Introduction to Probability

Introduction
• In our day-to-day life involving decision-making problems, we encounter two broad
types of problems.
• These problems can be categorized into two types of models:
• Deterministic Models
• Random or Probabilistic Models.

Deterministic Models cover those situations, where everything related to

the situation is known with certainty to the decision-maker, when decision is
to be made.
Probabilistic Models, the totality of the outcomes is known but it can not be
certain, which particular outcome will appear. So, there is always some
uncertainty involved in decision-making.
 Random Experiment
• However, no matter how carefully our experiment is designed and conducted,
the variation is almost always present.
• Conclusions from our experiment/finding are not obvious.
• Our goal is to understand, quantify, and model the type of variations that we
often encounter.
• When we incorporate the variation into our thinking and analyses, we can
make informed judgments from our results that are not invalidated by the
variation.
• An experiment that can result in different outcomes, even though it is repeated
in the same manner every time, is called a random experiment.
Key-points
• In Deterministic Models, frequency distribution or descriptive
statistics measures are used to arrive at a decision.
• In random situations, probability and probability distributions are used
to make decisions
• So, Probability can also defined as measure of uncertainty means that
something can never happen.
• A probability of 1 indicates that something will always happen.
Basic Concept
• To model and analyze a random experiment, we must understand the set of
possible outcomes from the experiment.
• In this introduction to probability, we make use of the basic concepts of sets
and operations on sets.

• The set of all possible outcomes of a random experiment is called the sample
space of the experiment. The sample space is denoted as S.
• S ={low, medium, high}
• S={yes, no}
• A sample space is discrete if it consists of a finite or countable infinite set of
outcomes.
• A sample space is continuous if it contains an interval (either finite or infinite)
of real numbers.
 Events
• An event is a subset of the sample space of a random experiment.
• We can also be interested in describing new events from combinations of existing events.
• Because events are subsets, we can use basic set operations such as unions, intersections,
and complements to form other events of interest.
• Some of the basic set operations are summarized below in terms of events:
• The union of two events is the event that consists of all outcomes that are contained in
either of the two events. We denote the union as . A U B
• The intersection of two events is the event that consists of all outcomes that are contained
in both of the two events. We denote the intersection as A  B.
• The complement of an event in a sample space is the set of outcomes in the sample space
that are not in the event. We denote the component of the event E as E’.
The event consists of the 40 samples for which scratch and shock resistances are high.
The event consists of the 9 samples in which the shock resistance is low.
The event consists of the 45 samples in which the shock resistance, scratch resistance,
or both are high.
Summarizing Probability
Ways to Assign Probabilities
There are three ways to assign probabilities:
1. CLASSICAL PROBABILITY
Based on the assumption that the outcomes of an experiment are equally
likely.
2. EMPIRICAL PROBABILITY or RELATIVE FREQUENCY METHOD
The probability of an event happening is the fraction of the time similar
events happened in the past.
3. SUBJECTIVE PROBABILITY
The likelihood (probability) of a particular event happening that is
assigned by an individual based on whatever information is available.
Classical Method
Relative Frequency Method

On February 1, 2003, the Space

Shuttle Columbia exploded. This
was the second disaster in 123
space missions for NASA. On the
basis of this information, what is
the probability that a future
mission is successfully completed?
 Subjective Probability - Example

• If there is little or no data or information to calculate a

probability, it may be arrived at subjectively.

• 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 a person 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.
Mutually Exclusive
The sample space consists of the 1,000 respondents.
Simple events are “planned to purchase,” “did not plan to purchase,” “purchased,” and “did not purchase.” The complement
of the event “planned to purchase” is “did not plan to purchase.” The event “planned to
purchase and actually purchased” is a joint event because in this joint event, the respondent must plan to purchase the
television and actually purchase it.
Contingency Tables
• There are several ways in which you can view a particular sample space.
• The method used a contingency table such as the one displayed.
• You get the values in the cells of the table by subdividing the sample space of
1,000 households according to whether someone planned to purchase and actually
purchased a large TV.
• For example, 200 of the respondents planned to purchase a large TV and
subsequently did purchase the large TV.
Interpretations of Probability
• We introduce probability for discrete sample spaces.
• Probability is used to quantify the likelihood, or chance, that an outcome of a
random experiment will occur.
• “The chance of rain today is 30%’’ is a statement that quantifies our feeling about
the possibility of rain.
• The likelihood of an outcome is quantified by assigning a number from the
interval [0, 1] to the outcome (or a percentage from 0 to 100%).
• Higher numbers indicate that the outcome is more likely than lower numbers.
• A 0 indicates an outcome will not occur.
• A probability of 1 indicates an outcome will occur with certainty.
 Example
• For example, suppose that we will select one laser diode randomly from a batch of 100.
• The sample space is the set of 100 diodes. Randomly implies that it is reasonable to assume
that each diode in the batch has an equal chance of being selected.
• Because the sum of the probabilities must equal 1, the probability model for this experiment
assigns probability of 0.01 to each of the 100 outcomes.
• We can interpret the probability by imagining many replications of the experiment.
• When the model of equally likely outcomes is assumed, the probabilities are chosen to be
equal.

Each time we start with all 100 diodes and select one at random.
The probability 0.01 assigned to a particular diode represents the proportion of replicates
in which a particular diode is selected.
 Definition
• Whenever a sample space consists of N possible outcomes that are
equally likely, the probability of each outcome is 1N.

For a discrete sample space, the probability of an event E, denoted as P(E), equals the sum of
the probabilities of the outcomes in E.
Axioms of Probability
 Evaluate
• The Finance Minister Mr. Singh is preparing to make his annual budget to the
parliament and is speculating about his chances of getting all or part of his
requested budget approved. From his 20 years of experience in making these
requests, he has deduced that his chances of getting between 50 and 74 percent
of his budget approved are twice as good as those of getting between 75 and 99
percent approved, and two and one-half times as good as those of getting
between 25 and 49 percent approved. Further, the general believes that there is
no chance of less than 25 percent of his budget being approved. Finally, the
entire budget has been approved only once during the general’s tenure, and the
general does not expect this pattern to change. What are the probabilities of 0–
24 percent, 25–49 percent, 50–74 percent, 75–99 percent, and 100 percent
approval, according to the general?
Sol
• Let the probabilities be:
• P(25−49%)=Y
• P(50−74%)=2.5Y
• P(75−99%)=(1/2)×2.5Y=1.25xY
• P(100%)=q = 1/20
PROBABILITIES UNDER CONDITIONS OF STATISTICAL
INDEPENDENCE
▪ When two events happen, the outcome of the first event may or may not have an
effect on the outcome of the second event.
▪ That is, the events may be either dependent or independent. In this section, we
examine events that are statistically independent:
▪ The occurrence of one event has no effect on the probability of the occurrence of
any other event.
▪ There are three types of probabilities under statistical independence:
1. Marginal
2. Joint
3. Conditional
CONDITIONAL PROBABILITY

• Example:
• Scenario: Suppose you have a standard deck of 52 playing cards. You
draw one card from the deck.
• Event A: The card drawn is a King.
• Event B: The card drawn is a face card (King, Queen, or Jack).
• Now, you want to find the conditional probability that the card drawn
is a King given that it is a face card.
Example: Conditional Probability in Marketing

• Scenario: A company is analyzing the purchasing behavior of its

customers to better target its marketing efforts. The company knows
the following:
• Event A: A customer buys a product after receiving a promotional
email.
• Event B: A customer opens the promotional email.
• From past data:
• The probability that a customer buys a product after opening the
promotional email (i.e., P(A∣B)P(A|B)P(A∣B)) is what we want to
determine.
 P (A and B) = joint probability of A and B
P(A) = marginal probability of A
P(B) = marginal probability of B

To select randomly implies that at each step of the sample, the items that remain
in the batch are equally likely to be selected.
A day’s production of 850 manufactured parts contains 50 parts that do not meet
customer requirements. Two parts are selected randomly without replacement
from the batch. What is the probability that the second part is defective given that
the first part is defective?

• Let A denote the event that the first part selected is defective, and let B denote the
event that the second part selected is defective. The probability needed can be
expressed as P(B|A).
• If the first part is defective, prior to selecting the second part, the batch contains
849 parts, of which 49 are defective,

Therefore, P(B|A) = 49/849

Determining Independence
Sol.
Two events are independent if any one of the following equivalent
statements is true:
Example
The south east regional manager of General Express, a private parcel
delivery firm, is worried about the likelihood of strikes by some of his
employees. He has learned that the probability of a strike by his pilots is
0.75 and the probability of a strike by his drivers is 0.65. Further, he
knows that if the drivers strike, there is a 90 percent chance that the
pilots will strike in sympathy.
• (a) What is the probability of both groups’ striking?
• (b) If the pilots strike, what is the probability that the drivers will
strike in sympathy?
 Sol.
• Given:
• Probability of pilots striking, P(P)=0.75P(P) = 0.75P(P)=0.75
• Probability of drivers striking, P(D)=0.65P(D) = 0.65P(D)=0.65
• Probability of pilots striking given that drivers strike, P(P∣D)=0.90
• (a) Probability of both groups striking:
• We need to find P(P∩D) = ? = 0.585
• (b) Probability of drivers striking given that the pilots strike:
• We need to find P(D∣P) = ? = 0.78.
 Practice

The probability of the complement of A is:

P(B∩A’)=P(B)×P(A’) P(A’)=1−P(A)=1−0.3=0.7

Now, to find P(B∩A’): Check if P(B∩A’) = P(B)×P(A’)

P(B∩A’)= P(B) − P(A∩B) =0.8−0.24=0.56 Calculate the right-hand side:
P(B)×P(A’)=0.8×0.7=0.56
Conclusion:
Since P(B∩A’)=0.56 and P(B)×P(A’)=0.56
the events B and A’ are independent.
Example
 Sol.
The total no. of boxes inspected is 10,000.
• Out of these, the number of boxes having damaged fruit from Malihabad is 200, and from Hyderabad, it's 365.
• Similarly, the number of boxes with overripe fruit from Malihabad is 840, and from Hyderabad, it's 295.
• Total boxes with damaged or overripe fruit = (number of boxes with damaged fruit) + (number of boxes with overripe
fruit) = (200 + 840) + (365 + 295) = 1,400 + 660 = 2,060
• So, the probability of selecting a box with overripe or damaged fruit will be:
• Satisfied Cases/Total = 2060/10,000 = 0.206 Or in terms of Percentage, it's 20.6%.
• Thus, the probability of getting a box with overripe or damaged fruit will be 0.206.
• Now, let's calculate the probability that a randomly selected box containing overripe fruit came from Hyderabad.
• Number of boxes with overripe fruit from Hyderabad = 295
• The probability of selecting a box from Hyderabad, given that it contains overripe fruit, can be calculated using
conditional probability:
• P(Hyderabad | Overripe) = (Number of boxes with overripe fruit from Hyderabad) / (Total number of boxes with overripe
fruit)
• = 295 / (840 + 295) = 295 / 1,135 ≈ 0.259
• Therefore, the probability that a randomly selected box containing overripe fruit came from Hyderabad is approximately
0.259 or 25.9%.
Bayes’ Theorem
• Bayes’ theorem is a method for revising a probability given additional
information.
• Often we begin probability analysis with initial or prior probabilities.
• Given this information, we calculate revised or posterior probabilities.
• Bayes’ theorem provides the means for revising the prior
probabilities.
Continued…
To find the posterior probability that event Ai will occur given that event B
has occurred, we apply Bayes’ theorem.

Bayes’ theorem is applicable when the events for which we want to compute
posterior probabilities are mutually exclusive and their union is the entire
sample space.
Example: Quality of Purchased Parts

We can apply Bayes’ theorem to a manufacturing firm that receives

shipments of parts from two different suppliers. Let A1 denote the
event that a part is from supplier 1 and A2 denote the event that a
part is from supplier 2.

Currently, 65% of the parts purchased by the company are from

supplier 1, and the remaining 35% are from supplier 2. Thus, if a part
is selected at random, we would assign the prior probabilities P(A1) =
0.65 and P(A2) = 0.35.
Continued…
New Information
Now suppose that the parts from the two suppliers are used in the
firm’s manufacturing process and that a bad part causes a machine to
break down. What is the probability that the bad part came from
supplier 1 and what is the probability that it came from supplier 2?
With the information in the probability tree, we can use Bayes’
theorem to answer these questions.
Posterior Probability
• Given that the part received was bad, we revise the prior
probabilities as follows:
Bayes’ Theorem - Example
Example
In a bolt factory, machine A, B and C manufacture, respectively, 25%,
35%, and 40% of the total output. Of their output, respectively, 5%, 4%,
and 2% are known to be defective. A bolt is picked up at random and is
found to be defective. Find the probability that
(1) It is produced by machine B
(2) It is produced by machine B or C
Let us define the following events:
A1 : a randomly selected bolt is produced by machine A.
A2 : a randomly selected bolt is produced by machine B.
A3 : a randomly selected bolt is produced by machine C.
E : a randomly selected bolt is defective.
P(A1) = 0.25, P(A2) = 0.35, P(A3) = 0.40, P(E/A1) = 0.05, P(E/A2) =
0.04, P(E/A3) = 0.02

(i) Given that the bolt is found to be defective i.e

P(A2/E) = ?
By using Bayes’ Theorem:-
P(A2/E) = (0.35*0.04)/(0.25*0.05 + 0.35*0.04 +
0.40*0.02) = 0.4058
Continued…
(ii) Required probability = 1-P(A1/E)

P(A1/E) = (0.25*0.05)/(0.0345) = 0.3623

Required probability = 1- 0.3623 = 0.6377

Example
• A company has two plants to manufacture scooters. Plant-1
manufacture 70% of the scooter and Plant-2 manufactures 30%. At
plant-1, 80% of scooters produced are of standard quality and at
Plant-2 90% of scooters produced are of standard quality. A scooter is
picked up at random and is found to be of standard quality. What is
the chance that it has come from Plant-2?
Sol
Let us define the following events:-
A1: Scooter manufactured by Plant-1
A2: Scooter manufactured by Plant-2
E: Scooter is of standard quality
Clearly the events A1 and A2 are mutually exclusive and from data
given in the problem, we have
P(A1) = 0.70
P(A2) = 0.30
P(E/ A1) = 0.80 P(E/ A2) = 0.90
 Given that the scooter chosen is of standard quality, we
want to find the probability that it came from Plant-2

• Using Bayes’ formula, we get

𝑃 𝐴2 𝑃(𝐸 Τ𝐴2)
𝑃(𝐴2Τ𝐸) =
𝑃 𝐴1 𝑃(𝐸 Τ𝐴1)+𝑃 𝐴2 𝑃(𝐸 Τ𝐴2)
(0.30)(0.90)
= 0.70 0.80 +(0.30)(0.90)
= 0.325
Example: Marketing Campaign Effectiveness
• Problem: A company runs an email marketing campaign to promote a
new product. The company knows from past experience that 20% of
their campaigns are usually successful (meaning they lead to a
significant increase in sales). To assess the effectiveness of the current
campaign, the company looks at customer response data:
• 70% of successful campaigns (from past data) result in a high
customer response rate.
• 30% of unsuccessful campaigns result in a high customer response
rate.
• After running the campaign, the company observes a high customer
response rate. What is the probability that this particular campaign was
successful?
Solution Using Bayes' Theorem:

• Let’s define the events:

• SSS: The campaign is successful.
• FFF: The campaign is not successful.
• HHH: High customer response rate observed.
• We are interested in finding the posterior probability P(S∣H), the
probability that the campaign is successful given a high customer
response rate.
Continued…
• We know:
• P(S)=0.20P(S) = 0.20P(S)=0.20 (Prior probability of the campaign
being successful)
• P(F)=0.80P(F) = 0.80P(F)=0.80 (Prior probability of the campaign
being unsuccessful)
• P(H∣S)=0.70P(H | S) = 0.70P(H∣S)=0.70 (Probability of a high response
rate given the campaign is successful)
• P(H∣F)=0.30P(H | F) = 0.30P(H∣F)=0.30 (Probability of a high response
rate given the campaign is not successful)