
ROUTE
L (late) P(B) x P(L|B) L’ (not late)
ARRIVE AT WORK
99
(a) P(L)=
P(A) x P( L | A +P(B) ) xP(L |B) 12 3 43 4
5
12 BAYES’
(b) P ( L |B )P ( ( B |L )
THEOREM
P ( L )
1
3
3
4 5
3 5 100
12
Example
Aishah, Siti and Muna pack biscuits in a factory. Aishah packs 55%, Siti 30% and Muna 15% from the batch allotted to them. The probability that Aishah breaks some biscuits in a packet is 0.7, and the respective probabilities for Siti and Muna are 0.2 and 0.1. What is the probability that a packet with broken biscuits found by the checker was packed by Aishah ?
101
Solution:
A – Aishah, S – Siti , M - Muna B – Broken Biscuits 0.7
0.55 0.3 0.15
A 0.3 0.2 S 0.8 M 0.1 0.9
B B’
B
P(A |B) P(B |A ) P ( P( B )
B’
B B’
102
BAYES’
THEOREM
P ( B |A ) ( P ( A |B ) P ( B )
0 . 7 0 . 5 5 ( 0 . 5 5 0 . 7 ) ( 0 . 3 0 . 2 ) ( 0 . 1 5 0 . 7 0 . 5 5 0 . 0 . 4 6
103
Example According to a firm’s internal survey, of those employees living more than 2 miles from work , 90% travel to work by car. Of the remaining employees, only 50% travel to work by car. It is known that 75% of employees live more than 2 miles from work. Find (i) the overall proportion of employees who travel to work by car. (ii) the probability that an employee who travels to work by car lives more than 2 miles from work.
104
Solution: Define the events C , B1 , B2 as follows : C : Travels to work by car B1 : Lives more than 2 miles from work B2 : Lives not more than 2 miles from work The events B1 and B2 are mutually exclusive and exhaustive. P(B1) = 0.75 , P(B2) = 0.25 P( C | B 1 ) = 0.9 and P( C | B
2
) = 0.5
105
Solution: C
0.
0 .7
B1
0.1
(i) P(C) = P(B1) x P( C | B 1 )
C’
+ P(B2) x P( C | B 2 ) 0. 0 .2
B2 0.
C
C’ = 0.8
P( C | B 1 ) = 0.9 P( C | B 2 ) = 0.5 P(B1) = 0.75,P(B2) = 0.25
= ( 0.75 x 0.9 ) + ( 0.25 x 0.5 )
80% of employees travel to work 106
by car.
BAYES’
(ii)
THEOREM
P ( C |B ) ( 1 P ( B |C ) 1 P ( C ) 0 .90 .7 0 .8
0 . 8 4 107
