Basic Probability - PowerPoint Presentation

Slide1

Basic Probability

With an Emphasis on Contingency

TablesSlide2

Students in PSYC 2101Skip to Slide

# 7. Slide3

Random Variable

A

random variable

is real valued function defined on a sample space

.

The

sample space

is the set of all distinct outcomes possible for an experiment

.

Function

: two sets’ (well defined collections of objects) members are paired so that each member of the one set (

domain

) is paired with one and only one member of the other set (

range

)Slide4

The domain is the sample space, the range is a set of real numbers

.

A random variable is the set of pairs created by pairing each possible experimental outcome with one and only one real number.Slide5

Examples

the

outcome of rolling a die:



= 1,





= 2,







= 3, etc. (Each outcome has only one number, and, vice versa

)



= 1,





= 2,







= 1, etc. (each outcome has (odd-even) only one number, but not

vica

versa

)

The

weight of each student in my statistics class.Slide6

Probability Distribution

Each value of the random variable is paired with one and only one probability.

More on this later.Slide7

Probability Experiments

A

probability experiment

is a well-defined act or process that leads to a single well defined outcome

.

Flip a coin, heads or tails.

Roll a die, how many spots up.

Stand on a digital scale, what number is displayed.Slide8

Probability

The

probability

of an event,

P(A)

is the fraction of times that event will occur in an indefinitely long series of trials of the experiment

.

Cannot be known, can be estimated.Slide9

Estimating Probability

Empirically – perform experiment many times, compute relative frequencies.

Rationally – make assumptions and then apply logic.

Subjectively – strength of individual’s belief regarding whether an event will or will not happen – often expressed in terms of odds.Slide10

Odds of Occurrence of Event A

If the experiment were performed (a & b) times, we would expect A to occur

a

times and B to occur

b

times.

There are 20 students in a class, 14 of whom are women. If randomly select one, what are the odds it will be a woman?

14 to 6 = 7 to 3.Slide11

Convert Odds to Probability

Probability = a/(a & b).

14 women, 6 men.

Odds = 7

to

3.

Probability =

7 out of

10.Slide12

Convert Probability to Odds

Odds = P(A)/P(not A)

Probability = .70

Odds = .70/(1 - .70) = 7 to 3Slide13

Independence

Two events are

independent

iff

(if and only if) the occurrence or non-occurrence of the one has no effect on the occurrence or non-occurrence of the other

.

I roll a die twice. The outcome on the first roll has no influence on the outcome on the second roll.Slide14

Mutual Exclusion

Two events are

mutually exclusive

iff

the occurrence of the one precludes occurrence of the other (both cannot occur simultaneously on any one trial

).

You could earn final grade of A in this class.

You could earn a B.

You can’t earn both.Slide15

Mutual Exhaustion

Two (or more) events are

mutually exhaustive

iff

they include all possible outcomes

.

You could earn a final grade of A, B, C, D, or F.

These are mutually exhaustive since there are no other possibilities.Slide16

Marginal Probability

The

marginal probability

of event A,

P(A)

, is the probability of A ignoring whether or not any other event has also occurred.

P(randomly selected student is female) =

.70Slide17

Conditional Probability of A

the probability that A will occur given that B has

occurred

P(A|B), the probability of A given B.

Given that the selected student is wearing a skirt, the probability that the student is female is .9999

Unless you are in Scotland

If P(A|B) = P(A), the A and B are independent of each other.Slide18

Joint Probability

The probability that both A and B will occur.

P(A



B) = P(A)



P(B|A) = P(B)



P(A|B)

If A and B are independent, this simplifies to

P(A



B) = P(A)



P(B

)

This is known as the Multiplication RuleSlide19

The Addition Rule

If A and B are mutually exclusive, the probability that one or the other will occur is the sum of their separate probabilities.

Grade

A

B

C

D

F

Probability

.2

.3

.3

.15

.05Slide20

If A and B are not mutually exclusive, things get a little more complicated.

P(A



B) = P(A) + P(B) - P(A



B)Slide21

Two-Way Contingency Table

A matrix where rows represent values of one categorical variable and columns represent values of a second categorical variable.

Can be use to illustrate the relationship between two categorical variables.Slide22

Survey Questions

We have asked each of 150 female college students two questions:

Do you smoke (yes/no)?

Do you have sleep disturbances (yes/no)?

Suppose that we obtain the following data (these are totally contrived, not real): Slide23

Marginal ProbabilitiesSlide24

Conditional Probabilities Show Absolute Independence Slide25

Multiplication Rule Given Independence

Sixty of 150 have sleep disturbance and smoke, so P (Sleep

 Smoke) = 60/150 = .40

P(A



B) = P(A) x P(B)Slide26

“Sleep” = Sexually Active

Preacher claims those who smoke will go to Hell.

And those who fornicate will go to Hell.

What is the probability that a randomly selected coed from this sample will go to Hell?Slide27

Addition Rule

A probability cannot exceed one.

Something is wrong here!Slide28

Welcome to Hell

The events (sleeping and smoking) are not mutually exclusive.

We have counted the overlap between sleeping and smoking (the 60 women who do both) twice.

30 + 40 + 60 = 130 of the women sleep and/or smoke.

The probability we seek = 130/150 = 13/15 = .87 Slide29

Addition Rule For Events That Are NOT Mutually ExclusiveSlide30

Sleep = Sexually Active,

Smoke = Use CannabisSlide31

Marginal ProbabilitiesSlide32

Conditional Probabilities Indicate NonindependenceSlide33

Joint Probability

What is the probability that a randomly selected coed is both sexually active and a cannabis user?

There are 60 such coeds, so the probability is 60/150 = .40.

Now let us see if the multiplication rule works with these data.Slide34

Multiplication Rule

Oops, this is wrong. The joint probability is .40. We need to use the more general form of the multiplication rule.Slide35

Multiplication Rule NOT Assuming Independence

Now that looks much better.Slide36

Actual Data From Jury Research

Castellow, Wuensch, and Moore (1990,

Journal of Social Behavior and Personality

, 5, 547-562

Male employer sued for sexual harassment by female employee.

Experimentally manipulated physical attractiveness of both litigantsSlide37

Effect of Plaintiff Attractiveness

P(Guilty | Attractive) = 56/73 = 77%.

P(Guilty | Not Attractive) = 39/72 = 54%.

Defendant found guilty more often if plaintiff was attractive.Slide38

Odds and Odds Ratios

Odds(Guilty | Attractive) = 56/17

Odds(Guilty | Not Attractive) = 39/33

Odds Ratio = 56/17

 39/33 = 2.79.

Odds of guilty verdict 2.79 times higher when plaintiff is attractive.Slide39

Effect of Defendant Attractiveness

P(Guilty | Not Attractive) = 53/70 = 76%.

P(Guilty | Attractive) = 42/75 = 56%.

The defendant was more likely to be found guilty when he was unattractive.Slide40

Odds and Odds Ratio

Odds(Guilty | Not Attractive) = 53/17.

Odds(Guilty | Attractive) = 42/33.

Odds Ratio = 53/17

 42/33 = 2.50

.

Odds of guilty verdict 2.5 times higher when defendant is unattractive.Slide41

Combined Effects of Plaintiff and Defendant Attractiveness

Plaintiff attractive, Defendant not = 83% guilty.

Defendant attractive, Plaintiff not = 41% guilty.

Odds ratio = 83/17



41/59 = 7.03.

When attorney tells you to wear Sunday best to trial, listen.Slide42

Odds Ratios and Probability Ratios

Odds of Success

90/10 = 9 for Antibiotic Group

40/60 = 2/3 for Homeopathy Group

Odds Ratio = 9/(2/3) =

13.5Slide43

Odds Ratios and Probability Ratios

Odds of Failure

10/90 = 1/9 for Antibiotic Group

60/40 = 1.5 for Homeopathy Group

Odds Ratio = 1.5/(1/9) =

13.5

Notice that the odds ratio comes out the same with both perspectives.Slide44

Odds Ratios and Probability Ratios

Probability of Success

90/100 = .9 for Antibiotic Group

40/100 = .4 for Homeopathy Group

Probability Ratio = .9/(.4) =

2.25Slide45

Odds Ratios and Probability Ratios

Probability of Failure

10/100 = .1 for Antibiotic Group

60/100 = .6 for Homeopathy Group

Odds Ratio = .6/(.1) =

6

Notice that the probability ratio differs across perspectives.Slide46

Another Example

According to Medscape, 0.5% of the general population has narcissistic personality disorder (NPD)

The rate is 20% among members of the US Military.Slide47

Odds Ratios

Odds of NPD

Military: .2/.8 = .25

General: .005/.995 = .005

Ratio: .25/.005 =

49.75

Odds of NOT NPD

Military: .8/.2 = 4

General: .995/.005 = 199

Ratio: 199/4 =

49.75Slide48

Probability Ratios

Probability of NPD

Military: 20%

General: 0.5%

Ratio: 20/0.5 =

40

.

Probability of NOT NPD

Military: 80%

General: 99.5%

Ratio: .995/.8 =

1.24Slide49

Probability Distributions

For a

discrete variable

, pair each value with the probability of obtaining that value.

For example, I flip a fair coin five times. What is the probability for each of the six possible outcomes?

May be a table, a chart, or a formula.Slide50

Probability Table

Number of Heads

Percent

0

3.1

1

15.6

2

31.2

3

31.2

4

15.6

5

3.1Slide51

Probability ChartSlide52

Probability Formulay

is number of heads,

n

is number of tosses,

p

is probability of heads,

q

is probability of tailsSlide53

Continuous VariableThere is an infinite number of values, so a table relating each value to a probability would be infinitely large.

The probability of any exact value is vanishingly small.

We can find the probability that a randomly selected case has a value between

a

and

b

.Slide54

Evolution of a Continuous Variable

I’ll start with a histogram for a discrete variable.

In each step I’ll double the number of values (and number of bars).

All the way up to an infinite number of values with each bar infinitely narrow.Slide55
Slide56
Slide57
Slide58
Slide59

Now one final step, to an uncountably large number of bars, each infinitely narrow, yielding a continuous, uniform distribution ranging from A to B.Slide60
Slide61

Now I do the same but I start with a binomial distribution with

p

= .5 and three bars.

Note that the bars are not all of equal height.

Each time I split one, I lower the height of the tail-wards one more than the center-wards one.Slide62
Slide63
Slide64
Slide65
Slide66
Slide67
Slide68

Now one final leap to a continuous (normal) distribution with an uncountably large number of infinitely narrow bars.Slide69
Slide70

Random Sampling

Sampling

N

data points from a population is random if every possible different sample of size

N

was equally likely to be

selected.

Random samples most often will be representative of the population.

Our stats assume random sampling.Slide71

Y Random, X Not

Probability

Sample

X

Y

AB

1/2

1/6

AC

0

1/6

AD

0

1/6

BC

0

1/6

BD

0

1/6

CD

1/2

1/6Slide72

Counting RulesPSYC 2101 students can skip the material in the rest of this slide show.Slide73

Arranging Y ThingsThere are

Y!

ways to arrange Y different things

.

I am getting a four scoop ice cream cone.

Chocolate, Vanilla, Coconut, and Mint.

How many different ways can I arrange these four flavors?

4! = 4(3)(2)(1) = 24.Slide74

PermutationsIf I have 10 different flavors, how many different ways can I select and arrange 4 different flavors from these 10?Slide75

Combinations

Same problem, but order of the flavors does not count.

The are Y! ways to arrange Y things, so just divide the number of permutations by Y!Slide76

Number of Different StringsC

L

= number of different strings

C

is the number of different characters

available

L

is the length of the

string.

Ten different characters (0 – 9) and two character strings

10

2

= 100 different stringsSlide77

Use letters instead (A through Z)



26

2

=

676 different strings

Use letters and numbers



36

2

=

1,296

different

strings

Use strings of length 1 or 2.

 36 + 1,296 = 1,332 different stringsSlide78

Use strings of length up to 3.

36

3

=

46,656 three character strings

+ 1,332 one and two character strings

 47,988 different strings.

Use lengths up to 4



1,679,616 + 47,988 =

1,727,604

Use lengths up to 5



60,466,176 + 1,727,604 = 62,193,780Slide79

Use strings of length up to 6



2,176,782,336 + 62,193,780 = 2,238,976,116 different

strings

That is over 2 BILLION different strings.