Course Title: Business StatisticsBBA (Hons)

2nd Semester

Course Instructor: Atiq ur Rehman ShahLecturer, Federal Urdu University of Arts,

Science & Technology, Islamabad

Structure of Probability

• Experiment• Event• Sample Space• Unions and Intersections• Mutually Exclusive Events

Experiment• Experiment: A planned activity or process

whose results yield a set of data.

• Random Experiment: An experiment which produces different results even though it is repeated under similar conditions.

– Tossing of a fair coin– Throwing of a balance die etc…

• Trial: A single performance of an experiment is called trial.

Event

• Event: An event is an individual outcome or any number of outcomes of a random experiment(under stated conditions).

• An event that contains exactly one sample point is called Simple event. e.g., occurrence of 6, when a die is thrown.

• An event which contains more than one sample points and is produced by union of sample points is called Compound event. e.g. occurrence of a sum of 10 when a die is thrown.

Sample Space

• A set consisting of all possible outcomes that can result from a random experiment.

• It is usually denoted by S. • The experiment of tossing a coin can result

in either a head (H) or a tail (T), so the sample space S= {H,T}

• The sample space for tossing a coin twice can be:

S= {HH, HT, TH, TT}

Sample Space: Tree Diagram for Random Sample

A B A*B1 (H, 1)2 (H, 2)

H 3 (H, 3)4 (H, 4) 5 (H, 5)6 (H, 6)

1 (T, 1)2 (T, 2)3 (T, 3)

T 4 (T, 4)5 (T, 5)6 (T, 6)

A= {H, T} , B= {1,2,3,4,5,6}

Union of Sets• The union of two sets contains an instance

of each element of the two sets.

X

Y

X Y

1 4 7 9

2 3 4 5 6

1 2 3 4 5 6 7 9

, , ,

, , , ,

, , , , , , ,

C IBM DEC Apple

F Apple Grape Lime

C F IBM DEC Apple Grape Lime

, ,

, ,

, , , ,

YX

Intersection of Sets• The intersection of two sets contains only

those element common to the two sets.

X

Y

X Y

1 4 7 9

2 3 4 5 6

4

, , ,

, , , ,

C IBM DEC Apple

F Apple Grape Lime

C F Apple

, ,

, ,

YX

Mutually Exclusive Events• Events with no

common outcomes• Occurrence of one

event precludes the occurrence of the other event

X

Y

X Y

1 7 9

2 3 4 5 6

, ,

, , , ,

C IBM DEC Apple

F Grape Lime

C F

, ,

,

YX

P X Y( ) 0

Methods of Assigning Probabilities

• Classical method of assigning probability (rules and laws)

• Relative frequency of occurrence (cumulated historical data)

• Subjective Probability (personal intuition or reasoning)

Classical Probability

• Number of outcomes leading to the event divided by the total number of outcomes possible

• Each outcome is equally likely• Determined a priori -- before

performing the experiment• Applicable to games of chance• Objective -- everyone correctly

using the method assigns an identical probability

P EN

Where

N

en( )

:

total number of outcomes

number of outcomes in Een

Relative Frequency Probability

• Based on historical data• Computed after

performing the experiment

• Number of times an event occurred divided by the number of trials

• Objective -- everyone correctly using the method assigns an identical probability

P EN

Where

N

en( )

:

total number of trials

number of outcomes

producing Een

Subjective Probability• Comes from a person’s intuition or

reasoning• Subjective -- different individuals may

(correctly) assign different numeric probabilities to the same event

• Degree of belief• Useful for unique (single-trial) experiments

– New product introduction– Site selection decisions

Counting sample points

• Rule of Multiplication• Rule of Permutation• Rule of Combination

Rule of Multiplication

• If a compound experiment consist of two experiments such that first experiment has exactly m distinct outcomes, and if corresponding to each outcome of the first experiment there can be n distinct outcomes of the second experiment, then the compound experiment has exactly mn outcomes.

Rule of Multiplication

• For example, tossing a fair coin and throwing a die together consists of two experiments.

• A= {H, T}; B= {1, 2, 3, 4, 5, 6}

• Total number of possible distinct outcomes of the compound experiment is therefore 2*6= 12

Rule of Permutation

• A Permutation is any ordered subset from a set of n distinct objects.

• The number of permutation of r objects selected in a definite order from n distinct objects is denoted by .

Rule of Permutation

• A club consist of four members. How many sample points are in the sample space when three officers: president, secretary and treasurer, are to be chosen?

n= 4; r= 3, 4P3= 4!/(4-3)! = 24

Rule of Permutation

Rule of Combination

• A combination is any subset of r objects, selected without regard to their order, from a set of n distinct objects.

• The total number of such combination is denoted by the symbol .

Rule of Combination

• A three members committee is to be formed from a list of four persons. How many sample points are associated with the experiment?

• Since the order is unimportant here, so it is a problem involving combination,Here: n=4, r= 3So 4C3= 4!/3! (4-3)!

= 4So the sample space associated with the experiment contains 4 sample points.

PROBABILITY!

What is probability?

• Probability is the measure of how likely an event or outcome is.

• Different events have different probabilities!

How do we describe probability?

• You can describe the probability of an event with the following terms:– certain (the event is definitely going to happen)– likely (the event will probably happen, but not

definitely)– unlikely (the event will probably not happen, but it

might)– impossible (the event is definitely not going to happen)

How do we express probabilities?

• Usually, we express probabilities as fractions.– The numerator is the number of ways the event

can occur.– The denominator is the number of possible

events that could occur.• Let’s look at an example!

What is the probability the spinner will land on the number 3?

1

43

2

Ask yourself the following questions:1. How many 3’s are on the spinner?

2. How many possible numbers could the spinner land on?

1

43

2 14

What is the probability the die will land on an even number?

Remember, a die has six sides. Numbers 1, 2, 3, 4, 5, and 6 are each depicted once on the die.

Ask yourself the following questions:1. How many even numbers are on the die?

2. How many possible numbers could the die land on?

36

What is the probability that I will choose a red marble?

• In this bag of marbles, there are:– 3 red marbles– 2 white marbles– 1 purple marble– 4 green marbles

Ask yourself the following questions:1. How many red marbles are in the bag?

2. How many marbles are in the bag IN ALL?

3

10

Example-1

• If a card is drawn from an ordinary deck of 52 playing cards, find the probability that:i) The card is a red cardii) The card is a diamond cardiii) The card is a 10

Solution:i) Let A represent the event that the card drawn is a red card, So the

number of outcomes favourable the event A is 26 since there are 26 red cards.

Hence, P(A)= m/n = 26/52 =1/2

ii) Let B represent that the card drawn is a diamond, so the number of favourable outcomes to the event B is 13 (Since there are 13 diamonds in an ordinary deck)

So, P(B)= m/n = 13/52 = 1/4

iii) ** Class activity **

Example-2

• A fair coin is tossed three times. What is the probability that at least one head will appear??

Solution: The sample space for this experiment is: S= {HHH, HHT, HTH, THH, HTT, THT, TTH, TTT}So, n(S)= 8Let A represent the event that at least one head appears, thenA= {HHH, HHT, HTH, THH, HTT, THT, TTH}Hence n(A)= 7So P(A)= m/n

= 7/8

Example-3

• A fair coin is tossed three times. What is the probability that more than one tail will appear??

Solution: ** Class activity **

Example-4

• A fair die is thrown twice, what is the probability of gettingi) A double fiveii) A sum of 8 or more dots

Solution: S= { (1,1), (1,2), (1,3), (1,4), (1,5), (1,6) (2,1), (2,2), (2,3), (2,4), (2,5), (2,6)

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

So n(S)= 36i) Let a represent the event that a double 5 occurs. then A= {(5,5)}So, n(A)= 1

Hence, P(A)= m/n= 1/36

ii) ** Class activity**