statistics lecture 5 (ch4)

49

Upload: jillmitchell8778

Post on 05-Dec-2014

792 views

Category:

Education


1 download

DESCRIPTION

Basic Probability

TRANSCRIPT

Page 1: Statistics lecture 5 (ch4)
Page 2: 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

Page 3: Statistics lecture 5 (ch4)

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.

3

Page 4: Statistics lecture 5 (ch4)

Sample space Collection of all possible outcomes of an

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

S = { }

S

4

Page 5: Statistics lecture 5 (ch4)

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

5

Page 6: Statistics lecture 5 (ch4)

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

6

Page 7: Statistics lecture 5 (ch4)

7

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

C7

Page 8: Statistics lecture 5 (ch4)

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

Page 9: Statistics lecture 5 (ch4)

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

9

Page 10: Statistics lecture 5 (ch4)

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

20

fP A

n

10

Page 11: Statistics lecture 5 (ch4)

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

11

Page 12: Statistics lecture 5 (ch4)

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

12

Page 13: Statistics lecture 5 (ch4)

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.

13

Page 14: Statistics lecture 5 (ch4)

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)

14

Page 15: Statistics lecture 5 (ch4)

15

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

Page 16: Statistics lecture 5 (ch4)

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

16

Page 17: Statistics lecture 5 (ch4)

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.

17

Page 18: Statistics lecture 5 (ch4)

18

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

Page 19: Statistics lecture 5 (ch4)

19

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)

19

Page 20: Statistics lecture 5 (ch4)

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

20

Page 21: Statistics lecture 5 (ch4)

21

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

21

Page 22: Statistics lecture 5 (ch4)

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

22

Page 23: Statistics lecture 5 (ch4)

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

23

Page 24: Statistics lecture 5 (ch4)

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

24

Page 25: Statistics lecture 5 (ch4)

Additional examples

25

Page 26: Statistics lecture 5 (ch4)

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

26

Page 27: Statistics lecture 5 (ch4)

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

27

Page 28: Statistics lecture 5 (ch4)

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

28

Page 29: Statistics lecture 5 (ch4)

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

29

Page 30: Statistics lecture 5 (ch4)

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

30

Page 31: Statistics lecture 5 (ch4)

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

31

Page 32: Statistics lecture 5 (ch4)

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

32

Page 33: Statistics lecture 5 (ch4)

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

33

Page 34: Statistics lecture 5 (ch4)

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

52

P C and BP C B

P B

B

C

Conditional rule

34

Page 35: Statistics lecture 5 (ch4)

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

52

P A and CP A C

P C

A

C

Conditional ruleConditional rule

35

Page 36: Statistics lecture 5 (ch4)

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

36

Page 37: Statistics lecture 5 (ch4)

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

37

Page 38: Statistics lecture 5 (ch4)

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

38

Page 39: Statistics lecture 5 (ch4)

39

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

39

Page 40: Statistics lecture 5 (ch4)

40

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

40

Page 41: Statistics lecture 5 (ch4)

41

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.

41

Page 42: Statistics lecture 5 (ch4)

42

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

42

Page 43: Statistics lecture 5 (ch4)

43

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

43

Page 44: Statistics lecture 5 (ch4)

44

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

44

Are the marketing and finance departments

mutually exclusive?

Page 45: Statistics lecture 5 (ch4)

45

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?

Page 46: Statistics lecture 5 (ch4)

46

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?

Page 47: Statistics lecture 5 (ch4)

47

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

47

Are the marketing and finance departments

mutually exclusive?

Page 48: Statistics lecture 5 (ch4)

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

48

Page 49: Statistics lecture 5 (ch4)

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?

49