•Business Analytics

•Chapter 6

•Probability

•Chapter outline

6.1 Assigning probabilities to events6.2 Joint, marginal and conditional probability

6.3 Rules of probability

•Learning objectives

Lecture 5A:LO1 Explain the importance of probability theory to statistical inference

LO2 Define the terms random experiment, sample space, elementary event, event, union of events, intersection of events, complement of an event and mutually exclusive events

LO3 Explain the different approaches to assigning probabilities to events

Lecture 5B:

LO4 Define the terms joint, marginal and conditional probabilities and independent events

LO5 Calculate probabilities using the basic probability rules – complement rule, addition rule and multiplication rule of probabilities

•6.1 Assigning Probabilities to Events

Random experimentA random experiment is a process or course of action whose outcome is uncertain.

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ExamplesExperiment Outcomes

Flip a coin Heads and tails

Record statistics test marks Numbers between 0 and 100

Measure the time to assemble Numbers from 0 and above

a computer

•Sample Space

A list of all possible outcomes of a random experiment is called a sample space.Sample Space: S = {O1, O2, …, Ok}

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•Sample Space: S = {O1, O2, …, Ok}

•Approaches to Assigning Probabilities…

There are three ways to assign a probability, P(Oi), to an outcome, Oi, namely:
1.Classical approach: based on equally likely events.

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2.Relative frequency: assigning probabilities based on experimentation or historical data.

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3.Subjective approach: Assigning probabilities based on the assignor’s (subjective) judgment.

•Classical Approach…

If an experiment has n possible outcomes, this method would assign a probability of 1/n to each outcome. It is necessary to determine the number of possible outcomes.Experiment 1: Rolling a die

Outcomes {1, 2, 3, 4, 5, 6}

Probabilities: Each sample point has a 1/6 chance of occurring.

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Experiment 2: Rolling two dice and observing the total (X)Outcomes: {X} = {2, 3, …, 12}

Examples:

P(X=2) = 1/36

P(X=6) = 5/36

P(X=10) = 3/36

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Bits & Bytes Computer Shop tracks the number of desktop computer systems it sells over a month (30 days):For example,

10 days out of 30 days

2 desktops were sold.

From this we can construct

the probabilities of an event

(i.e. the number of desktops sold on a given day)…

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‘There is a 40% chance that 3 desktops will be sold on any given day.’
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‘In the subjective approach we define probability as the degree of belief that we hold in the occurrence of an event.’Example: Weather forecasting’s ‘P.O.P.’

‘Probability of Precipitation’ (or P.O.P.) is defined in different ways by different forecasters, but basically it’s a subjective probability based on past observations combined with current weather conditions.

POP 60% – based on current conditions, there is a 60% chance of rain (say).

•Interpreting Probability…

No matter which method is used to assign probabilities all will be interpreted in the relative frequency approach.i.e. the chance of occurrence

e.g.

•A 40% chance that 3 desktops will be sold on any given day.’

•a 60% chance of rain

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•Assigning Probabilities

Given a sample space S = {O1, O2, …, On}, the following characteristics for the probability P(Oi) of the simple event Oi must hold:
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Probability of an event: the probability P(A) of event A is the sum of the probabilities assigned to the simple events contained in A.
•Complement, union and intersection of events

We study methods to determine probabilities of events that result from combining other events in various ways.There are several types of combinations and relationships between events:

•Complement of an event

•Intersection of events

•Union of events

•Mutually exclusive events

•Dependent and independent events

•Conditional event

•Probability of Combinations of Events

If A and B are two events, thenP(A È B) = P(A occur or B occur or both)

P(A Ç B) = P(A and B both occur)

P(Ā) = P(A does not occur)

P(A|B) = P(A occurs given that B has occurred)

•Complement of an Event

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•Complement of an Event…

For example, the rectangle stores all the possible tosses of 2 coins S = {(H,H), (H,T), (T,H), (T,T)}Let A = observing at least one head

= {(H,H), (H,T), (T,H)}

Ac = {(T,T)}

P(A) = ¾ , P(Ac) = ¼

P(A) + P(Ac) = P(S) = 1

P(Ac) = 1 – P(A)

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The intersection of events A and B is the set of all sample points that are in both A Ç B.The intersection is denoted: A Ç B.

The joint probability of

A and B is the probability of

the intersection of A and B,

i.e. P(A Ç B).

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For example, consider all possible tosses of two dice.S = {(1,1),(1,2),…,(6,6)}.

Let A = tosses where first toss is 1

= {(1,1),(1,2),(1,3),(1,4),(1,5),(1,6)}

and B = tosses where the second toss is 5

= {(1,5),(2,5),(3,5),(4,5),(5,5),(6,5)}

The intersection, A Ç B = {(1,5)}.

The joint probability of A and B is

The probability of the intersection

of A and B, i.e. P(A Ç B) = 1/36

•Union of two events

The union of two events A and B, is the event containing all outcomes that are in A or B or both:Union of A and B is denoted: A È B

•Union of two events…

For example, consider all possible tosses of two dice.S = {(1,1),(1,2), …,(6,6)}.

Let A = tosses where first toss is 1

= {(1,1),(1,2),(1,3),(1,4),(1,5),(1,6)}

and B = tosses where the second toss is 5

= {(1,5),(2,5),(3,5),(4,5),(5,5),(6,5)}

Union of A and B, A È B is

A È B = {(1,1),(1,2),(1,3),(1,4),(1,5), (1,6),(2,5),(3,5),(4,5),(5,5),(6,5)} (6,5)}

•Mutually exclusive events…

When two events are mutually exclusive (that is the two events have no outcomes in common), their joint probability is 0, hence:
•Basic relationships of probability…

•Example 1

The number of spots turning up when a 6-sided die is tossed is observed.Consider the following events:

A: The number observed is at most 2 i.e. A={1,2}

B: The number observed is an even number. i.e. B={,2,4,6}

C: The number 4 turns up. i.e. C={4}

Answer the following questions.

•Example 1

a.Define the sample space for this random experiment and assign probabilities to the simple events.

b.Find P(A).

c.Find P(AC).

d.Are events A and C mutually exclusive?

e.Find P(A È C).

f.Find P(A Ç B).

g.Find P(A È B).

h.Find P(C|B).

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•Example 1 - Solution

a)Define the sample space for this random experiment and assign probabilities to the simple events.

SolutionS = {1, 2, 3, 4, 5, 6}

Each simple event is equally likely to occur, thus,

P(1) = P(2) = … = P(6) = 1/6.

•Example 1 – Solution…

b) Find P(A).Solution

•Example 1 – Solution…

c) Find P(Ac).Solution

•Example 1 – Solution…

•Example 1 – Solution…

•Example 1 – Solution…

•Example 1 – Solution…

•Example 1 – Solution…

•6.2 Joint, Marginal and Conditional Probability

We have studied methods to determine probabilities of events that result from combining other events in various ways.We have also learnt that there are several types of combinations and relationships between events:

•intersection of events

•union of events

•dependent and independent events

•complement events

•conditional events

•Intersection

The intersection of event A and B is the event that occurs when both A and B occur.The intersection of events A and B is denoted by (A Ç B) or (A and B) .

The joint probability of A and B is the probability of the intersection of A and B, and is denoted by P(A Ç B).

•Example 2

A potential investor examined the relationship between the performance of mutual funds and the university where the fund manager earned his/her MBA. The relevant frequencies are given below
•Example 2

Dividing the frequencies by grand total givesjoint probabilities.

•Example 2 - Solution

The joint probability of[mutual fund outperforms …] and [… from a top 20 …] = 0.11.

P(A1 Ç B1) = 0.11

•Example 2 - Solution

The joint probability of[mutual fund outperforms …] and [… not from a top 20 …] = 0.06

P(A2 Ç B1) = 0.06

•Marginal probabilities

Marginal probabilities are computed by adding across rows and down columns; that is they are calculated in the margins of the table.
•Marginal probabilities

These probabilities are computed by adding across rows and down columns.
•Marginal probabilities

•Conditional probability

Conditional probability is used to determine how two events are related; that is, we can determine the probability of one event given the occurrence of another related event.Conditional probabilities are written as P(A|B) and read as ‘the probability of A given B’ and is calculated as:

•Conditional probability…

Again, the probability of an event given that another event has occurred is called a conditional probability…
•Example 2

Find the conditional probability that a randomly selected fund is managed by a top 20 MBA program graduate, given that it did not outperform the market.Solution:

A1 = Fund manager graduated from a top-20 MBA program

A2 = Fund manager did not graduate from a top-20 MBA program

B1 = Fund outperforms the market

B2 = Fund does not outperform the market

We need to find P(A1|B2).

•Example 2 – Solution …

We need to find P(A1|B2).For event A1|B2, the new information that B2 is given reduces the relevant sample space for A1|B2 to 83% of event B2..

•Conditional probability…

Before the new information becomes available, we haveP(A1) = 0.40

After the new information on B2 becomes available, P(A1) changes to

P(A1| B2) = 0.3494

Since the occurrence of B2 has changed the probability of A1, the two events are related (ie, A1 and B2 are not independent) and are called ‘dependent events’.

•Independence

One of the objectives of calculating conditional probability is to determine whether two events are related.In particular, we would like to know whether they are independent, that is, if the probability of one event is not affected by the occurrence of the other event.

Two events A and B are said to be independent if

P(A|B) = P(A)

or

P(B|A) = P(B)

•Example 2

Is ‘Fund outperforms the market’ dependent on the event that the ‘Manager graduated from a top-20 MBA program’?Solution:

A1 = Fund manager graduated from a top-20 MBA program

A2 = Fund manager did not graduate from a top-20 MBA program

B1 = Fund outperforms the market

B2 = Fund does not outperform the market

In order to answer this question, it suffices to show that, for at least one i and j, P(Bi| Aj)¹ P(Bj)

If B2 and A2 are independent, then P(B2|A2)=P(B2)

•Example 2 - Solution

We have already seen the dependency between A1 and B2, as P(A1|B2) = 0.3494 ¹ P(A1) = 0.40Now, let us check the independence of A2 and B2.

P(B2) = 0.83

P(B2|A2)=P(B2 Ç A2)/P(A2) = 0.54/0.60 = 0.90

Therefore, P(B2|A2) ¹ P(B2)

Conclusion: A2 and B2 are dependent. That is, the probability of event B2

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•Example 3: Try this yourself

The personnel department of an insurance company has compiled data on promotion, classified by gender. Are promotion and gender dependent on each another?
•Example 3 - Solution

If gender has no effect on promotion, then R and M are independent and P(R|M) = P(R). If this equalityholds, there is no difference in the probability of promotion between male and female managers.

•Example 3 – Solution…

•Union

As discussed earlier, the union event of A and B is the event that occurs when either A or B or both occur.It is denoted (A È B) or (A or B).

•Example 3

Determine the probability that a randomly selected fund outperforms the market or the manager graduated from a top 20 MBA program.That is, we want to calculate P (A1 È B1).

•Example 3 - Solution

•6.3 Rules of Probability

For any two event A and BComplement rule:

Addition rule:

Multiplication Rule:

General:

When A and B are independent

•Example 4

A stock market analyst feels that
•The probability that a certain mutual fund will receive increased contributions from investors is 0.6.

•The probability of receiving increased contributions from investors becomes 0.9 if the stock market goes up.

•There is a probability of 0.5 that the stock market rises.

The events of interest are:A: The stock market rises

B: The company receives increased contribution

•Example 4…

Calculate the following probabilities
•The probability that both A and B will occur, P(A Ç B) [sharp increase in earnings].

•The probability that either A or B will occur, P(A È B)

[at least moderate increase in earnings].Solution

•P(A) = 0.5; P(B) = 0.6; P(B|A) = 0.9

•P(A Ç B) = P(B|A) P(A) = (0.9)(0.5) = 0.45

•P(A È B) = P(A) + P(B) – P(A and B)

= 0.5 + 0.6 – 0.45 =0.65
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