Statistics lecture 5 (ch4)

What is probability? Probability is a numerical value used to

express the chance that a specific event will occur.

Probability is always in the interval 0 to 1 and can be expressed as percentages.

The greater the chance that an event will occur, the closer the probability is to 1.

The smaller the chance that an event will occur, the closer the probability is to 0.

Probabilities is needed to make generalisations about the population based on a sample drawn from the population. 2

Experiment Process that results in obtaining observations

from an experimental unit. Experimental unit is the object on which the

observations are made. Results of experiment are called outcomes. Stochastic experiment

The results are a definite set of two or more possible outcomes.

The outcome can not be determined in advance. Can be repeated under stable conditions.

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Sample space Collection of all possible outcomes of an

experiment. It is denoted by S. List all possible outcomes inside braces.

S = { }

S

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Event Collection of some outcomes of the sample

space. It is denoted by A, B, C, etc. List all possible outcomes inside braces.

A = { } Can have one or more

outcomes. S

A

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Event Can define more than one event for the same

sample space. Two events are mutually exclusive if they can

not occur at the same time. Events A and B are mutually exclusive.

S

AB

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Event Can define more than one event for the same

sample space. Two events are non mutually exclusive if they

can occur at the same time. Events A and C are non mutually exclusive.

Probability of an event Probability of Event A. P(A)

S

A

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Properties of probability: 0 ≤ P(A) ≤ 1 P(B) = 0

B impossible event P(C) = 1

C certain event P(S) = 1 Compliment of Event A:

P(Ā) = 1 – P(A) Events A and B mutually exclusive.

P(A or B) = P(A) + P(B)8

Three approaches to probability 1. Relative frequency approach

The probabilities of the outcomes differ. Counting the number of times that an event

occurs when performing an experiment a large number of times.

number of times event A occurredP(event A)

number of times the experiment was repeated

( )f

P An

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Three approaches to probability Relative frequency approach

Example In a group of 20 tourists staying in a hotel,

nine prefer to pay cash for their accommodation.

The probability that a tourist will pay cash for accommodation is:

9( ) 0,45

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fP A

n

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Three approaches to probability 2. Classic approach

The outcomes all have the same probabilities. Not necessary to performing an experiment a

large number of times.

number of outcomes of experiment favourable to the eventP(eventA)

total number of outcomes of experiment

( )f

P An

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Three approaches to probability Classic approach

Example Chance to get an uneven number on a dice:

S = {1; 2; 3; 4; 5; 6} F = {1; 3; 5}

1 1 1 3( ) 0,5

6 6 6 6P F

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Three approaches to probability 3. Subjective approach

The probabilities assigned to the outcomes of the experiment is subjective to the person who performs the experiment.

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The word “or” in probability is an indication of addition

P(A or B)

The word “and” in probability is an indication of multiplication P(A and B)

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Addition rules for calculating probabilities Events are mutually exclusive when they

have no outcomes in common. For mutually exclusive events:

P(A or B) = P(A) + P(B) = 4/21 + 3/21 = 7/21

SA B

Events A and B are mutually

exclusive 15

Addition rules for calculating probabilities Events are mutually exclusive when they

have no outcomes in common If events are not mutually exclusive

P(C or D) = P(C) + P(D) – P(C and D) P(C and D) Outcomes are included in C and D.

Events C and D are not mutually

exclusive S

= 5/21 + 5/21 – 2/21 = 8/21

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Conditional probability Need to know the probability of an event

given another event has already occurred The two events must be dependent. Probability of A, given B has occurred:

P(A|B) = , P(B) ≠ 0

Multiplication rule: P(A and B) = P (B)P(A|B)

Probability of B, given A has occurred: P(B|A) = , P(A) ≠ 0

( )

( )

P A and B

P B

( )

( )

P A and B

P A

The outcome of one event affects the probability of the occurrence of another event.

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Conditional probability Need to know the probability of an event

given another event has already occurred. If sampling without replacement takes place.

The events are considered dependent. Example – Three students need to pick a

biscuit from a plate with 10 biscuits. 1st student can pick any of the 10 biscuits. 2nd student has only 9 biscuits to pick from. 3rd student has only 8 biscuits to pick from. The probability to pick the 1st biscuit is 1/10, the 2nd

is 1/9 and the 3rd is 1/8. 18

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Independent Events Events are statistically independent if the

outcome of one event does not affect the probability of occurrence of another event. P(A|B) = P(A) P(B|A) = P(B)

Multiplication rule for dependent events: P(A and B) = P(B)P(A|B)

Multiplication rule for independent events: P(A and B) = P(A)P(B)

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Counting rules Multi-step experiments

The number of outcomes for ‘k’ trails each with the same ‘n’ possible outcomes. The number of outcomes in S = nk.

Example How many ways can 10 multiple choice question

with 4 possible answers be answered: 410 = 1 048 576 ways

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Counting rules Multi-step experiments

The number of outcomes for ‘j’ trails each with a different number of ‘n’ outcomes. The number of outcomes in S = n1 x n2 … X nj

Example Need to order a meal where you can pick ‘1’

burger from ‘8’, ’1’ cool drink from ’10’, ‘1’ ice cream from ‘5’. Number of possible orders: 8×10×5 = 400

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Counting rules The factorial

The number of ways in which ‘r’ objects can be arranged in a row, without replacement. r! = r×(r – 1)×(r – 2)× …×3×2×1

Note r! = 0! = 1 Example

Six athletes compete in a race. The number of order arrangements for completing the race. 6! = 720 different ways

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Counting rules Combination

Select r objects without replacement from a larger set of n objects, order of selection not important.

Example Six lotto numbers should be selected form a

possible 49 – order of selection not important.

!

!( )!n r

nC

r n r

49 6

49!13 983 816

6!(49 6)!n rC C

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Counting rules Permutation

Select r objects without replacement from a larger set of n objects, order of selection is important.

Example Six athletes in a race, how many ways to

compete for the gold, silver and bronze medals.

!

( )!n r

nP

n r

6 3

6!120

(6 3)!n rP P

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Additional examples

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Sample space:

S={ ♠ 2 3 4 5 6 7 8 9 10 J Q K A ♣ 2 3 4 5 6 7 8 9 10 J Q K A ♥ 2 3 4 5 6 7 8 9 10 J Q K A ♦ 2 3 4 5 6 7 8 9 10 J Q K A }

52 Cards in the pack

Experiment: Draw a card from a play pack of cards

Experimental unit: Play pack of cards

Outcome of experiment: Any card from the pack

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S={ ♠ 2 3 4 5 6 7 8 9 10 J Q K A ♣ 2 3 4 5 6 7 8 9 10 J Q K A ♥ 2 3 4 5 6 7 8 9 10 J Q K A ♦ 2 3 4 5 6 7 8 9 10 J Q K A }

Event A – get a ‘4’ if you pick one card from the pack of cards

A = {♠4 ♣4 ♥4 ♦4}

P(A) = 4/52

A

Event - some outcomes of the sample space

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S={ ♠ 2 3 4 5 6 7 8 9 10 J Q K A ♣ 2 3 4 5 6 7 8 9 10 J Q K A ♥ 2 3 4 5 6 7 8 9 10 J Q K A ♦ 2 3 4 5 6 7 8 9 10 J Q K A }

Event B – get a picture card if you pick one card from the pack of cards

B = {♠JQKA ♣ JQKA ♥ JQKA ♦ JQKA}

P(B) = 16/52

B

Event - some outcomes of the sample space

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S={ ♠ 2 3 4 5 6 7 8 9 10 J Q K A ♣ 2 3 4 5 6 7 8 9 10 J Q K A ♥ 2 3 4 5 6 7 8 9 10 J Q K A ♦ 2 3 4 5 6 7 8 9 10 J Q K A }

Event B – get a picture card if you pick one card from the pack of cards

What is the probability not to get a picture card?

P( ) = 1 – P(B) = 1 – 16/52 = 36/52B

B

Complement of an event

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P(A or B) = P(A) + P(B) = 4/52 + 16/52 = 20/52

S={ ♠ 2 3 4 5 6 7 8 9 10 J Q K A ♣ 2 3 4 5 6 7 8 9 10 J Q K A ♥ 2 3 4 5 6 7 8 9 10 J Q K A ♦ 2 3 4 5 6 7 8 9 10 J Q K A }

Are events A and B mutually exclusive?

BA

Additional rule

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S={ ♠ 2 3 4 5 6 7 8 9 10 J Q K A ♣ 2 3 4 5 6 7 8 9 10 J Q K A ♥ 2 3 4 5 6 7 8 9 10 J Q K A ♦ 2 3 4 5 6 7 8 9 10 J Q K A }

Event C – get a red card if you pick one card from the pack of cards

C = { ♥ 2 3 4 5 6 7 8 9 10 J Q K A ♦ 2 3 4 5 6 7 8 9 10 J Q K A }

P(C) = 26/52

C

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S={ ♠ 2 3 4 5 6 7 8 9 10 J Q K A ♣ 2 3 4 5 6 7 8 9 10 J Q K A ♥ 2 3 4 5 6 7 8 9 10 J Q K A ♦ 2 3 4 5 6 7 8 9 10 J Q K A }

Are events B and C mutually exclusive?

B

C

Additional rule

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P(B or C) = P(B) + P(C) – P(B and C)

= 16/52 + 26/52 – 8/52 = 34/52

S={ ♠ 2 3 4 5 6 7 8 9 10 J Q K A ♣ 2 3 4 5 6 7 8 9 10 J Q K A ♥ 2 3 4 5 6 7 8 9 10 J Q K A ♦ 2 3 4 5 6 7 8 9 10 J Q K A }

B

C

Are events B and C mutually exclusive?

Additional rule

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S={ ♠ 2 3 4 5 6 7 8 9 10 J Q K A ♣ 2 3 4 5 6 7 8 9 10 J Q K A ♥ 2 3 4 5 6 7 8 9 10 J Q K A ♦ 2 3 4 5 6 7 8 9 10 J Q K A }

What is the probability to get a red card if the card must be a picture card?

8( ) 852( | )16( ) 16

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P C and BP C B

P B

B

C

Conditional rule

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S={ ♠ 2 3 4 5 6 7 8 9 10 J Q K A ♣ 2 3 4 5 6 7 8 9 10 J Q K A ♥ 2 3 4 5 6 7 8 9 10 J Q K A ♦ 2 3 4 5 6 7 8 9 10 J Q K A }

What is the probability to get a ‘4’ if the card must be a red card?

2( ) 252( | )26( ) 26

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P A and CP A C

P C

A

C

Conditional ruleConditional rule

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226 2( ) ( ) ( | ) 52 26 52P A and C P C P A C

S={ ♠ 2 3 4 5 6 7 8 9 10 J Q K A ♣ 2 3 4 5 6 7 8 9 10 J Q K A ♥ 2 3 4 5 6 7 8 9 10 J Q K A ♦ 2 3 4 5 6 7 8 9 10 J Q K A }

We know: P(C) = 26/52 and P(A|C) = 2/26

Sampling with replacement takes place

If we pick two cards, what is the probability that the first one is red and the second one is ‘4’

A

C

Conditional rule

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S={ ♠ 2 3 4 5 6 7 8 9 10 J Q K A ♣ 2 3 4 5 6 7 8 9 10 J Q K A ♥ 2 3 4 5 6 7 8 9 10 J Q K A ♦ 2 3 4 5 6 7 8 9 10 J Q K A }

P(B) = 16/52 and P(A|B) = 4/51 6416 4( ) ( ) ( | ) 52 51 2652

P B and A P B P A B

Sampling without replacement takes place If we pick two cards, what is the probability that the first one is a picture card and the second one is ‘4’

A B

Conditional rule

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A soccer team is playing two matches on a specific day.•The chance of winning the first match is:

•P(1st) = ½•The chance of winning the second match is:

•P(2nd) = ½•Will the outcome of the first match influence the outcome of the second match?•The chance of winning both matches is:

•P(1st and 2nd) = P(1st) x P(2nd |1st) •P(1st and 2nd) = P(1st) x P(2nd) = ½ X ½ = ¼

Independent events

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In chapter 2 we looked at an example to organise qualitative data into a frequency distribution table.

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Example:The data below shows the gender of 50 employees and the department in which they work at ABC Ltd.

Organising and graphing qualitative data in a frequency distribution table.

Emp. no. Gender Dept. Emp. no. Gender Dept …..

1 M HR 6 M Fin. …..

2 F Mark. 7 M Mark. …..

3 M Fin. 8 M Fin. …..

4 F HR 9 F HR …..

5 F Fin. 10 F Fin. …..

M – Male F – Female

HR – Human resourcesMark. – MarketingFin. – Finance

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HR Marketing Finance Total

M 4 10 5 19

F 10 16 5 31

Total 14 26 10 50

Organising and graphing qualitative data in a frequency distribution table.

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HR Marketing Finance Total

M 4 10 5 19

F 10 16 5 31

Total 14 26 10 50

If one employee is chosen from the 50 employees, what is the probability that the employee will be male?

P(M) = 19/50

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HR Marketing Finance Total

M 4 10 5 19

F 10 16 5 31

Total 14 26 10 50

If one employee is chosen from the 50 employees, what is the probability that the employee will be from the Marketing department?

P(Mark) = 26/50

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HR Marketing Finance Total

M 4 10 5 19

F 10 16 5 31

Total 14 26 10 50

If one employee is chosen from the 50 employees, what is the probability that the employee will be from the Marketing or Finance departments?

P(Mark or Fin) = 26/50 + 10/50 = 36/50

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Are the marketing and finance departments

mutually exclusive?

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HR Marketing Finance Total

M 4 10 5 19

F 10 16 5 31

Total 14 26 10 50

If one employee is chosen from the 50 employees, what is the probability that the employee will be Female and from the Finance department?

P(F and Fin) = 5/50 Are female and the finance department mutually exclusive?

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HR Marketing Finance Total

M 4 10 5 19

F 10 16 5 31

Total 14 26 10 50

If one employee is chosen from the 50 employees, what is the probability that the employee will be Female or from the Finance department?

P(F or Fin) = 31/50 + 10/50 – 5/50 = 36/50Are female and the finance department mutually exclusive?

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HR Marketing Finance Total

M 4 10 5 19

F 10 16 5 31

Total 14 26 10 50

If one employee is chosen from the 50 employees, what is the probability that the employee will be from the Marketing and Finance departments?

P(F and Fin) = 0

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Are the marketing and finance departments

mutually exclusive?

P( ) = 1 - P(HR) = 1 - 14/50 = 36/50

HR Marketing Finance Total

M 4 10 5 19

F 10 16 5 31

Total 14 26 10 50

If one employee is chosen from the 50 employees, what is the probability that the employee will not be from the HR department?

HR

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P(F|HR) = P(F and HR)/P(HR)

= 10/50 / 14/50 = 10/14

HR Marketing Finance Total

M 4 10 5 19

F 10 16 5 31

Total 14 26 10 50

If one employee is chosen from the 50 employees, what is the probability that the employee will female if she is from the HR department?

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