History of probability CHAPTER 5 Engineering

History of probability
 The foundations of probability were laid by two French
mathematicians of 17th
century- Blaise Pascal (1623-1662) and Pierre
de Fermat (1601-1665) in connection with gambling problems.
 Later on, it was developed by
 Jacob Bernoulli(1654-1705)
 Abraham de moivre (1667-1754)
 Pierre Simon Laplace (1749-1827)
modern probability theory which consists of few assumptions and rules
was developed in 20s
and 30s of this century.

Probability
 The word probability has two basic meanings,
1. a quantitative measure of uncertainty
2. A measure of degree of belief in a particular statement or problem.
There are abundant examples in every day life where probability is used as a
degree of belief or measure of uncertainty.
 What are the chances of rain showers tomorrow?
 What is chance of Pakistan to win the toss in todays match?
 What are the chances X brand of PC will survive 10,000 hours of operation without
repair?
 what are the chances of heart attack of a patient?

Some basic term or concepts
Experiment
means a planned activity or process which yield some results or outcomes.
Random experiments
An experiment which produces different results even though, it is repeated a large
number of times under essentially similar conditions is called random experiment.
Examples:
1. Tossing a coin
2. Throwing a die
3. Drawing a card from 52 cards
4. Measuring the current in a thin copper wire
5. call durations in a day
6. Examining a fuse

Properties of a random experiment
1. The experiment can be repeated , practically or theoretically any
number of time.
2. The experiment always has two or more possible outcomes.
3. The outcome of each repetition is unpredictable in advance.

 Trial
It is single performance of an experiment.
 Outcome
the result obtained from an experiment or a trial is called an
outcome.
 Sample space
The set of all possible outcomes of a random experiment is
called a sample space .
It is denoted as S.
The elements of sample space are called sample points.

Examples of sample space
 Tossing a coin S= {H,T}
 Throwing a die S={ 1,2,3,4,5,6}
 tossing two coin
 Throwing two die

 Event;
An event is an individual outcome or a set of outcomes of a random experiment. Or
any subset of a sample space is an event.
Simple event
Compound event
Exhaustive events
Mutually exclusive events.

An event that contains exactly one sample
point is defined as simple event or elementary
event.
Example
two coins are tossed, the event A={HH} is a simple
event.
Simple event

Compound event
An event consists of more than one sample points, is defined as
compound event.
Example
1: let B be the event that at least one tail appears in tossing two coins
B={HT,TH,TT}

Mutually Exclusive events;
Two events A and B of a single experiment are said to be
mutually exclusive or disjoint events if and only if they cannot
both occur at the same time. That is they have no points in
common.
Example:
A product can either be a defective or non defective.
 a student either qualifies or fail
When a die is rolled , outcomes are mutually exclusive as we get
one and only one of six possible outcomes 1,2,3,4,5,6.

Exhaustive events;
Events are said to be exhaustive events, when
union of mutually exclusive events is the entire
sample space.
Example:
Tossing a coin

Equally likely events
 Events are said to be equally likely events, when they have equal chance
of occurrence at each trial.
 Example,
1. Examining three fuse in sequence
2. Tossing a coin

Multiplication rule
 If an experiment consists of two sub-experiments such that one has
n1 outcomes and other has n2 outcomes, then the combined
sample space consists of n1×n2 outcomes.
 Example,
1. if a die and a coin are thrown at the same time the number of
possible outcomes is 2×6 = 12.
Similarly, if an experiment has K possible outcomes in a single trial then
in n trails it will have K x n possible outcomes.
 Example,
1. Tossing a coin twice time , we have 2×2 = 4 possible outcomes.
2. Similarly, if the die is thrown twice, we have 6×6 = 62 = 36 possible
outcomes.

Permutation rule
 A permutation is an “arrangement” of n objects in a “speciﬁc order”.
 The arrangement of n objects in a speciﬁc order using r objects at a time is
called a permutation. It is written as n
Pr, and the formula is:
n
Pr = n!/(n-r)!

Combination rule
 A “selection” of distinct objects “without any regard to order” is called a
combination.
The number of combinations of r objects selected from n objects is denoted
n
Cr and is given by the formula:
n
Cr = n!/r!(n-r)!
Example:
 A bridge hand of any any 13 cards selected from a 52 cards without
regards to order. There are:
 52!/13!*39!

Question:1
(permutation or combination)
A student club consists of 6 students of which a
three person committee consisting of a
president, a secretary, and a treasure is to be
chosen. In how many ways we can choose this
committee?

Question:2
(permutation or combination)
 Now suppose a student has 3 friends among these 6
students and wants them to be elected at these
positions but does not care which friend is elected at
which position. That’s the order does not matter. In how
many ways 3 students can be chosen out of 6 students?

Definition of probability
Classical definition:
 If a random experiment can produce n mutually exclusive and equally
likely outcomes, and if m out to these outcomes are considered favorable
to the occurrence of a certain event A, then the probability of the event A,
denoted by P(A), is defined as the ratio
P(A) = =
=
Drawback:
It's the simplest definition of probability. However, this definition is not
applicable if the assumption of equally likely does not hold.

Empirical or relative definition of
probability
 If a random experiment is repeated a large number of times, say n, under
identical conditions and if an event A is observed to occur m times, then
the probability of A is deﬁned as,
 P(A) =
 As n → ∞, the ratio m/ n becomes stable at its numerical value.

Assumptions of probability
1. For any event A,
0 ≤ P(A) ≤ 1; the probability of any thing lies
between 0 and 1. Negative probability has no
meanings.
2. P(S) = 1; the probability of the entire sample
space is 1

Questions
Q1:A coin is tossed twice, what is the probability that
i) a head appears on both?
ii) at least one head occurs?
Q2:
Two fair dice are thrown, what is the probability that a sum of 8 or more dots
occurs (i.e. the sum of dots on both dice is at least 8.)?

Tree diagram
 Example 2:
an automobile manufacturer provides vehicles equipeed with selected
option. Each vehicle is ordered
 With and without automatic transmission
 With and without air conditioning
 With one of three choices of a stereo system
 One of four exterior colors
tree diagram for possible no. of outcomes.

Laws of probability
 There are some probability rules, that make a probability problem easier to handle. Sometimes the
probabilities of certain events are known to us and we can use them to calculate the probability
of another event. Some events can be written as the union or intersection of two or more events
and sometimes an event can be split into two or more events.
 For example
1. If a die is thrown twice, we may be interested in ﬁnding the probability of getting a total of 7 or 11.
Now here we have two outcomes 7 or 11, if either one occurs we consider it our favorite outcome.
⇒ We can consider them as separate events that’s let A denotes the event that a sum of 7 occurs and
let B be the event that a sum of 11 occurs.
2. If a die is thrown twice, we may be interested in ﬁnding the probability of getting a 3 on one die
and 5 on the other die. (beware here, we are not talking about a total or sum of dots)
⇒ Here again we have two outcomes 3 and 5, if both occurs we consider it our favorite outcome.
⇒ Again we can consider them as separate events that’s let A denotes the event that 3 occurs and
let B be the event that 5 occurs

Addition law
 This law has two forms:
1. If A and B are any two events then
P(A B) = P(A) + P(B)−P(A∩B)
∪
2. If A and B are two mutually exclusive events then
P(A B) = P(A) + P(B)− P(A∩B)
∪
P(A∩B) = 0
So
P(A B) = P(A) + P(B)
∪

Example.
 Consider a sample space S = {1,2,3,4,5,6,9,15}. A number is chosen at random,
what is the probability that the number is an even number or a prime number?
(prime number: a natural number greater than 1, has no divisor other than itself
and 1.)
 Solution:. Let A be event of an even number so A = {2,4,6} and
 B be the event of a prime number so B = {2,3,5}.
 We need P(A or B) = P(A B).
∪
 Here P(A) = 3/ 8 and P(B) = 3/ 8.
 P(A∩B) = 1/
 P(A B) =P(A) + P(B)−P(A∩B) = 3/ 8 + 3/ 8 − 1 /8 = 5 /8
∪
 Which is the ﬁrst form of the law of addition in probability, as mentioned above.

Questions
Q1:
A card is drawn at random from a pack of 52 playing cards. What is the
probability that the card is face card or a club ♣ card.
Hint: We know that there are 12 face cards (the cards with faces of King,
Queen and Jack) and 13 cards for each suit (diamond ♦, club ♣, heart ♥,
spade ♠)
Q2:Consider a sample space S = {1,2,3,4,5,6,9,15}.
A number is chosen at random, what is the probability that the number is an
even number or a prime number? (prime number: a natural number greater
than 1, has no divisor other than itself and 1.)

 Q3:
 a card is drawn at random from a pack of 52 playing cards. What is
the probability that the card is a diamond ♦ card or a club ♣ card.
Q4:What is the probability of the occurrence of a number that is odd or less
than 5 when a fair die is rolled.

 Solution
 Let the event of the occurrence of a number that is odd be ‘A’ and the
event of the occurrence of a number that is less than 5 be ‘B’. We need to
find P(A or B).
 P(A) = 3/6 (odd numbers = 1,3 and 5)
 P(B) = 4/6 (numbers less than 5 = 1,2,3 and 4)
 P(A and B) = 2/6 (numbers that are both odd and less than 5 = 1 and 3)
 Now, P(A or B) = P(A) + P(B) – P(A or B)
 = 3/6 + 4/6 – 2/6
 P(A or B) = 5/6.

QUESTION
 At a political rally, there are 20 Republicans (R), 13 Democrats (D), and 6
Independents (I). If a person is selected, find the probability that he or she
is either a Democrat or an Independent.
 = 19/39.

Conditional Probability
We often have some information available before ﬁnding the probability of a
favorable event. For example,
 An experiment consists of rolling a die once. Let A be the event that a 6
occurs. When the die was thrown, we were told that anything greater than 4
(i.e, 5, 6) has occurred. Let B be the event that anything greater than 4 (i.e,
5, 6) has occurred. We know that the probability of every face of the die is
1 6 , thus, P(A) = 1/ 6 .
Now suppose that the die is rolled and we are told that the event B has
occurred. This leaves only two possible outcomes: 5 and 6. Based on this
information, the probability of A becomes 1/ 2 ,
 The probability of A given that B has already occurred =1 /2
 What we have done is that we just conditioned the probability of our
favorable event on available information. This is called conditional
probability.

Definition of conditional probability
The mathematical formula that helps in calculating such
conditional probabilities is deﬁned as:
 for any two events A and B in a sample space S with P(B) > 0, the
probability of event A given that event B has already occurred,
written as P(A|B) (here the vertical bar “|” means “given that”),
is given by,
P(A|B) = P(A∩B) /P(B) ,
 provided P(B) > 0 If P(B) = 0, the probability becomes undeﬁned.

Example:
We toss a fair coin three successive times. We wish to ﬁnd the conditional
probability that more heads occurs than tails, GIVEN that the ﬁrst toss resulted
in a head.

solution: Let A and B are the events where
A = {more heads than tails come up},
B = {1st toss is a head}.
The sample space for this experiment is,
S = {HHH, HHT, HTH, HTT, THH, THT, TTH, TTT}.
The ﬁrst head occurs at B = {HHH, HHT, HTH, HTT},
so its probability is P(B) = 4 /8 Now the event
A consists of the four elements HHH, HHT, HTH, THH, so, event A∩B consists of the
three elements HHH, HHT, HTH so its probability is
 P(A∩B) =3/ 8
 Thus, the conditional probability is
 P(A|B) =P(A∩B) P(B)
 =3 /8 / 4/ 8
 =3 /4

Question
 A man tosses two fair dice. What is the conditional probability that the sum
of the two dice is 6, given that two di erent numbers appear.
ﬀ
 Ans: 2/15

Multiplication law
 P(A|B) = P(A∩B)/ P(B)
⇒ P(A∩B) = P(A|B)P(B)
= P(B|A)P(A)
By this rule we can find the probability of the simultaneous occurrence of two or
more events.
 In the addition rule we used OR and here we use AND.
 For example in the case of addition rule we said “what is the probability that A
occurs or B occurs”. Whereas here we say that “what is the probability that A and B
occur”.
 We also notice that here we use the definition of conditional probability because
in most experiments the events occur in such a way that the occurrence of one
event a ects the occurrence of other events.
ff

Example : multiplication law
Two items are drawn at random from a box, which contains 10 items 4 of which are
defective. What is the probability that one item is defective and the other is good.
Solution: Lets assume that the ﬁrst item drawn is good and the second is defective.
Let A denotes that a good item is drawn and B denotes that a defective is selected.
We know that out of 10 items there, 6 good items so the probability of A is
P(A) =6/10
 Now since one good item is selected, we are left with 9 items in the box, the
probability of selecting a defective item (event B) given that a good item (event
A) has already been selected is now a conditional probability, P(B|A). Which is,
P(B|A) = 4 /9
Hence the probability that one item is good and the other is defective is given
as, P(B ∩A) = P(A)P(B|A) = 6 /10 × 4 /9=4 /15

Independent and dependent events
 Two events A and B are said to be independent if and only if,
 P(B|A) = P(B) or P(A|B) = P(A)
 Otherwise A and B are dependent.
 Thus we can say that two events A and B are said to be independent if and
only if, P(A∩B) = P(A)P(B)
 Therefore the probability that two independent events will both occur, we
simply ﬁnd the product of their individual probabilities.

Example: Independent and
dependent events
 A town has two fire engines operating independently. The probability that a
specific fire engine is available when needed is 0.96. Find the following
probabilities:
 1 Probability that neither is available when needed.
 2 Probability that both are available when needed.

Solution: Let A and B denote the availability of the two ﬁre engines.
Thus ¯ A and ¯ B are events of non availability of the ﬁre engines.
We have P(A) = P(B) = 0.96 and
hence P( ¯ A) = P( ¯ B) = 0.04 ( ¯ A = 1−A P( ¯ A) = 1−P(A)).
∵ ⇒
P(neither is available) = P( ¯ A∩ ¯ B) = P( ¯ A)P( ¯ B) = 0.04×0.04 = 0.0016
P(both are available) = P(A∩B) = P(A)P(B) = 0.96×0.96 = 0.9216

Bayes’ Rule:
This is a rule which revolutionized the theory of statistics. Before learning it, lets
learn about the total probability ﬁrst.
 Total Probability Theorem:
Let A1, A2, A3,...,An be mutually exclusive and exhaustive events that form a
partition of the sample space. Assume that P(Ai) > 0, for all i. Then for any event
B, we have
P(B) = P(A1 ∩B) +···+ P(An ∩B)
= P(A1)P(B|A1) +···+ P(An)P(B|An)
=

Statement : Bayes theorem
Let A1, A2, A3,...,An be mutually exclusive and exhaustive events that form a partition of
the sample space. Assume that P(Ai) > 0, for all i.
Then for any event B such that P(B) > 0, we have
 P(Ai|B) =
=
=
Bayes’ rule enables us to combine our own belief about certain event Ai with the
observed event B. Here P(Ai) are called a priori probabilities, whereas P(Ai|B) are
called the a posteriori probabilities. A prior probability is an initial probability value
originally obtained before any additional information is obtained. A posterior
probability is a probability value that has been revised by using additional information
that is later obtained.

Example : Bayes theorem:
An aircraft emergency locator transmitter (ELT) is a device designed to transmit a signal in the case of a crash.
The A Manufacturing Company makes 80% of the ELTs,
the B Company makes 15% of them, and
the C Company makes the other 5%.
The ELTs made by A have a 4% rate of defects, the B ELTs have a 6% rate of defects, and the C ELTs have a 9% rate of
defects.
 If an ELT is randomly selected from the general population of all ELTs, ﬁnd the probability that it was made by the
A Manufacturing Company. If a randomly selected ELT is then tested and is found to be defective, ﬁnd the
probability that it was made by the A Manufacturing Company.
Solution: Lets use the following notation: A = ELT manufactured by A
B = ELT manufactured by B
C = ELT manufactured by C
D = ELT is defective
¯ D = ELT is not defective (or it is good)
1. Now ¬ If an ELT is randomly selected from the general population of all ELTs, the probability that it was made by A
is 0.8 (because A manufactures 80% of them).

2. If we now have the additional information that the ELT was tested and was found to
be defective. We want to ﬁnd the value of P(A|D), which is the probability that the ELT
was made by the A company given that it is defective.
Based on the given information, we know these probabilities:
P(A) = 0.80 because A makes 80% of the ELTs
P(B) = 0.15 because B makes 15% of the ELTs
P(C) = 0.05 because C makes 5% of the ELTs
P(D|A) = 0.04 because 4% of the A ELTs are defective
P(D|B) = 0.06 because 6% of the B ELTs are defective
P(D|C) = 0.09 because 9% of the C ELTs are defective
Here is Bayes’ theorem extended to include three events corresponding to the
selection of ELTs from the three manufacturers (A, B, C):
 P(A|D) =
= 0 .0703

History of probability CHAPTER 5 Engineering

History of probability CHAPTER 5 Engineering

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History of probability CHAPTER 5 Engineering