Download DEPARTMENT OF MATHEMATICS AND STATISTICS

DEPARTMENT OF MATHEMATICS AND STATISTICS
Memorial University of Newfoundland
CANADA A1C 5S7
St. John's, Newfoundland
ph. (709) 737-8075 fax (709) 737-3010
email: [email protected]
Alwell Julius Oyet, Phd
STATISTICS FOR PHYSICAL SCIENCES - LECTURE
NOTES
2. PROBABILITY
There are several actions in life that have several possible outcomes. For instance, if I take a drug I
can get better, or worse or it can lead to other complications. If I take an exam, my grade will be A,
or B or C or D or E or F . However before I take the exam., I cannot say for sure what the outcome
of the examination will be. Although at the back of my mind or mentally, I may have a vague idea
of the chance of making each of the grades. For instance, I may feel that I can never fail the course.
So, in numerical terms, I am actually saying that my chance of failing is 0%. I may sense that I have
a very good chance of making an A in the course. So, I assign a chance of 70% to my making an
A. The remaining 30% is then distributed amongst the other grades depending on the information I
have about the course. After the exam has been taken and before the result is published I may then
be able to revise the values assigned to my chances of making each of the grades. These updated
values may then be more accurate than the previous values because I now have more information
about the exam. Once the result is release, the grade I made will now be the observed value of my
action.
These values which represent the chance of making a particular grade are what we will refer to as
probability. In quality control departments, the decision to accept or reject a batch of raw materials
is usually based on the probability that the batch contains an unacceptable number of defective
items. If that probability is very high based on a technique called acceptance sampling, the batch
is rejected and accepted otherwise. So, probability is a measure of the degree of con¯dence we have
in the occurence of an outcome or group of outcomes of an action. It is a measure of uncertainty.
Under the topic of probability, we will study random and uncertain events and attempt to assign
speci¯c numbers to the chance that they will occur.
The di±culty in being certain of the outcome of an action or experiment arises from the fact that
the outcomes occur in a completely random manner and there are several factors we cannot control
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that can in°uence the outcome of an action. Sometimes, some particular outcomes of an action
or experiment may have very signi¯cant impact on our lifes, positive or negative. For instance, in
August 2000 I said to an American Statistician whom I met at a conference that I will never °y
American Airlines because of my assessment of the very high risk of an attack against the airline.
For me, the chance of an attack against AA compared to Air Canada or British Airways was so
high that I was not willing to take the risk. He °ew Korean Airline while I °ew Air Canada and
Japan Airline. The September 11 attacks had very signi¯cant impact on everybody. If there was a
person like me travelling on that date, the choice of airline based on the chance of an attack (risk
assessment) may have saved their life. My assessment was based on information on the actions of the
US government around the world and the nature of its relationship with groups around the world.
I am certain that most of us are taking this course today after an assessment of our chance of passing
this course. You are here because you feel you have a fair or a very good chance of passing. As
you gather more information about the course, twice a week, you also mentally review your chance
of passing the course. If at some point you feel that your chances are not good you will then take
action (e.g. spend more time on it or move to drop the course) to correct the problem.
Thus based on probability (chance) we are able to make informed decisions on a daily basis whether
we recognise it or not. It therefore seems appropriate to start our statistics course with a study of the
concept of probability. Some ¯nd this concept very di±cult to follow because we are called upon to
use our imagination for risk assessment and because of the language that is normally used to discuss
the subject. In this course, we will lay the foundation necessary for understanding the subject of
statistics and its application in the real world.
We begin our study by establishing notations and de¯ning terms that will be needed.
x2.1 Sample Spaces and Events
De¯nition (Experiment): An Experiment is any action or process that generates observations
(This de¯nition is su±cient for our purposes in this class).
Experiment: Prof. delivers lecture.
Outcomes: \Boring" Lecture, \Interesting" Lecture.
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Experiment: Wait for metrobus.
Outcomes: Waiting time(in mins.): 0, 0.5, 0.58, 1, 2, 3,¢ ¢ ¢
As you can see, we cannot list all the outcomes of certain experiments. What I want to know is the
chance of waiting for 0, 1, 2, 3, ¢ ¢ ¢ mins. so that I can plan my trip and not wait for too long - even
if I can't list the entire outcomes. In the ¯rst part of this course, we will concentrate on studying
techniques for calculating these probabilities.
De¯nition (Sample Space): The set of all possible outcomes of an experiment, denoted by S, is
called the sample space of that experiment.
Example 1.2: If a new type-D °ashlight battery has a voltage that is outside certain limits, that
battery is characterized as a failure (F); if the battery has a voltage within the prescribed limits, it
is a success (S). Suppose an experiment consists of testing each battery as it comes o® an assembly
line until we ¯rst observe a success. What are the possible outcomes of this experiment ?
Solution: First battery examined can be a success - S;
First, a failure; Second, a success - FS;
First, a failure; Second, a failure; Third, a success - FFS; and so on.
Therefore S = fS, FS, FFS, FFFS,¢ ¢ ¢g.
Example 1.3: PROBLEM 2, PAGE 57
Suppose that vehicles taking a particular freeway exit can turn right (R), turn left (L), or go straight
(S). Consider observing the direction for each of three successive vehicles.
(a) List the possible outcomes of this experiment.
(b) List all outcomes with the property that all three vehicles take di®erent directions.
(c) List all outcomes with the property that exactly two vehicles go in the same direction.
Solution: (a) Using a tree diagram, it is easy to ¯nd that
S = f(R,R,R), (R,R,L), (R,R,S), (R,L,R), (R,L,L), (R,L,S), (R,S,R), (R,S,L), (R,S,S),
(L,R,R), (L,R,L), (L,R,S), (L,L,R), (L,L,L), (L,L,S), (L,S,R), (L,S,L), (L,S,S),
(S,R,R), (S,R,L), (S,R,S), (S,L,R), (S,L,L), (S,L,S), (S,S,R), (S,S,L), (S,S,S)g.
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(b) A = f(R,L,S), (R,S,L), (L,R,S), (L,S,R), (S,R,L), (S,L,R)g
(c) B = f(R,R,L), (R,R,S), (R,L,R), (R,L,L), (R,S,R), (R,S,S), (L,R,R), (L,R,L), (L,L,R),
(L,L,S), (L,S,L), (L,S,S), (S,R,R), (S,R,S), (S,L,L), (S,L,S), (S,S,R), (S,S,L)g
Example 1.3 clearly illustrates the fact that sometimes, we are not just interested in the individual
outcomes of an experiment but rather in outcomes with a certain desirable property. For instance,
the property of interest in part (b) is that the vehicles turn in diferent directions. Notice that A and
B are both subsets of the sample space S. When this happens, we say that A and B are events of
the experiment. Events can be represented graphically in what is called a Venn diagram.
De¯nition: An event is any collection or subset of outcomes contained in the sample space of an
experiment.
Types of Events
(a) Simple Events: Is any event containing only one outcome.
(b) Compound Events: Is any event containing more than one outcome.
Example 1.5: PROBLEM 4, PAGE 57
Each of a sample of four home mortgages is classi¯ed as ¯xed rate (F) or variable rate (V).
(a) What are the 16 outcomes in S ?
(b) Which outcomes are in the event E1 that exactly three of the selected mortgages are ¯xed rates ?
(c) Which outcomes are in the event E2 that all four mortgages are of the same type ?
(d) Which outcomes are in the event E3 that at most one of the four is a variable rate mortgage ?
Solution: Let VFFV represent ¯rst mortgage is variable rate; second is ¯xed rate; third is ¯xed and
fourth is variable rate. Then,
(a) S = fFFFF, FFFV, FFVV, FVVV, VVVV, VVVF, VVFF, VFFF, VFFV, VFVV, VVFV,
FFVF, FVVF, FVFV, VFVF, FVFFg
(b) E1 = fFFFV, VFFF, FFVF, FVFFg
(c) E2 = fFFFF, VVVVg
(d) E3 = fFFFF, FFFV, VFFF, FFVF, FVFFg.
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Suppose, we perform the experiment in Example 1.5 by taking one sample of four home mortgages
and ¯nd that the ¯rst was a variable rate mortgage (V); the second, third and fourth were ¯xed
(F) (i.e VFFF). Since the sample VFFF belongs to E1 and E3 , we say that events E1 and E3 have
occurred. When an experiment is performed, a particular event E is said to have occurred if the
resulting outcome is an element of E.
Question: Can two simple events occur simultaneously ?
Example 1.6: PROBLEM 6, PAGE 57
A college library has ¯ve copies of a certain text on reserve. Two copies (1 and 2) are ¯rst printings,
and the other three (3, 4 and 5) are second printings. A student examines these books in random
order, stopping only when a second printing has been selected. One possible outcome is 5, and
another is 213.
(a) List the outcomes in S.
(b) Let A denote the event that exactly one book must be examined. What outcomes are in A ?
(c) Let B be the event that book 1 is not examined. What outcomes are in B ?
Solution:
(a) S = f3, 4, 5, 13, 14, 15, 23, 24, 25, 123, 124, 125, 213, 214, 215g
(b) A = f3, 4, 5g
(c) B = f5, 15, 25, 125, 215g
(d) C = f3, 4, 5, 23, 24, 25g
Some Relations From Set Theory: Observe that both the sample space of an experiment and
events are sets. So, we adopt some operations of set theory in studying events.
Union: The union of A and B denoted by A [ B is the event consisting of all outcomes that are
either in A or in B or in both A and B (without repeating any outcomes).
Example: Consider the experiment described in Problem 6, Page 57. Here, A [ B = f3,4,5,15,25,125,215g;
B [ C = f3,4,5,15,23,24,25,125,215g.
Intersection: The intersection of A and B denoted by A \ C is the event consisting of all outcomes
that are in both A and B.
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Example: For the experiment in Problem 6, Page 57, A \ B = f5g; B \ C = f5g and A \ C =
f3,4,5g.
Compliment: The compliment of the event A denoted by A0 is the event consisting of all outcomes
in S that are not in A.
Example: For the experiment in Problem 6, Page 57, C0 = f13,14,15,123,124,125,213,214,215g
Example 1.7: Refer to PROBLEM 2, PAGE 57.
Solution:
B 0 = f(R,R,R), (R,L,S), (R,S,L), (L,R,S), (L,L,L), (L,S,R), (S,R,L), (S,L,R), (S,S,S)g.
A [ B = f(R,R,L), (R,R,S), (R,L,R), (R,L,L), (R,L,S), (R,S,R), (R,S,L), (R,S,S),
(L,R,R), (L,R,L), (L,R,S), (L,L,R), (L,L,S), (L,S,R), (L,S,L), (L,S,S),
(S,R,R), (S,R,L), (S,R,S), (S,L,R), (S,L,L), (S,L,S), (S,S,R), (S,S,L)g.
A \ B = f g or ; - the empty set.
The events A and B are said to be mutually exclusive or disjoint because they have no common
outcomes or their intersection is empty. Note that these operations apply to more than two events.
x2.2 Axioms and Properties of Probability
So far, we have only studied how to construct sample spaces of experiments and events. Beginning
from this section, we will study how to assign a number P (A), to an event A. This number P (A)
measures the chance that A will occur. First, we state some basic properties, called axioms, which
all assigned numbers P (A) must satisfy, for consistency.
Axiom 1: For any event A, P (A) ¸ 0. That is, negative probabilities are not allowed.
Axiom 2: P (S) = 1. That is, the sample space is an event that must occur when the experiment
is performed.
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Axiom 3: If A1 ; A2 ; ¢ ¢ ¢ ; An is a ¯nite collection of mutually exclusive events, then
P (A1 [ A2 [ ¢ ¢ ¢ [ An ) = P (A1 ) + P (A2 ) + ¢ ¢ ¢ + P (An ) =
n
X
P (Ai ):
i=1
If A1 ; A2 ; An ; ¢ ¢ ¢ is an in¯nite collection of mutually exclusive events, then
P (A1 [ A2 [ An [ ¢ ¢ ¢) = P (A1 ) + P (A2 ) + P (A3 ) + ¢ ¢ ¢ =
1
X
P (Ai ):
i=1
Properties of Probability
For any event A,
P (A) = 1 ¡ P (A0 ):
If A and B are mutually exclusive events, then
P (A \ B) = 0:
For any two events, A and B,
P (A [ B) = P (A) + P (B) ¡ P (A \ B):
Examples
1. Problem 13, Pg 65
A computer consulting ¯rm presently has bids out on three projects. Let Ai = fawarded
projectig, for i = 1; 2; 3, and suppose that P (A1 ) = 0:22, P (A2 ) = 0:25, P (A3 ) = 0:28,
P (A1 \ A2 ) = 0:11, P (A1 \ A3 ) = 0:05, P (A2 \ A3 ) = 0:07, P (A1 \ A2 \ A3 ) = 0:01. Compute
the probability of each event.
(a) A1 [ A2 (the probability that at least one of the ¯rst two projects was awarded).
(b) The probability that neither of the ¯rst two projects was awarded.
(c) The probability that at least one of the three projects was awarded.
Solution
(a) P (A1 [ A2 ) = P (A1 ) + P (A2 ) ¡ P (A1 \ A2 ) = .22 + .25 - .11 = .36.
(b) P (A01 \ A02 ) = 1 ¡ P (A1 [ A2 ) = 1 - 0.36 = 0.64.
(c) P (A1 [ A2 [ A3 ) = P (A1 ) + P (A2 ) + P (A3 ) ¡ P (A1 \ A2 ) ¡ P (A1 \ A3 ) ¡ P (A2 \ A3 ) +
P (A1 \ A2 \ A3 ) = .22 + .25 + .28 - .11 - .05 - .07 + .01 = 0.53
This example is an illustration of the fact that the properties of probability which we stated
for two events can be extended to more than two events.
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2. Problem 15, Page 65
Consider the type of clothes dryer (gas or electric) purchased by each of ¯ve di®erent customers
at a certain store.
(a) If the probability that at most one of these purchases an electric dryer is 0.087, what is the
probability that at least two purchase an electric dryer ?
(b) If P (all ¯ve purchase gas) = 0.0768 and P (all ¯ve purchase electric) = 0.0102, what is the
probability that at least one of each type is purchased ?
Solution
(a) Let A = fat most one of ¯ve purchases an electric dryerg; B = A0 = fat least two purchase
an electric dryerg. Then, P(A) = P(0 or 1 customer will purchase a dryer) = 0.087. The
probability of interest is
P(B) = P(A0 ) = P(at least two will purchase an electric dryer)
= P(2 or 3 or 4 or 5 will purchase a dryer).
The experiment here is noting the number of customers that buy clothes dryer at the store. So,
the sample space S = f0,1,2,3,4,5g. Observe that S = A [ B = f0; 1g [ f2; 3; 4; 5g = A [ A0 .
Now,
P (S) = P (A) + P (A0 ) = 1: So;
P (B) = P (A0 ) = 1 ¡ P (A) = 1 ¡ 0:087 = 0:913:
(b) At least one of each type is purchased means that all ¯ve will not purchase a gas dryer and
all ¯ve will not purchase an electric dryer. Now, if A = fall ¯ve purchase gasg and B = fall
¯ve purchase electricg, then the required probability is
P (A0 \ B 0 ) = 1 ¡ P (A [ B) = 1 ¡ P (A) ¡ P (B) = 1 - 0.087 = 0.913.
3. Problem
A family that owns two cars is selected at random. Let A1 = fthe older car is Canadiang and
A2 = fthe newer car is Canadiang. If P(A1 ) = 0.7, P(A2 ) = 0.5 and P(A1 \ A2 ) = 0.4, compute
the following:
a. P(A1 [ A2 ) (the probability that at least one car is Candian)
b. The probability that neither car is Canadian.
c. The probability that exactly one of the two cars is Canadian.
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Solution
a. Using the Addition Law,
P (A1 [ A2 ) = P (A1 ) + P (A2 ) ¡ P (A1 \ A2 )
= 0:7 + 0:5 ¡ 0:4 = 0:8:
b. P(neither car is Canadian) = P(not A1 and not A2 ) = P(A01 \ A02 ).
Since no formula exist for this probability, we use a venn diagram to develop one. Based on
the venn diagram,
P (A01 \ A02 ) = 1 ¡ P (A1 [ A2 )
= 1 ¡ 0:8 = 0:2:
c. P(exactly one car is Canadian) = P[(A1 and not A2 ) or (A2 and not A1 )]. Based on a venn
diagram, we have that
P [(A1 \ A02 ) [ (A01 \ A2 )] = P (A1 [ A2 ) ¡ P (A1 \ A2 )
= 0:8 ¡ 0:4 = 0:4:
4. Problem
A construction ¯rm is working on two di®erent projects. Let A be the event that the ¯rst one is
completed by the contract date and de¯ne B analogously for the second project. If P(A1 [ A2 )
= 0.9 and P(A1 \ A2 ) = 0.5, what is the probability that exactly one project is completed by
the contract date ?
Solution
The probability of interest is
P(exactly one project is completed) = P[(A1 and not A2 ) or (A2 and not A1 )].
Since there is no formula for calculating this probability, we will use a venn diagram of the
sample space and events to develop a formula. Based on the venn diagram, the probability of
interest is
P [(A1 \ A02 ) [ (A01 \ A2 )] = P (A1 [ A2 ) ¡ P (A1 \ A2 )
= 0:9 ¡ 0:5 = 0:4:
5. Problem
A video store sells two di®erent brands of VCRs, each of which comes with either two heads
or four heads. The accompanying table gives the percentages of recent purchasers buying each
type of VCR:
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Number of Heads
Brand
2
4
M
25%
16%
Q
32%
27%
Suppose a recent purchaser is randomly selected and both the brand and the number of heads
are determined.
a. What are the four simple events ?
b. What is the probability that the selected purchaser bought brand Q, with two heads ?
c. What is the probability that the selected purchaser bought brand M ?
Solution
a. S = fM 2; M 4; Q2; Q4g.
b. P(purchaser bought brand Q, with two heads) = P(Q2) = 0.32.
c. P(purchaser bought brand M ) = P(M 2 or M 4)
P (M 2 or M 4) = P (M 2 [ M 4)
= P (M 2) + P (M 4)
= 0:25 + 0:16 = 0:41:
Note the strategy we adopt in solving these problems
(1) Write down the information provided in the problem.
(2) Write down the probability you are asked to calculate.
(3) Obtain a formula based on known formulae and a venn diagram for computing (2).
(4) Use the formula to calculate the required probability.
This approach may not work all the time but it is very useful - going from the known to the unknown.
The Relative Frequency Interpretation of Probability
Consider the experiment of tossing a balanced two-sided coin under identical and independent conditions. Independent, in the sense that the outcomes of previous trials does not in°uence, in any
way, the outcome of current or future trials. Let A be the event that the outcome in a particular
toss is the tail. If we toss the coin repeatedly n number of times, the outcome will be tails sometimes
and heads at other times. Let n(A) be the number of times the tail turns up. Then the ratio
n(A)
n
-
the proportion of tails in n trials - is called the relative frequency of the event A. As the number of
trials n becomes larger and larger, the relative frequency
10
n(A)
n
will tend to stabilize to a particular
value called the limiting relative frequency of the event A. It is this limiting relative frequency that
we interpret as the probability of the event A, P (A). In practice, this limiting relative frequency of
an event may not be known. Thus, we will have to assign probabilities based on our beliefs about
the limiting relative frequency of events under study.
Examples
1. Experiment: toss a fair two-sided coin
Limiting relative frequency interpretation suggests assigning P(H) = 0.5 and P(T ) = 0.5. That
is, if the number of times the experiment is performed is large enough, the limiting relative
frequency of head and tail is 50%.
2. Experiment: toss a fair six-sided die
Limiting relative frequency interpretation suggests assigning P (1) = P (2) = ¢ ¢ ¢ = P (6) = 16 .
That is, if the number of times the experiment is performed is large enough, the limiting relative
frequency of 1, 2,¢ ¢ ¢, 6 is 1=6 for each of the six possible outcomes.
3. Problem 16, page 65
An individual is presented with three di®erent glasses of cola, labeled C, D, and P. He is asked
to taste all three and then list them in order of preference. Suppose that the same cola has
actually been put into all three glasses.
(a) What are the simple events in this ranking experiment, and what probability would you
assign to each one ?
(b) What is the probability that C is ranked ¯rst ?
(c) What is the probability that C is ranked ¯rst and D is ranked last ?
Solution
(a) Using a tree diagram, it is easy to ¯nd that the sample space of the experiment is
fCDP,CPD,DCP,DPC,PCD,PDCg. P (Ei ) = 1=6, (i = 1; : : : ; 6), where Ei is any simple event
in the sample space.
(b) Let A = fC is ranked ¯rstg = fCDP,CPDg. Then P (A) = 2=6.
(c) Let B = fC is ranked ¯rst and D is ranked lastg = fCPDg. Then P (B) = 1=6.
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Equally Likely Outcomes
Consider an experiment with N possible outcomes. That is, S = fO1 ; O2 ; : : : ; ON g. Let Ei , (i =
1; 2; ¢ ¢ ¢ ; N ) be the ith simple event (i:e event with only one outcome) of the experiment. That is,
E1 = fO1 g, E2 = fO2 g, and so on. If
P (Ei ) =
1
; i = 1; 2; ¢ ¢ ¢ ; N;
N
then each outcome of the experiment is said to be equally likely. That is, the outcomes are equally
likely if they have equal probability of occurence.
Now, let A be a compound event of an experiment with N equally likely outcomes. Let N (A) be the
number of outcomes contained in A. Then,
P (A) =
N(A)
:
N
For instance, if A = fO3 ; O6 ; O8 g, then
P (A) =
N(A)
3
= :
N
N
Examples
1. Problem 19, page 65
Human visual inspection of solder joint on printed circuit boards can be very subjective. Consequently, even highly trained inspectors can disagree on the disposition of a particular joint. In
one batch of 10,000 joints, inspector A found 724 that were judged defective, inspector B found
751 such joints, and 1159 of the joints were judged defective by at least one of the inspectors.
Suppose that one of the 10,000 joints is randomly selected.
(a) What is the probability that the selected joint was judged defective by neither of the two
inspectors ?
(b) What is the probability that the selected joint was judged to be defective by inspector B
but not by inspector A ?
Solution
Let A = fjudged defective by inspector Ag and B = fjudged defective by inspector Bg. Now,
N = 10000;
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N(A) = number of joints judged defective by A = 724;
N(B) = number of joints judged defective by B = 751;
N(A [ B) = number of joints judged defective by at least one of the two inspectors = 1159.
(a) The required probability is P (A0 \ B 0 ) = 1 ¡ P (A [ B).
Now, P (A [ B) =
N(A[B)
N(S)
= 0:1159. Therefore, P (A0 \ B 0 ) = 1 ¡ 0:1159 = 0:8841.
(b) The required probability is P (A0 \ B) = P (B) ¡ P (A \ B) = P (B) ¡ [P (A) + P (B) ¡
P (A [ B)] = P (A [ B) ¡ P (A) = 0.1159 - 0.0724 = 0.0435.
2. If a standard deck of 52 cards (spades, hearts, diamonds and clubs; with ace, 2,....,10, jack,
queen, and king in each suit) is well mixed before a single card is drawn, all 52 outcomes are
equally likely. De¯ne events A = fheartg, B = fblack cardg, C = fjack, queen, or kingg and
D = fselected card is at most a 5g and use elementary counting along with the axioms and
properties to compute the following probabilities.
a. P(A), P(B), P(C), and P(D).
b. P(A [ B) and P(A [ C).
c. P(A [ B [ C [ D).
Solution
a. Since the outcomes are equally likely with N = 52,
N(A) = number of hearts = 13,
N(B) = number of black cards = 26,
N(C) = number of jacks, queens and kings (4 each) = 12,
N(D) = number of cards between ace and 5 = 20, then
N(A)
13
1
N (B)
26
1
=
= ; P (B) =
=
= ;
N
52
4
N
52
2
N(C)
12
3
N (D)
20
5
P (C) =
=
= ; P (D) =
=
= :
N
52
13
N
52
13
b. From part (a) and the addition law,
P (A) =
P (A [ B) = P (A) + P (B); since P (A \ B) = 0;
1 1
3
=
+ = :
4 2
4
P (A [ C) = P (A) + P (C) ¡ P (A \ C);
where P (A \ C) = P (jack, queen and king of hearts) =
Therefore, P (A [ C) =
13 12
3
22
+
¡
= :
52 52 52
52
13
3
;
52
Alternatively, note that N (A [ C) = number of hearts, jacks, queens and kings in suit without
repetition. = N (A) + N (C) - number of jack, queen and king of hearts = 13 + 12 - 3 = 22.
Therefore,
P (A [ C) =
N (A [ C)
22
= :
N
52
c. First, we simplify the expression P (A [ B [ C [ D) to the form we can recognize. To do
this, we let E = A [ B and F = C [ D. Then, using the addition law, we have
P (A [ B [ C [ D) = P (E [ F ) = P (E) + P (F ) ¡ P (E \ F )
= P (A [ B) + P (C [ D) ¡ P [(A [ B) \ (C [ D)]
= P (A) + P (B) + P (C) + P (D) ¡ P [(A [ B) \ (C [ D)]
13 26 12 20
=
+
+
+
¡ P [(A [ B) \ (C [ D)]
52 52 52 52
Note that (A [ B) = fheart, black cardg; (C [ D) = fjack, queen, or king, selected card is at
most a 5g. So that, (A [ B) \ (C [ D) = fjack, queen and king of heart, black jack, black queen
and black king, heart and black cards that are at most 5g. So, N[(A [ B) \ (C [ D)] = 24. It
follows that
P (A [ B [ C [ D) =
13 26 12 20 24
47
+
+
+
¡
= :
52 52 52 52 52
52
An alternative and simpler approach is to note that the only outcomes not in A [ B [ C [ D
are the 6, 7, 8, 9 and 10 of diamonds. So, using the fact that P (A) = 1 ¡ P (A0 ) we have
P (A [ B [ C [ D) = 1 ¡
5
47
= :
52
52
x2.3 Counting Techniques
We have seen that if the N outcomes of an experiment are equally likely, then the probability of a
compound event A is simply
P (A) =
N(A)
;
N
where N(A) = the number of outcomes in A. That means, we need to be able to count the number of
outcomes in S and in A. Sometimes, this is not possible because of the complexity of the experiment
or the outcomes are too many for us to be able to list them before counting. So, it is important for
us to develop or study some rules for counting the number of outcomes in any event of an experiment
without necessarily listing the outcomes. The ¯rst of these rules is the product rule for ordered pairs.
14
METHOD I: Product Rule for Ordered Pairs
De¯nition: By an ordered pair, we mean that if, O1 and O2 are objects, then the pair (O1 ; O2 ) is
di®erent from the pair (O2 ; O1 ). We will use an example to illustrate this rule before stating it.
Examples:
1. Problem 31, Page 73
(a) Beethoven wrote 9 symphonies and Mozart wrote 27 piano concertos. If a university radio
station announcer wishes to play ¯rst a Beethoven symphony and then a Mozart concerto, in
how many ways can this be done ? In other words, what is the number of elements N in the
sample space of this experiment ?
Solution:
This experiment involves playing a Beethoven, then a Mozart concertos. Note that there is an
ordering involved - Beethoven ¯rst before a Mozart. So, (M1 ; B1 ) is di®erent from (B1 ; M1 ).
In fact (M1 ; B1 ) does not qualify to be an outcome of this experiment. The problem is to
determine the number of outcomes in the sample space of the experiment.
Let n1 = 9, n2 = 27. Then, from the tree diagram, this number is
N = n1 £ n2 = 9 £ 27 = 243.
Note that in this example, the selection is made from two di®erent sets. The ¯rst element must
come from the set of Beethovens and the second from the set of Mozarts. This is an important
component of any problem which requires the application of the product rule in its solution.
We now state the product rule.
Product Rule for Ordered Pairs:
If the ¯rst element or object of an ordered pair can be selected in n1 ways, and for each of
these n1 ways the second element of the pair can be selected in n2 ways, then the total number
of pairs is N = n1 £ n2 .
We use part (b) of Problem 31, Page 73 to show that the product rule can be extended to an
ordered sequence consisting of more than two objects (e.g. an ordered sequence of 3 objects).
Such ordered sequences consisting of k objects is called a k-tuple. A 2-tuple is an ordered
15
pair; A 3-tuple an ordered sequence of 3 objects or elements; etc. Suppose, I'm °ying from St.
John's to Edmonton. The sequence of my trip could be (St. John's, Toronto, Edmonton) and
my return trip is (Edmonton, Toronto, St. John's). These two ordered sequences of 3 cities are
not the same although they contain the same cities.
Problem 31, Page 73
(b) The station manager decides that on each successive night (7 days per week), a Beethoven
symphony will be played, followed by a Mozart piano concerto, followed by a Schubert string
quartet (of which there are 15). For roughly how many years could this policy be continued
before exactly the same program would have to be repeated ?
Solution:
N = n1 £ n2 £ n3 = 9 £ 27 £ 15 = 3645:
Since there are approximately 365 days in a year, the policy can be continued for
without repeating exactly the same program.
3645
365
¼ 10years
2. A homeowner doing some remodelling requires the services of both a plumbing contractor and
an electrical contractor. If there are 12 plumbing contractors and 9 electrical contractors available in the area, in how many ways can the contractors be chosen ?
Solution: Denote the plumbers by P1 ; ¢ ¢ ¢ ; P12 and the electricians by E1 ; ¢ ¢ ¢ ; E9 . Then we
seek the number of pairs of the form (Pi ; Ej ), (i = 1; ¢ ¢ ¢ ; 12; j = 1; : : : ; 9). From the tree
diagram, let n1 = 12, n2 = 9, then N = 12 £ 9 = 108.
METHOD II: Permutation
Consider a ¯xed set A consisting of 4 distinct objects:
A = fa; b; c; dg
Suppose that we form a 3-tuple by selecting successively from the set A without replacement so that
an element can appear in at most one of the 3 positions. The question is, how many 3-tuples can we
construct from set A ?
16
Using a tree diagram, we see that the ¯rst element can be selected in 4 ways; for each of the 4 ways
in which the ¯rst can be selected, the second element can be selected in 3 ways and the third in 2
ways. So, the total number of 3-tuples possible is 4 £ 3 £ 2 = 24.
De¯nition: Any ordered sequence of k objects taken from a set of n distinct objects is called a
permutation of size k of the objects. The number of permutations of size k that can be constructed
from the n objects is denoted by Pk;n .
In general, suppose that A contains n distinct objects or elements. How many k-tuples can we construct from the set A ?
As before, 1st element can be chosen in n ways;
2nd element in n ¡ (2 ¡ 1) = n ¡ 1 ways;
3rd element in n ¡ (3 ¡ 1) = n ¡ 2 ways; and so on
The kth element in n ¡ (k ¡ 1) ways.
Therefore, the total number of k-tuples we can construct from the set A is
Pk;n = n £ (n ¡ 1) £ (n ¡ 2) £ ¢ ¢ ¢ £ (n ¡ k + 1) =
Examples
n!
(n ¡ k)!
1. Problem 30, Page 73
A real estate agent is showing homes to a propective buyer. There are ten homes in the desired
price range listed in the area. The buyer has time to visit only three of them.
(a) In how many ways could the three homes be chosen if the order of visiting is considered ?
Solution
(a) The problem here is to select three homes to visit in a set consisting of ten homes H =
fh1 ; h2 ; : : : ; h10 g without replacement - order being important. Thus, n = 10 and k = 3.
Implying that
Pk;n = P3;10 =
10!
= 10 £ 9 £ 8 = 720:
(10 ¡ 3)!
(b) In how many ways could the three homes be chosen if the order is disregarded ?
This question requires that we use a method called combination.
17
METHOD III: Combination
We have ten houses in the set H = fh1 ; h2 ; : : : ; h10 g. Now we wish to select 3 elements from
H without replacement. If order were to be important, we would have P3;10 = 720 possible
permutations of size 3 of the 10 homes in H. Since order is no longer important, the selection
(h1 ; h5 ; h8 ) ´ (h8 ; h1 ; h5 ):
So, the number of possible selections or arrangements will be smaller. When order is not
important, any k selection or arrangement of objects or elements from a set, say A, consisting
of n objects is called a combination. The number of combinations of size k that can be formed
from n distinct objects denoted by
Ã
n
k
!
=
n!
:
k!(n ¡ k)!
Thus the solution to part (b) of the problem is as follows.
Solution:
(b) Here, n = 10 and k = 3
Ã
n
k
!
=
n!
10!
10 £ 9 £ 8
=
=
= 120:
k!(n ¡ k)!
3!(10 ¡ 3)!
3£2£1
Next we use the idea of combination to answer part (c).
(c) If four of the homes are new and six have previously been occupied and if the three homes
to visit are randomly chosen, what is the probability that all three are new ?
Solution:
(c) In this problem the set of homes has been split into two disjoint sets A = fnew homesg; B
= fused homesg. If E = fthe three selected homes are newg, then
P (E) =
N (E)
N
where N = the total number of ways of selecting three homes from ten. Now, N (E) = n1 £ n2 ,
where n1 = total number of ways of selecting three new homes from four; n2 = total number
of ways of selecting zero used homes from six. Thus,
N(E)
n1 £ n2
P (E) =
=
=
N
N
Ã
18
4
3
!
Ã
£
10
3
Ã
!
6
0
!
=
4
1
= :
120
30
Alternatively, we can use the concept of permutation to solve the problem
P (E) =
N(E)
n1 £ n2
P3;4 £ P0;6
1
24
=
=
=
= :
N
N
P3;10
720
30
In solving this problem, we have used the product rule and combinations or permutations.
2. Problem 29, Page 73
The student Engineering Council at MUN has one student representative from each of the ¯ve
engineering majors (civil, electrical, industrial, materials and mechanical). In how many ways
can both a council president and a vice-president be selected ?
Solution The problem is to ¯ll two positions by selecting from a set consisting of ¯ve department representatives without replacement. Thus, n = 5 and k = 2. Implying that
Pk;n = P2;5 =
5!
= 5 £ 4 = 20:
(5 ¡ 2)!
3. Problem 33, page 74
Shortly after being put into service, some buses manufactured by a certain company have
developed cracks on the underside of the main frame. Suppose a particular city has 20 of these
buses, and cracks have actually appeared in 8 of them.
(a) How many ways are there to select a sample of 5 buses from the 20 for a thorough inspection
?
(b) In how many ways can a sample of 5 buses contain exactly 4 with visible cracks?
(c) If a sample of 5 buses is chosen at random, what is the probability that exactly 4 of the 5
will have visible cracks ?
(d) If buses are selected as in part (c), what is the probability that at least 4 of those selected
will have visible cracks ?
4. A rental car service facility has 10 foreign cars and 15 domestic cars waiting to be serviced on
a particular Saturday morning. Because there are so few mechanics working on Saturday, only
six can be serviced. If the six are chosen at random, what is the probability that
a. 3 of the cars selected are domestic and the other three are foreign ?
b. at least 2 domestic cars are selected ?
19
Solution:
a. Again, let A = fforeign carsg; B = fDomestic carsg; E the event f3 selected cars are
domestic; 3 are foreign g. As before,
P (E) =
N (E)
N
where N total number of ways of selecting or arranging 6 cars out of 25. Now, N(E) = n1 £ n2 ,
where n1 = total number of ways of selecting 3 domestic cars from 15; n2 = total number of
ways of selecting 3 foreign cars from 10. Thus,
N (E)
n1 £ n2
P (E) =
=
=
N
N
Ã
15
3
!
Ã
£
25
6
Ã
!
10
3
!
=
78
= 0:3083:
253
b. Note that, P (at least 2 domestic cars are selected) = P (2, 3, 4, 5, or 6 domestic cars are
selected) = 1 - P (0 or 1 domestic car is selected). Now, P (0 or 1 domestic car is selected) =
Ã
15
0
!
Ã
£
25
6
Ã
!
10
6
!
+
Ã
15
1
!
Ã
£
Ã
25
6
!
10
5
!
3990
! = 0:0225296:
=Ã
25
6
Therefore, P (at least 2 domestic cars are selected) = 1 - 0.0225296 ¼ 0.9774.
x2.4 Conditional Probability
The probability assigned to various events is based on information on the experiment that we have
when the assignment is made. Later on, we can revise the assigned probability if extra information
becomes available. Lets take for instance, that your friend visits from out of town to search for you
on Campus. His chance of ¯nding you will be very small because of the large area he/she has to
search. Lets assume that during the search, the information that you are attending a STAT 2510
class at that time becomes available. This extra information then improves his/her chance of ¯nding
you.
Let A = f¯nd a friend in MUN g;
Let B = fin STAT 2510 classg.
Then, P (A) is the unconditional probability of A.
P (AjB) is called the conditional probability of A given that B has occured.
20
With this extra information B, my sample space for the search reduces from the entire university
campus to the two sections of STAT 2510. So, conditioning on a given event has the e®ect of reducing
our sample space from a larger one to a smaller one.
Example
1. Consider the example on purchasers of brands of VCRS. Suppose the number of customers
purchasing di®erent brands can be summarized as follows.
Number
Brand
M
Q
of Heads
2
4
3
2
4
5
Let A = fcustomer purchased brand M g; and B = fcustomer purchased a VCR with 2 headsg.
Now,
N (A)
5
= :
N
14
3
N (A \ B)
P (A \ B)
P (AjB) = =
=
=
7
N (B)
P (B)
P (A) =
3
14
7 :
14
In general, we have the following de¯nition.
De¯nition: For any two events A and B with P (B) > 0, the conditional probability of A given
that B has occurred is de¯ned by
P (AjB) =
P (A \ B)
:
P (B)
Observe, from this de¯nition, that
P (A \ B) = P (AjB) £ P (B):
This is called the multiplication rule for P (A \ B).
Examples
1. Problem 59, Page 84
At a certain gas station, 40% of the customers use regular unleaded gas (A1 ), 35% use extra
unleaded gas (A2 ), and 25% use premium unleaded gas (A3 ). Of those customers using regular
21
gas, only 30% ¯ll their tanks (B). Of those customers using extra gas, 60% ¯ll their tanks,
while of those using premium, 50% ¯ll their tanks.
(a) What is the probability that the next customer will request extra unleaded gas and ¯ll the
tank (A2 \ B) ?
(b) What is the probability that the next customer ¯lls the tank ?
(c) If the next customer ¯lls the tank, what is the probability that regular gas is requested ?
Extra gas ? Premium gas ?
Solution:
De¯ne A1 = fcustomer uses regular unleaded gasg; A2 = fcustomer uses extra unleaded gasg;
A3 = fcustomer uses premium unleaded gasg; B = fcustomer ¯lls tankg. The events A1 , A2
and A3 are mutually exclusive and collectively exhaustive in the sense that A1 \ A2 \ A3 = fg
(the empty set) and A1 [ A2 [ A3 = S. Now, P (A1 ) = 0:4; P (A2 ) = 0:35; P (A3 ) = 0:25;
P (BjA1 ) = 0:3; P (BjA2 ) = 0:6; P (BjA3 ) = 0:5.
(a) P (A2 \ B) = P (BjA2 ) £ P (A2 ) = 0:6 £ 0:35 = 0:21:
(b)
P (B) = P (A1 \ B) + P (A2 \ B) + P (A3 \ B)
= P (BjA1 ) £ P (A1 ) + P (BjA2 ) £ P (A2 ) + P (BjA3 ) £ P (A3 )
= 0:3 £ 0:4 + 0:6 £ 0:35 + 0:5 £ 0:25 = 0:455:
We generalize the solution to part (b). The generalization is called the law of total probability.
Law of Total Probability: Let A1 , A2 ; ¢ ¢ ¢ ; An be a collection of mutually exclusive and
exhaustive events. Then, for any other event B
P (B) = P (BjA1 ) £ P (A1 ) + P (BjA2 ) £ P (A2 ) + ¢ ¢ ¢ + P (BjAn ) £ P (An )
=
n
X
i=1
P (BjAi ) £ P (Ai ):
(c)
P (A1 jB) =
P (A1 \ B)
P (B)
P (BjA1 ) £ P (A1 )
P (BjA1 ) £ P (A1 ) + P (BjA2 ) £ P (A2 ) + P (BjA3 ) £ P (A3 )
0:12
=
= 0:2637:
0:455
=
22
From the solution to part (c), we obtain a general formula or rule called Bayes' Theorem.
Bayes' Theorem: Let A1 ; A2 ; ¢ ¢ ¢ ; An be a collection of mutually exclusive and exhaustive
events with P (Ai ) > 0 for i = 1; ¢ ¢ ¢ ; n. Then, for any other event B for which P (B) > 0
P (Ak jB) =
2. Problem 62, Page 85
P (Ak \ B)
P (BjAk )P (Ak )
= Pn
; k = 1; 2; ¢ ¢ ¢ ; n
P (B)
i=1 P (BjAi )P (Ai )
A company that manufactures video cameras produces a basic model and a deluxe model. Over
the past year, 40% of the cameras sold have been of the basic model. Of those buying the basic
model, 30% purchase an extended warranty, whereas 50% of all deluxe purchasers do so. If
you learn that a randomly selected purchaser has an extended warranty, how likely is it that
he/she has a basic model ?
Solution: B = fcustomer purchases a basic modelg; D = fcustomer purchases a deluxe
modelg; W = fcustomer purchases extended warrantyg. Now, P (B) = 0:4; P (D) = 0:6;
P (W jB) = 0:3; P (W jD) = 0:5.
P (B \ W )
P (W jB) £ P (B)
=
P (W )
P (W jB) £ P (B) + P (W jD) £ P (D)
0:12
=
= 0:2857:
0:42
P (BjW ) =
3. Problem 58, Page 84
Show that for any three events A, B and C with P (C) > 0, P (A [ BjC) = P (AjC) + P (BjC) ¡
P (A \ BjC).
Solution:
P [(A [ B) \ C]
P (A \ C) + P (B \ C) ¡ P (A \ B \ C)
=
P (C)
P (C)
= P (AjC) + P (BjC) ¡ P (A \ BjC):
P (A [ BjC) =
4. Problem 60, Page 85
Seventy percent of the light aircraft that disappear while in °ight in a certain country are
subsequently discovered. Of the aircraft that are discovered, 60% have an emergency locator,
whereas 90% of the aircraft not discovered do not have such a locator. Suppose a light aircraft
has disappeared.
(a) If it has an emergency locator, what is the probability that it will not be discovered ?
23
(b) If it does not have an emergency locator, what is the probability that it will be discovered
?
Solution:
A = faircraft discoveredg; B = faircraft has emergency locatorg. Now, P (A) = 0:7; P (BjA) =
0:6; P (B 0 jA0 ) = 0:6.
a.
P (A \ B)
P (B)
P (BjA) £ P (A)
= 1¡
P (BjA) £ P (A) + P (BjA0 ) £ P (A0 )
0:6 £ 0:7
= 1¡
= 0:067:
0:6 £ 0:7 + 0:1 £ 0:3
P (A0 jB) = 1 ¡ P (AjB) = 1 ¡
b.
P (A \ B 0 )
P (B 0 jA) £ P (A)
P (AjB ) =
=
P (B 0 )
1 ¡ P (B)
0:28
=
= 0:509:
0:55
0
x2.5 Independence
In the presence of extra information, we have seen that the unconditional probability of an event,
say P (A), is sometimes not equal to the conditional probability of A given that B has occurred
P (AjB). There are situations where P (A) = P (AjB). That is, the extra information that B has
occurred does not a®ect the chance that A will occur. In such cases, we say that the events A and
B are independent. Thus the two events, A and B are said to be independent if
P (AjB) = P (A); P (B) > 0:
They are otherwise said to be dependent. From the multiplication rule, we have that
P (A \ B) = P (AjB) £ P (B):
Implying that if A and B are independent
P (A \ B) = P (AjB) £ P (B) = P (A) £ P (B):
24
Examples
1. Problem 69, Page 89
An executive on a business trip must rent a car in each of two di®erent cities. Let A denote
the event that the executive is o®ered a free upgrade in the ¯rst city and B represent the
analogous event for the second city. Suppose that P (A) = 0:2, P (B) = 0:3, and that A and B
are independent events.
(a) If the executive is not o®ered a free upgrade in the ¯rst city, what is the probability of not
getting a free upgrade in the second city ?
(b) What is the probability that the executive is o®ered a free upgrade in at least one of the
two cities ?
(c) If the executive is o®ered a free upgrade in at least one of the two cities, what is the probability that such an o®er was made only in the ¯rst city ?
Solution:
Let A = fexecutive is o®ered free upgrade in city 1g
and B = fexecutive is o®ered free upgrade in city 1g.
Now, P (A) = 0:2 and P (B) = 0:3. Therefore, using independence we have that P (A \ B) =
0:2 £ 0:3 = 0:06.
(a)
P (B 0 jA0 ) =
Now,
P (B 0 \ A0 )
1 ¡ P (A [ B)
=
:
0
P (A )
1 ¡ P (A)
P (A [ B) = 0:2 + 0:3 ¡ 0:06 = 0:44:
Therefore,
P (B 0 jA0 ) =
1 ¡ 0:44
= 0:7:
1 ¡ 0:2
(b) From part (a), P (A [ B) = 0:2 + 0:3 ¡ 0:06 = 0:44:
(c)
P [(A \ B 0 ) \ (A [ B)]
P (A \ B 0 )
=
P (A [ B)
P (A [ B)
P (A) ¡ P (A \ B)
0:14
=
=
¼ 0:318:
P (A [ B)
0:44
P [(A \ B 0 )j(A [ B)] =
25
2. Problem 72 Page 90
Suppose that the proportions of blood phenotypes in a particular population are as follows:
A
B
AB
O
0.42
0.10
0.04
0.44
Assuming that the phenotypes of two randomly selected individuals are independent of one
another, what is the probability that
(a) both phenotypes are O ?
(b) both phenotypes match ?
Solution:
Let A = fPhenotype 1 is Og and B = fPhenotype 2 is Og.
(a) P (A \ B) = P (A) £ P (B) = 0:44 £ 0:44 = 0:1936:
(b) P (both phenotypes match) = P (both are A or both are B or both are AB or both are O)
= 0:42 £ 0:42 + 0:1 £ 0:1 + 0:04 £ 0:04 + 0:44 £ 0:44 = 0.3816.
3. Problem 78, Page 90.
Solution:
Let Ai = fcomponent i worksg, (i = 1; 2; 3; 4). So that
System works = (A1 or A2 ) or (A3 and A4 ):
P (System works) = P [(A1 [ A2 ) [ (A3 \ A4 )]
= P (A1 ) + P (A2 ) + P (A3 \ A4 ) ¡ P (A1 \ A2 ) ¡ P (A1 \ A3 \ A4 )
¡P (A2 \ A3 \ A4 ) + P (A1 \ A2 \ A3 \ A4 )
= 0:9 + 0:9 + 0:92 ¡ 0:92 ¡ 0:93 ¡ 0:93 + 0:94 = 0:9981:
26