chapter 3

CHAPTER 3PROBABILITY TOPICS

PROBABILITY

• Probability is a measure that is associated with how certain we are of outcomes of a particular experiment or activity.• An experiment is a planned

operation carried out under controlled conditions

• A result of an experiment is called an outcome

• The sample space of an experiment is the set of all possible outcomes

Suppose your experiment is flipping a coin; there are two possible outcomes:

So, we say that the sample space, or S, of this experiment is:

S = {H, T}

We can say that the size of the sample space, n(S) = 2

How many ways? 1 1

PROBABILITY

• Probability is a measure that is associated with how certain we are of outcomes of a particular experiment or activity.• An experiment is a planned

operation carried out under controlled conditions

• A result of an experiment is called an outcome

• The sample space of an experiment is the set of all possible outcomes

Suppose your experiment is rolling a die; there are six possible outcomes:

So, we say that the sample space, or S, of this experiment is:

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

We can say that the size of the sample space, n(S) = 6

How many ways? 1 1 1 1 1 1

PROBABILITY

•Suppose your experiment is picking a card from a standard 52-card deck? •Here, you need to specify what the experiment is.• Are you looking at which suit you

got?• There are four, indicated by

different shapes• S = {Heart, Club, Diamond, Spade}• n(S) = 4

Suit

No of: 13 13 13 13

PROBABILITY

•Suppose your experiment is picking a card from a standard 52-card deck? • Are you looking at which rank you

got?• There are thirteen• S = {Ace, 2, 3, 4, 5, 6, 7, 8, 9, 10,

Jack, Queen, King}• n(S) = 13

Rank 2 3 4 5 6 7 8 9 10 Jack Queen King Ace

No of: 4 4 4 4 4 4 4 4 4 4 4 4 4

A 2 3 4 5 6 7 8 9 10 J Q K

1 1 1 1 1 1 1 1 1 1 1 1 1

1 1 1 1 1 1 1 1 1 1 1 1 1

1 1 1 1 1 1 1 1 1 1 1 1 1

1 1 1 1 1 1 1 1 1 1 1 1 1

PROBABILITY

•Suppose your experiment is picking a card from a standard 52-card deck? • Are you looking at which suit

AND Rank?• There are fifty-two• S = {Ace of Clubs, 2 of Clubs,

3 of Clubs…Queen of Diamonds, King of Diamonds}

• n(S) = 52

PROBABILITY

• An event, E, is some set or collection of the outcomes in a sample space. • The event could also be none of the

outcomes• For example, with rolling a die, we could

talk about the event “rolling an even number.”• Then E = {2, 4, 6}• n(E)=3

• Another event could be “rolling at least a 5”• E = {5, 6}• n(E)=2

The event could be “rolling less than a 7”. What is E and n(E) then?

What of the event is “rolling more than a 6”?

Here, we say that E is { }

Or, we say that E is ó

Both of which mean the empty set

How many ways? 1 1 1 1 1 1

PROBABILITY

• Now let’s consider the experiment of flipping two coins• We’ll need to distinguish the two

coins from each other• Perhaps we’ll use a dime and a

quarter• We could also use a silver and a

gold coin• We could simply call one ‘Coin 1’

and the other ‘Coin 2’Coin 1

Coin 2

PROBABILITY

• So, what does the sample space look like?• S = {HH, HT, TH, TT} and n(S) = 4

• Let’s consider the event “getting two heads” and call this 2H• 2H = {HH} and n(2H) = 1

• Similarly, we could have• 1H = {HT, TH}, n(1H) = 2• 0H = {TT}, n(0H) = 1

• An outcome of HH means both coins landed heads.

• HT means Coin 1 landed heads, and Coin 2 landed tails

• TH means Coin 1 landed tails, Coin 2 landed heads• These two are different events, and we

need to keep this in mind

Coin 1

Heads

Coin 1

Tails

Coin 2

Heads

HH TH

Coin 2

TailsHT TT

PROBABILITY

Having set up the structure of the sample space, outcomes, and events, we can finally get to probability.

• Classical Probability – all outcomes are equally likey to occur each time you perform the experiment.

• The coins are fair, so each time you flip them, they have an equal chance of coming up heads or tails

• The dice are fair…• The deck is shuffled fairly…

• This does NOT mean that every two flips of a coin you will get one head and one tail

• Rather, it means that in the long run the outcomes will even out in frequency

Coin 1

Heads

Coin 1

Tails

Coin 2

Heads

HH TH

Coin 2

TailsHT TT

PROBABILITY

The closer the probability is to 1, the

more confident we are of it happening

We talk about the probability of an event, E, occurring• This is a number that is between 0 and

1 – always!• We write P(E), pronounced “P of E”, in

the manner of function notation • Thus, we can write

• This means that the probability of an event is between 0 and 1

• The probability is 0 if the event is impossible (rolling a 7 on one die)

• The probability is 1 if it’s a sure thing (rolling a number less than 7 on one die)

• If it might or might not happen, it’s somewhere in between

Coin 1

Heads

Coin 1

Tails

Coin 2

Heads

HH TH

Coin 2

TailsHT TT

PROBABILITY• Simply put, the probability of an event

occurring is the number of ways the event can occur divided by the total possible number of events• So, P(2H) = …

• P(2H) = • P(1H) = …

• P(1H) = • P(0H) = …

• P(0H) = • You’ll note that all of the possibilities

are represented, and that the sum of the probabilities is 1

• That is, the probability of flipping 0, 1, or 2 heads is 1

Coin 1

Heads

Coin 1

Tails

Coin 2

Heads

HH TH

Coin 2

TailsHT TT

PROBABILITY

Suppose your experiment is rolling two dice…

Now, how many possible outcomes are there?

Let A be the event that a 7 is rolled; what is P(A)?

Next, let B be the event that an 11 is rolled; What is P(B)?

Now, what is the probability that either a 7 or an 11 is rolled?

We could call this event C, and ask what P(C) is, or we could also ask what is P(A OR B)?

(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)

𝑃 (𝐸 )=𝑛 (𝐸)𝑛(𝑆)

636

236

836

.17

.06

.25

17.67%

5.56%

25%

PROBABILITY

Next, let D be the event that one of the rolled dice is a 3; what is P(D)?• Be careful not to ‘double count’.

Now, we can ask what is the probability that a 7 is rolled and that one of the dice is a 3. We ask what is P(A AND D)

Next, let F be the event that the first die rolled is a 1 or 2. If we say that we know that F happens, what is the probability of A now?

The way that we say this is ‘What is the probability of A given F’, and we write it as P(A|F).• This ‘given’ condition has changed n(S).• Because we know that the first die is either a 1 or a 2,

n(S) = 12• What is n(A)?• So, P(A|F) =

𝑃 (𝐸 )=𝑛 (𝐸)𝑛(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)

1136

.31 30.56%

236

.06 5.56%

A = a sum of 7 is rolled

CONTINGENCY TABLES

• A Contingency Table provides a way of portraying data that can facilitate calculating probabilities• The table helps in determining conditional

probabilities quite easily• The table displays sample values in relation

to two different variables that may be dependent or contingent on another

• Let’s call Y the event that an athlete always stretches before exercising

• We would like to know if an athlete is selected at random, what is the probability that they always stretch before exercising?

Injury in Last year

No Injury in Last Year

Total

Stretches (Y)

55 295 350

Does Not Stretch

231 219 450

Total 286 514 800That is, what is P(athlete stretches before exercising)?

Or, what is P(Y)?

• n(Y) = 350• n(S) = 800• P(Y) =

CONTINGENCY TABLES

• Let’s call NI the event that an athlete was not injured last year

• What is the probability that an athlete, chosen at random, stretches before exercising given that they did not have an injury in the last year?

• Written in Probability notation we are looking for: P(Y|NI)

• Here, n(S) has changed to only those that did not have an injury in the last year.• n(S) = n(NI)

• n(Y) = 295• n(NI) = 514• P (Y|NI) =

Injury in Last year

No Injury in Last Year (NI)

Total

Stretches (Y)

55 295 350

Does Not Stretch

231 219 450

Total 286 514 800

YOU TRY ONE!

• Let’s call I the event that an athlete was not injured last year

• Let’s call N the event that an athlete does not stretch before exercising

• If you pick an athlete at random, what is the probability that the athlete was injured last year given that they do not stretch before exercising?• Write the probability statement• State the sample space using n(S) language• State the size of the event using n(E)

language• State the probability

• P(I|N)• n(N) = 450• n(I) = 231• P (I|N) =

Injury in Last year (I)

No Injury in Last Year (NI)

Total

Stretches (Y)

55 295 350

Does Not Stretch (N)

231 219 450

Total 286 514 800

CREATING CONTINGENCY TABLES

• Suppose that you have a sample of 100 people, and you are told that you have 48 females. Of the males, 43 are right handed. Also, there are 4 left handed females. • Assume that every subject is either

male or female, and that none of them are ambidextrous.

• How do we complete the table?

Right Hand

ed

Left Hand

edTotal

Male 43Female 4 48

Total 100

CREATING CONTINGENCY TABLES

• Suppose that you have a sample of 100 people, and you are told that you have 48 females. Of the males, 43 are right handed. Also, there are 4 left handed females. • Assume that every subject is either

male or female, and that none of them are ambidextrous.

• How do we complete the table?

Right Hand

ed

Left Hand

edTotal

Male 43 9 52Female 44 4 48

Total 87 13 100

CREATING CONTINGENCY TABLES• Now, how would we do this with

percentages?• The voting on Proposition X occurred

with the following breakdown• 52.29% of the voters live in city limits,

and 47.71% of the voters did not (they lived in the County limits)

• Of the city voters• 51.33% voted yes, 47.33% voted

no, and 1.33% did not vote on this proposition

• Of the county voters• 34.24% voted yes, 63.99% voted


YES on X

NO on X

No Respo

nseTotal

s

City VotersCounty VotersTotals

The totals columns should always add up to (close to) 100%

What happens if we just try and plug in what we know?

YES on X

NO on X

No Respo

nseTotal

s

CityVoters .5133 .4733 .0133 1

County Voters .3424 .6399 .0177 1

Totals .8557 1.1132 .0310

CREATING CONTINGENCY TABLESYES on X

NO on X

No Respo

nseTotal

s

CityVoters .5133 .4733 .0133 1

County Voters .3424 .6399 .0177 1

Totals .8557 1.1132 .03101.999

92

1, 2, 1.9999?

What did we get wrong?

• Now, how would we do this with percentages?

• The voting on Proposition X occurred with the following breakdown• 52.29% of the voters live in city limits,

and 47.71% of the voters did not (they lived in the County limits)

• Of the city voters• 51.33% voted yes, 47.33% voted




CREATING CONTINGENCY TABLES• 52.29% of the voters live in city limits,

and 47.71% of the voters did not (they lived in the County limits)• Of the city voters

• 51.33% voted yes, 47.33% voted no, and 1.33% did not vote on this proposition



YES on X

NO on X

No Respo

nseTotal

s

City Voters .5133 .4733 .0133 1

County Voters .3424 .6399 .0177 1

Totals .8557 1.1132 .03101.999

92

We need to remember that, while there were 51.33% of the city residents in favor, that was only 52.29% of the polled respondents.

Jason Edington






YES on X

NO on X

No Respo

nseTotal

s

City VotersCounty VotersTotals

To find actual percentage, multiply 51.33% by 52.29%

Jason Edington






YES on X

NO on X

No Respo

nseTotal

s

City Voters .2684

County VotersTotals

To find actual percentage, multiply 51.33% by 52.29%

Jason Edington






YES on X

NO on X

No Respo

nseTotal

s

City Voters .2684

County VotersTotals

To find actual percentage, multiply 51.33% by 52.29%Then, 47.33% by 52.29%

Jason Edington






YES on X

NO on X

No Respo

nseTotal

s

City Voters .2684 .2475

County VotersTotals

To find actual percentage, multiply 51.33% by 52.29%Then, 47.33% by 52.29%

Jason Edington






YES on X

NO on X

No Respo

nseTotal

s

City Voters .2684 .2475

County VotersTotals

To find actual percentage, multiply 51.33% by 52.29%Then, 47.33% by 52.29%Then, 1.33% by 52.29%

Jason Edington






YES on X

NO on X

No Respo

nseTotal

s

City Voters .2684 .2475 .0070

County VotersTotals

To find actual percentage, multiply 51.33% by 52.29%Then, 47.33% by 52.29%Then, 1.33% by 52.29%

Jason Edington






YES on X

NO on X

No Respo

nseTotal

s

City Voters .2684 .2475 .0070

County VotersTotals

To find actual percentage, multiply 51.33% by 52.29%Then, 47.33% by 52.29%Then, 1.33% by 52.29%Finally, do the same for the County Voters, using the proper numbers

Jason Edington






YES on X

NO on X

No Respo

nseTotal

s

City Voters .2684 .2475 .0070 .522

9County Voters .1634 .3053 .0084 .477

1Totals .4318 .5528 .0154

11

To find actual percentage, multiply 51.33% by 52.29%Then, 47.33% by 52.29%Then, 1.33% by 52.29%Finally, do the same for the County Voters, using the proper numbers

Jason Edington

CREATING CONTINGENCY TABLESNow we can ask questions such as:• If 1000 residents are randomly

selected:• How many would you expect to be city

voters?• 523

• What is the probability that out of the voters that voted no, a randomly selected voter lives in the county?• How many would you expect to have

voted no out of 1000?• 553

• How many of these would you expect to live in the county?

• 305• What is the probability that a randomly

selected voter, from the ‘no’ votes, lives in the county?

YES on X

NO on X

No Respo

nseTotal

s

City Voters .2684 .2475 .0070 .522

9County Voters .1634 .3053 .0084 .477

1Totals .4318 .5528 .0154 1

Jason Edington

HOMEWORK

• BEGINNING ON PAGE 210• 85 (A-E), 86 (A-D), 88 (A, B), 89 (A, B), 97, 98, 100, 101-106, 112, 124, 125

chapter 3

Education