2.
2
5-7
Empirical Probability
The empirical approach to probability is based on
what is called the law of large numbers. The key
to establishing probabilities empirically is that
more observations will provide a more accurate
estimate of the probability.
EXAMPLE:
On February 1, 2003, the Space Shuttle
Columbia exploded. This was the second
disaster in 113 space missions for NASA. On
the basis of this information, what is the
probability that a future mission is successfully
completed?
98.0=
113
111
=
=
flightsofnumberTotal
flightssuccessfulofNumber
flightsuccessfulaofyProbabilit
Limitations
If the number
of outcomes
is not
sufficiently
larger
It past
occurrences
are scarce
5-8
Law of Large Numbers
Suppose we toss a fair coin. The result of each toss is either a
head or a tail. If we toss the coin a great number of times, the
probability of the outcome of heads will approach 0.5.
The following table reports the results of an experiment of flipping
a fair coin 1, 10, 50, 100, 500, 1,000 and 10,000 times and then
computing the relative frequency of heads
5-9
Subjective Probability - Example
If there is little or no past experience or information on which to
base a probability, it may be arrived at subjectively.
Due to differences in individual judgment, the probability of the
same event might be different to different people
Illustrations of subjective probability are:
1. Estimating the likelihood the New England Patriots will play in the
Super Bowl next year.
2. Estimating the likelihood you will be married before the age of 30.
3. Estimating the likelihood the U.S. budget deficit will be reduced by
half in the next 10 years.
5-10
Summary of Types of Probability
5-11
Rules of Addition
Rules of Addition
Special Rule of Addition - If two
events A and B are mutually
exclusive, the probability of one or
the other event’s occurring equals
the sum of their probabilities.
P(A or B) = P(A) + P(B)
The General Rule of Addition - If A
and B are two events that are not
mutually exclusive, then P(A or B) is
given by the following formula:
P(A or B) = P(A) + P(B) - P(A and B)
EXAMPLE:
An automatic Shaw machine fills plastic bags with a mixture
of beans, broccoli, and other vegetables. Most of the bags
contain the correct weight, but because of the variation in
the size of the beans and other vegetables, a package might
be underweight or overweight.Acheck of 4,000 packages
filled in the past month revealed:
What is the probability that a particular package will be
either underweight or overweight?
P(A or C) = P(A) + P(C) = .025 + .075 = .10
Union= Or = U
Intersection= and= ∩ 5-12
Special Addition Rule – Example:1
New England Commuter Airways recently
supplied the following information on their
commuter flights from Boston to New York:
Arrival Frequency
Early 100
Late 75
On Time 800
Canceled 25
Total 1000
3.
3
5-13
Special Addition Rule – Example:1
The probability that a flight is either early or late is:
P(A or B) = P(A) + P(B) = 0.10 + 0.075 = 0.175
If A is the event that a
flight arrives early, then
P(A) = 100/1000 = 0.10
If B is the event that a flight
arrives late, then P(B) =
75/1000 = 0.075
5-14
General Addition Rule – Example:1
What is the probability that a card chosen at
random from a standard deck of cards will be
either a king or a heart?
P(A or B) = P(A) + P(B) - P(A ∩ B)
= 4/52 + 13/52 - 1/52
= 16/52, or 0.3077
5-15
General Addition Rule –Example:2
In a sample of 500 students, 320 said they had a stereo,
175 said they had a TV, and 100 said they had both. 5
said they had neither.
If a student is selected at random, what is the probability
that the student has only a stereo or TV? What is the
probability that the student has both a stereo and TV?
P(ST or TV) = P(ST) + P(TV) - P(ST ∩ TV)
= 320/500 + 175/500 – 100/500 = 0.79.
P(ST and TV) = 100/500 = 0.20
5-16
The Complement Rule
The complement rule is used to determine
the probability of an event occurring
by subtracting the probability of the
event not occurring from 1.
P(A) + P(~A) = 1
or P(A) = 1 - P(~A).
EXAMPLE
An automatic Shaw machine fills plastic bags with a
mixture of beans, broccoli, and other vegetables. Most of
the bags contain the correct weight, but because of the
variation in the size of the beans and other vegetables, a
package might be underweight or overweight. Use the
complement rule to show the probability of a satisfactory
bag is .900
P(B) = 1 - P(~B)
= 1 – P(A or C)
= 1 – [P(A) + P(C)]
= 1 – [.025 + .075]
= 1 - .10
= .90
5-17
If D is the event that a flight
is canceled, then
P(D) = 25/1000 = 0.025.
Recall Airways example. Use the complement rule to
find the probability of an early (A) or a late (B) flight
If C is the event that a
flight arrives on time, then
P(C) = 800/1000 = 0.8
The Complement Rule-Example
5-18
P(C) =0.8
P(D) =0.025
~(C or D) = (A or B)
0.175
P(A or B) = 1 - P(C or D)
= 1 - [0.8 +0.025]
= 0.175
The Complement Rule-Example
4.
4
5-19
The General Rule of Addition and Joint Probability
The Venn Diagram shows the result of a
survey of 200 tourists who visited Florida
during the year. The survey revealed that 120
went to Disney World, 100 went to Busch
Gardens and 60 visited both.
What is the probability a selected person
visited either Disney World or Busch
Gardens?
P(Disney or Busch) = P(Disney) + P(Busch) - P(both Disney and Busch)
= 120/200 + 100/200 – 60/200
= .60 + .50 –.30= .80
JOINT PROBABILITY Aprobability that measures
the likelihood two or more events will happen
concurrently.
5-20
Special Rules of Multiplication
• The special rule of multiplication requires that two events A and B are independent.
• Two events A and B are independent if the occurrence of one has no effect on the
probability of the occurrence of the other.
• This rule is written: P(A and B) = P(A)P(B)
EXAMPLE
A survey by the American Automobile Association (AAA) revealed 60% of its
members made airline reservations last year. Two members are selected at
random. Since the number of AAA members is very large, we can assume that R1
and R2 are independent. What is the probability both made airline reservations
last year?
SOLUTION:
The probability the first member made an airline reservation last year is .60, written as P(R1) = .60
The probability that the second member selected made a reservation is also .60, so P(R2) = .60.
Since the number of AAA members is very large, you may assume that R1 and R2 are independent.
P(R1 and R2) = P(R1)P(R2) = (.60)(.60) = .36
5-21
The health department routinely conducts two
independent inspection of each restaurant, with the
restaurant passing only if both inspectors pass it.
Inspector A is very experienced and hence, passes only
2% of restaurants that actually do have health code
violations. Inspector B is less experienced and passes 7%
of restaurants with violations. What is the probability that
(a) Inspector A passes a restaurant, given that inspector B has found
a violation?
(b)Inspector B passes a restaurant with a violation, given that
inspector A passes it?
(c)The health department passes a restaurant with a violation?
Special Multiplication Rule (Levin & Rubin)
5-22
a) P(A/B) = P(A) = 0.02
b) P(B/A) = P(B) = 0.07
c) P(A,B) = P(A)×P(B)
=(0.02)×(0.07)=0.0014
Special Multiplication Rule (Levin & Rubin)
SOLUTION:
5-23
Mr. Bill Borde has just launched a publicity campaign for a new
restaurant in a town. Bill has just installed four billboards on a highway
outside of town, and he knows from experience the probabilities that
each will be noticed by a randomly chosen motorist. The probability of
the first billboard’s being noticed by a motorist is 0.75. The probability
of the second’s being noticed is 0.82, the third has a probability of 0.87
of being noticed, and the probability of the fourth sign’s being noticed is
0.9. Assuming that the event that a motorist notices any particular
billboard is independent of whether or not he notices the others, what is
the probability that
(a) All four billboards will be noticed by a randomly chosen motorist?
(b) The first and fourth, but not the second & third billboards will be
noticed?
(c) Exactly one of the billboards will be noticed?
(d) None of the billboards will be noticed?
(e) The third and fourth billboards won’t be noticed?
Special Multiplication Rule(Levin & Rubin)
5-24
Special Multiplication Rule(Levin & Rubin)
Given us, P(A)=0.75, P(B)=0.82, P(C)=0.87 and P(D)=0.90
P(~A)=0.25, P(~B)=0.18, P(~C)=0.13 and P(~D)=0.10
(a) P(A,B,C,D)=P(A).P(B).P(C).P(D)= (.75)(.82)(.87)(.90)=0.4815
(b) P(A, ~B, ~C,D)=P(A).P(~B).P(~C).P(D)
= (.75)(.18)(.13)(.90)=0.015795
(c) P(A,~B,~C,~D) +P(~A,B,~C,~D) +P(~A,~B,C,~D) +P(~A,~B,~C,D)
=(.75)(.18)(.13)(.10) + .25)(.82)(.13)(.10)+(.25)(.18)(.87)(.10)
+ (.25)(.18)(.13)(.90)
=0.001755+0.002665+0.003915+0.005265=0.0136
(d) P(~A, ~B,~C, ~D)= P(~A).P(~B).P(~C).P(~D)
=(0.25)(0.18)(0.13)(0.10)= 0.000585
(e) P(~C, ~D)= P(~C).P(~D)= (0.13)(0.10)=0.013
SOLUTION:
5.
5
5-25
Suppose you are tossing a coin which is not
fair. It has P(H)=0.8 and P(T)=0.2. Find the
probability that 3 heads will appear in 3
successive tosses?
P(H1, H2, H3,) = P(H1) × P(H2) × P(H3)
= (0.80) ×(0.80) ×(0.80)
= 0.512
Special Multiplication Rule (Levin & Rubin)
SOLUTION:
5-26
Suppose you are tossing a coin which is fair. It has
P(H)=0.5 and P(T)=0.5. What is the probability of (a)
at least two heads of a fair coin on 3 tosses? and (b)
one tail on 3 tosses?
(a) Outcomes satisfying conditions are:
(H1, H2, H3),(H1,H2,T3), P(H1,T2,H3), P(T1,H2,T3)
The probability of each outcome is
=(0.5)×(0.5)×(0.5)= 0.125
The required probability will be= 4× 0.125=0.50
Special Multiplication Rule (Levin & Rubin)
SOLUTION:
5-27
Special Multiplication Rule (Levin & Rubin)
b) Possible outcomes of 7 events except for the event
(H1,H2, H3)
Thus, the probability is = 7x0.125 = 0.875
Or
The probability is= 1-0.125 = 0.875
5-28
Toss-1 Toss-2 Toss-3
Outcome Probability Outcomes Probability Outcomes Probability
H1,H2,H3 0.125
H1,H2,T3 0.125
H1,H2 0.25 H1,T2,H3 0.125
H1 0.5 H1,T2 0.25 H1,T2,T3 0.125
T1 0.5 T1,H2 0.25 T1,H2,H3 0.125
T1,H2 0.25 T1,H2,T3 0.125
T1,T2,H3 0.125
T1,T2,T3 0.125
Special Multiplication Rule (Levin & Rubin)
5-29
Chris owns two stocks, IBM and GE. The probability
that IBM stock will increase in value next year is 0.5
and the probability that GE stock will increase in
value next year is 0.7. Assume the two stocks are
independent. What is the probability that both stocks
will increase in value next year?
P(IBM and GE) = P(IBM) × P(GE)
=(0.5)×(0.7)
=0.35
Special Multiplication Rule
5-30
Special Multiplication Rule
What is the probability that at least one of these
stocks increases in value in the next year? This
means that either can increase or both
Outcomes satisfying conditions are:
(IBM and ~GE), (~IBM and GE) and (IBM and GE)
The probabilities of these outcomes are:
=(0.5)×(1-0.7)+ (1-0.5)×(0.7)+(0.5)×(0.7)
=0.15 + 0.35 + 0.35
=0.85
SOLUTION:
6.
6
5-31
General Rules of Multiplication
The general rule of multiplication is used to find the joint probability that two events are
not independent and they will occur under conditions of statistical dependence.
EXAMPLE
A golfer has 12 golf shirts in his closet. Suppose 9 of these shirts are white and
the others blue. He gets dressed in the dark, so he just grabs a shirt and puts it
on. He plays golf two days in a row and does not do laundry.
What is the likelihood both shirts selected are white?
The event that the first shirt selected is white is W1. The probability is P(W1) = 9/12
The event that the second shirt (W2 )selected is also white. The conditional probability that the second
shirt selected is white, given that the first shirt selected is also white, is
P(W2 | W1) = 8/11.
To determine the probability of 2 white shirts being selected we use formula: P(AB) = P(A) P(B|A)
P(W1 and W2) = P(W1)P(W2 |W1) = (9/12)(8/11) = 0.55
)A(P
)BandA(P
=)A/B(P=obabilityPrlConditiona
5-32
At a soup kitchen, a social worker gathers the following
data. Of those visiting the kitchen, 59% are men, 32%
are alcoholics, and 21% are male alcoholics. What is
the probability that a random male visitor to the kitchen
is an alcoholic?
Given us: P(M) = 0.59, P(A) = 0.32 and
P(M and A) = 0.21
General Multiplication Rule (Levin and Rubin)
356.0=
0.59
.210
=
)M(P
)AandM(P
=
P(M)
M)andP(A
=P(A/M)
SOLUTION:
5-33
If a hurricane forms in the eastern half of the Gulf of
Mexico, there is a 76% chance that it will strike the
western coast of Florida. From the data gathered over the
past 50 years, it has been determined that the probability
of a hurricane’s occurring in this area in a given year is
0.85.
(a) What is the probability that a hurricane will occur in
the Eastern Gulf of Mexico and strike Florida this year?
(b) If a hurricane in the eastern Gulf of Mexico is seeded
(induced to rain by addition of chemicals from aircraft), its
probability of striking Florida’s West Coast is reduced by
one-fourth. If it is decided to seed any hurricane in the
eastern gulf, what is the new value for the probability in
part (a)?
General Multiplication Rule (Levin and Rubin)
5-34
Given us, P(S/H)= 0.76 and P(H) = 0.85
We have to find out P(H and S)=?
P(H and S) = P(S and H) = P(S/H)×P(H)
=0.76×0.85=0.646
If seeded chemicals, the revised probability is:
P(H and S) = P(S and H) = P(S/H)×P(H)
=0.57×0.85=0.4845
General Multiplication Rule(Levin and Rubin)
SOLUTION:
5-35
Contingency Tables
A CONTINGENCY TABLE is a table used to classify sample observations according to two or more identifiable
characteristic
EXAMPLE:
A sample of executives were surveyed about
their loyalty to their company. One of the
questions was, “If you were given an offer
by another company equal to or slightly
better than your present position, would
you remain with the company or take the
other position?” The responses of the 200
executives in the survey were cross-
classified with their length of service with
the company. What is the probability of
randomly selecting an executive who is
loyal to the company (would remain) and
who has more than 10 years of service?
Event A1 happens if a randomly selected
executive will remain with the company
despite an equal or slightly better offer
from another company. Since there are
120 executives out of the 200 in the survey
who would remain with the company
P(A1) = 120/200, or .60.
Event B4 happens if a randomly selected
executive has more than 10 years of
service with the company. Thus, P(B4| A1)
is the conditional probability that an
executive with more than 10 years of
service would remain with the company. Of
the 120 executives who would remain 75
have more than 10 years of service, so
P(B4| A1) = 75/120.
5-36
The Dean of the School of Business at Owens
University collected the following information
about undergraduate students in her college:
Major Male Female Total
Accounting 170 110 280
Finance 120 100 220
Marketing 160 70 230
Management 150 120 270
Total 600 400 1000
Contingency Tables
7.
7
5-37
P(A|F) = P(A and F)/P(F)
= [110/1000]/[400/1000] = 0.275
If a student is selected at random, what is the
probability that the student is a female (F)
accounting major (A)?
P(F and A) = 110/1000 = 0.11
Given that the student is a female, what is
the probability that she is an accounting
major?
Contingency Tables
SOLUTION:
5-38
Tree Diagrams
A tree diagram is useful for
portraying conditional and
joint probabilities. It is
particularly useful for
analyzing business
decisions involving several
stages.
A tree diagram is a graph
that is helpful in organizing
calculations that involve
several stages. Each
segment in the tree is one
stage of the problem. The
branches of a tree diagram
are weighted by
probabilities.
5-39
Bayes’ Theorem
A revised probability or posterior probability
which is based on additional information
The basic formula for conditional probability
under statistical dependence:
P(B/A) = P(B and A)/P(A)
is called Bayes’ Theorem
5-40
Bayes’ Theorem Example-1
We have equal numbers of two types of deformed
(biased or weighted) dice in a bowl. On half of them,
ace (or one dot) comes up 40% of the time; therefore,
P(ace) = 0.40. On the other half, ace comes up 70%
of the time; P(ace) = 0.7. Let us call the former type-1
and the latter type-2. One die is drawn, rolled once,
and comes up ace.
(a) What is the probability that it is type-1 or type-2 dice?
(b) Assume that the same die is rolled a second time and
again comes up ace. What is the further revised probability
that the die is type-1 or type-2?
5-41
Bayes’ Theorem Example-1
Elementary Event
(E)
P(E) P(Ace/E) P(Ace and E)
Type-I 0.5 0.4 0.4x0.5=0.20
Type-II 0.5 0.7 0.7x0.5=0.35
P(E)=1.00 P(Ace)=0.55
364.0=55.0/20.0=
P(Ace)
Ace)andP(T
=/Ace)P(T 1
1
636.0=55.0/35.0=
P(Ace)
Ace)andP(T
=/Ace)P(T 2
2
5-42
Bayes’ Theorem Example-1
246.0=325.0/080.0=
P(2Aces)
2Aces)andP(T
=/2Aces)P(T 1
1
Event (E) P(E) P(Ace/E) P(2Aces/E) P(2 Aces, E)
Type-I 0.5 0.4 0.16 0.16x0.5=0.080
Type-II 0.5 0.7 0.49 0.47x0.5=0.245
P(E)=1.00 P(2Aces)=0.325
754.0=325.0/245.0=
P(2Aces)
2Aces)andP(T
=/2Aces)P(T 1
2
8.
8
5-43
Bayes’ Theorem Example-2
EconOcon is planning its company picnic. The only
thing that will cancel the picnic is a thunderstorm. The
Weather Service has predicted dry conditions with
probability 0.2, moist conditions with probability 0.45,
and wet conditions with probability 0.35.
If the probability of a thunderstorm given dry conditions is
0.3, given moist conditions is 0.6, and given wet conditions
is 0.8, what is the probability of a thunderstorm?
If we know the picnic was indeed canceled, what is the
probability moist conditions were in effect?
5-44
Bayes’ Theorem Example-2
Event (E) P(E) P(Th/E) P(Th and E)
Dry 0.20 0.3 0.3x0.2=0.06
Moist 0.45 0.6 0.6x0.45=0.27
Wet 0.35 0.8 0.8x0.35=0.28
P(T)=0.61
443.0=
61.0
27.0
=
)Th(P
)ThandM(P
=)Th/M(P
5-45
Bayes’ Theorem Example-3
We A doctor has decided to prescribe two new drugs to
200 heart patients as follows: 50 get drug A, 50 get
drug B, 100 get both. The 200 patients were chosen so
that each had an 80% chance of having a heart attack
if given neither drug. Drug A reduces the probability of
a heart attack by 35%, drug B reduces the probability
by 20%, and the two drugs, when taken together, work
independently.
If a randomly selected patient in the program has a heart
attack, what is the probability that the patient was given both
drugs?
5-46
Bayes’ Theorem Example-3
Event P(E) P (H|E) P (H and E) P (E|H)
Drug-A 0.25 (0.8) (0.65) = 0.520 0.130 0.130/0.498 = 0.2610
Drug-B 0.25 (0.8) (0.80) = 0.640 0.160 0.160/0.498 = 0.3213
Both 0.50 (0.8)(0.65)(0.80) = 0.416 0.208 0.208/0.498 = 0.4177
P (H) = 0.498
Thus, P(A and B)| P(H)
= 0.208/0.498= 0.417
5-47
Bayes’ Theorem Example-4
A physical therapist at Enormous State University
knows that the football team will play 40% of its game
on artificial turf this season. He also Knows that a
football player’s chances of incurring a knee injury are
50% higher if he is playing on artificial turf instead of
grass. If a player’s probability of knee injury on artificial
turf is 0.42 what is the probability that
(a) A randomly selected football player incurs a knee
injury.
(b) A randomly selected football player with a knee injury
the incurred the injury playing on grass?
5-48
Bayes’ Theorem Example-4
Elementary Event
(E)
P(E) P(I/E) P(I and E)
Artificial (A) 0.40 0.42 0.168
Grass (G) 0.60 0.28 0.168
1.00 P(I) = 0.336
(b) P(G/I) = P(G and I)/P(I) = (0.168)/0.336 = 0.50
9.
9
5-49
Bayes’ Theorem Example-5
A baseball team has been using an automatic pitching
machine. If the machine is correctly set-up, it will pitch
strikes 85 percent of the time. If incorrectly set-up, it will
pitch strikes only 35 percent of the time. Past experience
indicates that 75% of the set-ups of the machine are
correctly done.
(a) After the machine has been set-up at batting practice one
day, it throws three strikes on the first three pitches. What is
the revised probability that the set-up has been done correctly?
(b) In the case of pitching machine, we might find the five
pitches to be: strike, ball, strike, strike, strike. What is the
revised probability that the set-up has been done correctly?
5-50
Bayes’ Theorem Example-5
Event (E) P(E) P(Strike/E) P(3 Strikes/E) P(3 Strikes and E)
Correct 0.75 0.85 .85x.85x.85=.6141 .6141x.75=.4606
Incorrect 0.25 0.35 .35x.35x.35=.0429 .0429x.25=.0107
P(E)=1.0 P(3 Strikes)=.4713
9773.=
4713.
4606.
=
)Strikes3(P
)Strikes3andCorrect(P
=)strikes3/Correct(P
0227.=
4713.
0107.
=
)Strikes3(P
)Strikes3andIncorrect(P
=)strikes3/Incorrect(P
5-51
Bayes’ Theorem Example-5
Event (E) P(E) P(Strike/E) P(SBSSS/E) P(SBSSS,E)
Correct 0.75 0.85 .85x.15x.85x.85x.85=.07830 .07830x.75=.05873
Incorrect 0.25 0.35 .35x.65x.35x.35x.35=.00975 .00975x.25=.00244
P(E)=1.0 P(SBSSS)=.06117
9601.=
06117.
05873.
=
)SBSSS(P
)SBSSSandCorrect(P
=)SBSSS/Correct(P
0399.=
06117.
00244.
=
)SBSSS(P
)SBSSSandIncorrect(P
=)SBSSS/Incorrect(P
5-52
Permutation and Combination
A permutation is any arrangement of r
objects selected from n possible
objects. The order of arrangement is
important in permutations.
EXAMPLE
Suppose that in addition to selecting the
group, he must also rank each of the
players in that starting lineup according to
their ability.
A combination is the number of ways to
choose r objects from a group of n objects
without regard to order.
EXAMPLE
There are 12 players on the Carolina Forest High
School basketball team. Coach Thompson
must pick five players among the twelve on
the team to comprise the starting lineup.
How many different groups are possible?
792
)!512(!5
!12
512
C040,95
)!512(
!12
512
P
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