WHAT IS PROBABILITY? prof. Renzo Nicolini I.M.”G.CARDUCCI” - Trieste

WHAT IS PROBABILITY?

prof. Renzo Nicolini I.M.”G.CARDUCCI” - Trieste


CLIL projectClass II C

LESSON 1



The word probability derives

from the Latin probare (to prove, or to test).


PROBABLE is almost synonym of

• likely

•hazardous

• risky

•uncertain

•doubtful

theory of probability attempts to quantify the notion of probable.


HOW PROBABLE SOMETHING IS?

To answer, we need a number!!!!!

/LIKELY

HISTORICAL REMARKS

The scientific study of probability is a modern development.

Gambling

shows that there has been an interest in quantifying the ideas of probability for millennia,

but exact mathematical descriptions of use in those problems only arose much later.

HISTORICAL REMARKS

The doctrine of probabilities starts with the works of

Pierre de Fermat Blaise Pascal (1654) Christian Huygens(1657) Daniel Bernoulli (1713) Abraham de Moivre (1718) Blaise PascalBlaise Pascal

Vocabulary

An An experimentexperiment is a situation involving chance or probability is a situation involving chance or probability that leads to results called outcomes.that leads to results called outcomes.

An An outcomeoutcome is the result of a single trial of an experiment.is the result of a single trial of an experiment.

An An eventevent is one or more outcomes of an experiment.is one or more outcomes of an experiment.

ProbabilityProbability is the measure of how likely an event is.is the measure of how likely an event is.

Experiments!

Rolling a single 6-sided die Running an horse race Driving a car race Picking a card from a deck Tossing a coin.

EXPERIMENT

OUTCOME

EVENT

Outcomes!

Rolling a single 6-sided die:

“a number six was drawn” (to be drawn = uscire);

“a number three was drawn” ;

“a number eight can’t be drawn”

EXPERIMENT

OUTCOME

EVENT

Possible outcomes in the experiment

Impossible outcome in the experiment

Outcomes!

Experiment: Driving a car race

Outcome: “Schumacher wins”

EXPERIMENT

OUTCOME

EVENT

Outcomes!

Experiment: Picking a card from a deck

Outcome: “A king is drawn”

EXPERIMENT

OUTCOME

EVENT

Outcomes!

Experiment: Tossing a coin

Outcome: “a tail has been tossed”

EXPERIMENT

OUTCOME

EVENT

Outcomes!

THE SET OF ALL THE POSSIBLE OUTCOMES IS CALLED SAMPLE

SPACE and is denoted by S.

EXPERIMENT

OUTCOME

EVENT

Outcomes! Examples

Experiment: Rolling a die once: • Sample space S = {1,2,3,4,5,6}

Experiment: Tossing a coin: • Sample space S = {Heads,Tails}

Experiment: Measuring the height (cms) of a girl on her first day at school: • Sample space S = the set of all (?) possible

real numbers

EXPERIMENT

OUTCOME

EVENT

Event!

It’s the particular outcome or set of outcomes I’m interested to study:

“How possible is that a Queen is picked up from a deck of cards”?

“How possible is that a Jack OR a King are picked up from a deck of card”?

“Rolling a die once, how possible is it that the score is < 4?”

EXPERIMENT

OUTCOME

EVENT

This is OUR event!

This is OUR event!

Probability!

We call probability the value we estimate for a single event:

“What is the probability that a Queen is picked up from a deck of card”?

“ What is the probability that a Jack OR a King is picked up from a deck of card”?

LET’S REPEAT!!

WHEN I DO SOMETHING I SAY THAT I CARRY OUT AN

EXPERIMENT

EXAMPLES?EXAMPLES?

LET’S REPEAT!!

ANY POSSIBLE SITUATION THAT OCCURS WHEN I CARRY OUT THE EXPERIMENT IS AN

OUTCOME

EXAMPLES?EXAMPLES?

LET’S REPEAT!!

ALL THE POSSIBLE OUTCOMES THAT CAN OCCUR WHEN I EXECUTE THE EXPERIMENT, FORM THE

SAMPLE SPACE

EXAMPLES?EXAMPLES?

LET’S REPEAT!!

THE PARTICOLAR OUTCOME or SET OF OUTCOMES WE’RE INTERESTED IN IS AN

EVENT

EXAMPLES?EXAMPLES?

LET’S REPEAT!!

THE MEASURE OF HOW LIKELY AN EVENT IS, IS CALLED

PROBABILITY



LESSON 2


HOW TO EVALUATE PROBABILITY?

Probability is a number!

We need a formula or a procedure to find it!


CLASSICAL DEFINITION SUBJECTIVE PROBABILITY FREQUENTIST DEFINITION

THERE ARE THREE POSSIBLE WAYS TO FIND THIS VALUE


A CURIOSITY!!

SUBJECTIVE PROBABILITY was proposed in XX century by Bruno De Finetti, who worked in Triest (Generali, University) from 1931 to 1954

We will not talk about this type of probability


WE’LL SEE ONLY THE CLASSICAL DEFINITION OF

PROBABILITY.

by

SIMON DE LAPLACE (1749-1827)

CLASSICAL PROBABILITY

SIMON DE LAPLACE (1749-1827) gave the most famous definition of probability.It’s called

CLASSICAL DEFINITION OF PROBABILITY

Mathematics need fomulas!

In order to measure probabilities, he has proposed the following formula for finding the probability of an event.

THE FORMULA FOR THE CLASSICAL PROBABILITY

Probability Of An Event P(A) =

The Number Of Ways an Event A Can Occur

The Total Number Of Possible Outcomes

The number of elements of the sample space


The probability of event A is the number of ways event A can occur divided by the total number of possible outcomes.


The probability of event A is the number of favorable cases (outcomes) divided by the total number of possible cases (outcomes).

EXAMPLE/EXERCISEEXAMPLE/EXERCISE

•What is the probability of each What is the probability of each outcome? outcome?

•What is the probability of rolling an What is the probability of rolling an even number? even number?

•Of rolling an odd number?Of rolling an odd number?

A single 6-sided die is rolled. A single 6-sided die is rolled.

Outcomes: The possible outcomes of this experiment are 1, 2, 3, 4, 5 and 6.

P(1) = P(1) = number of ways to roll a 1number of ways to roll a 1 = = 11 total number of sides 6 total number of sides 6 P(2) = P(2) = number of ways to roll a 2number of ways to roll a 2 = = 11 total number of sides 6 total number of sides 6 P(3) = P(3) = number of ways to roll a 3number of ways to roll a 3 = = 11 total number of sides 6 total number of sides 6 P(4) = P(4) = number of ways to roll a 4number of ways to roll a 4 = = 11 total number of sides 6 total number of sides 6 P(5) = P(5) = number of ways to roll a 5number of ways to roll a 5 = = 11 total number of sides 6 total number of sides 6 P(6) = P(6) = number of ways to roll a 6number of ways to roll a 6 = = 11 total number of sides 6total number of sides 6

All the values All the values are the are the same!!! The same!!! The outcomes are outcomes are equally likelyequally likely

..

EQUALLY LIKELY EQUALLY LIKELY EVENTS HAVE THE SAME EVENTS HAVE THE SAME PROBABILITY TO OCCURPROBABILITY TO OCCUR

EQUALLY LIKELY EVENTS

What is the probability of rolling an What is the probability of rolling an even number? even number?

P(even) = # ways to roll an even number

# total number of sides

probability of rolling an probability of rolling an even number is even number is

one half = 0,5one half = 0,5

2

1

6

3

What is the probability of rolling an What is the probability of rolling an odd number?odd number?

probability of rolling an odd probability of rolling an odd number isnumber is

one halfone half

5.06

3

NOTE: classical probability is a priori

It’s interesting to note that, in order to calculate the probability in the classical way, it’s necessary to know EVERYTHING about the experiment.

We need to know the possible outcomes (the whole sample space),

we need to know the EVENT we are interest in. In few words, WE HAVE TO KNOW EVERYTHING BEFORE

RESULTS COME OUT.

That’s why we say that CLASSICAL PROBABILITY IS

A PROBABILITY “A PRIORI”.


The Number Of Ways Event A Can Occur The Total Number Of Possible Outcomes

Some more about the formula for Some more about the formula for probabilityprobability

The “impossible” event The “certain event”

Is it possible that there are no ways event Is it possible that there are no ways event A A can can occur?occur?

SURE!SURE!In this case the formula for probability In this case the formula for probability


has numerator equal to 0!has numerator equal to 0!P(A) =0P(A) =0

THE EVENT IS IMPOSSIBLE!THE EVENT IS IMPOSSIBLE!

Is it possible that there are no ways event Is it possible that there are no ways event A A can occur?can occur?

THE EVENT IS THE EVENT IS IMPOSSIBLE?IMPOSSIBLE?

P(A) =0P(A) =0

It has no probability to happen!It has no probability to happen!

EXAMPLE OF PROBABILITY = 0

Which is the probability of rolling number 7 on a 6 sided die?


The Number Of Ways Event A Can Occur is 0

P(A) =0P(A) =0

because number 7 doesn’t exist in such a die!!!

Is it possible that event Is it possible that event A A certainly will occur?certainly will occur?

SURE!SURE!In this case the formula for probability In this case the formula for probability


has numerator equal to the denominator.has numerator equal to the denominator.

The fraction values 1The fraction values 1

THE EVENT IS CERTAIN!THE EVENT IS CERTAIN!


WHICH IS THE PROBABILITY THAT, ROLLING A DIE, A NUMBER BETWEEN 0 AND 7 COMES OUT ?

P(A)= The Number Of Ways an Event A Can Occur = 6 = 1 The Total Number Of Possible Outcomes 6

P(A) =1P(A) =1


THE EVENT IS CERTAIN!THE EVENT IS CERTAIN!

P(A) =1P(A) =1

It willIt will certainly happen!certainly happen!

LET’S REPEAT!!

is a positive real number, between 0 and 1

PROBABILITY

Zero for the impossible event

One for the certain event

LET’S REPEAT!!

TO FIND THE CLASSICAL PROBABILITY (Laplace) we need the following definition:

P(A)= The Number Of Favorable Cases The Total Number Of Possible Cases


LET’S REPEAT!!

WHEN TWO EVENTS HAVE THE SAME PROBABILITY, WE SAY THAT THEY ARE EQUALLY LIKELY.

Heads and tails are equally likely!!!

LAST QUESTION!

Which is the probability that next time you’ll appreciate our CLIL lesson?

LAST ANSWER!

Classical probability doesn’t give any answer to this question, because it’s not a problem solving “a priori”.

LAST ANSWER!

It’s a typical situation of subjective probability, which depends on your particular feeling about the event “We come next time”. It’s a result of you own sensation!!

LAST ANSWER!

I HOPE THIS PROBABILITY IS NOT

ZERO!



LESSON 3


LET’S REPEAT!!

WHEN I DO SOMETHING I SAY THAT I EXECUTE AN

EXPERIMENT

LET’S REPEAT!!

ALL THE POSSIBLE OUTCOMES THAT CAN OCCUR WHEN I EXECUTE THE EXPERIMENT, FORM THE

SAMPLE SPACE

ANY POSSIBLE SITUATION THAT OCCURS WHEN I EXECUTE THE EXPERIMENT IS AN

OUTCOME

LET’S REPEAT!!

THE PARTICOLAR OUTCOME or SET OF OUTCOMES WE’RE INTERESTED IN, IS AN

EVENT

THE MEASURE OF HOW LIKELY AN EVENT IS, IS CALLED

PROBABILITY



The Number Of Ways an Event A Can Occur

The Total Number Of Possible Outcomes



The probability of event A is the number of favorable cases (outcomes) divided by the total number of possible cases (outcomes).

SOMETHING MORE ABOUT THE THEORY OF PROBABILITY

PROBLEM

Imagine to be asked to solve the following exercise:

Rolling a die, which is the probability of rolling any number except 2?

PROBLEM Rolling a die, which is the probability of rolling any number except 2.

The statement “any number except number 2” is the negation of the statement “number 2”

The EVENT A “any number except 2” is the negation of the EVENT “rolling a number 2”

PROBLEM Rolling a die, which is the probability of rolling any number except 2?

We say that the EVENT “any number except number 2” is the COMPLEMENT OF THE EVENT A “rolling a number 2”

COMPLEMENT OF AN EVENT A

It’s the opposite statement of the EVENT A

We use to indicate it with

Ā

Ā (A bar) is the complement of A

Let’s calculate the probability of Ā

Rolling a die, which is the probability of rolling any number except 2?


What do we need?

Favorable cases.

Possible cases.

1,2,3,4,5,6

All the elements of subset 1,3,4,5,6


The probability of this event is

5

6


Can we solve this problem in another way?

YES!

HOW?

PROBLEM Rolling a die, which is the probability of rolling a 2?

Let’s start considering the problem of the EVENT A

“Probability of rolling a 2”

PROBLEM Rolling a die, which is the probability of rolling a 2?

IT’S OBVIOUSLYIT’S OBVIOUSLY

1

6

PROBLEM Rolling a die, which is the probability of rolling WHICHEVER number BUT A 2?

Now, we have two data:

Probability one-sixth for EVENT A

Probability five-sixth for EVENT Ā

OBSERVE THAT ONE-SIXTH

PLUS FIVE-SIXTH IS EQUAL TO

1

PROBABILITY OF THE COMPLEMENT

The probability we found is 1.

In other words, the sum

between P(A) and P (Ā) is 1.

P(A) + P (Ā) = 1


P(A) + P (Ā) = 1

IS IT A IS IT A FORTUITOUSFORTUITOUS CASE?

NO! It’s a general rule!!!!

LET’S PROVE IT!!!!!

P(A) + P (Ā) = 1

Let n(A) be the number of favorable cases for the event A

Let n(Ā) be the number of favorable cases for the event Ā

Let n(E) be all the number of all possible cases for the experiment (the number of elements in the sample space

P(A) + P (Ā) = 1

It’s quite obvious that

n(A) + n(Ā)= n(E)

Let’s divide the equality by n(E)

In fact, the number of cases for A and In fact, the number of cases for A and the number of cases for the number of cases for Ā exhaust all the possibilities, i.e. all the possible

cases.

EnEn

En

AnAn

P(A) + P (Ā) = 1

and than, with an obvious (?) passage,

1En

An

En

An 1 APAP


WE GOT THE RESULT: WE GOT THE RESULT: P(A) + P (Ā) = 1P(A) + P (Ā) = 1

OR, IN OTHER TERMS,

The probability P (Ā) of the complement of an event A is given by the subtracting from 1 the probability P (A) of the event A

P (Ā) = 1- P(A)


In general, if we know the probability of an event, we can immediately calculate the probability of its complement !


A single card is chosen from a standard deck of 52 cards.

What is the probability of choosing a card that is not a King?

PP (not a king) = 1 – (not a king) = 1 – PP(king)(king)13

12

52

48

52

41

LET’S REPEAT!!

When an event B has the opposite requirements of an event A, we say that the event B is the

COMPLEMENT OF EVENT A

We can also indicate it as Ā

LET’S REPEAT!!

P(A) + P (Ā) = 1

We have proved that the following formula for finding the value of the

complement of an event:



LESSON 4


LET’S REPEAT!!

P(A) + P (Ā) = 1

We have proved the following formula for the complement :

LET’S REPEAT!! PROBABILITY OF THE COMPLEMENT

A single card is chosen from a standard deck of 52 cards.

What is the probability of choosing a card that is not a King?

PP (not a king) = 1 – (not a king) = 1 – PP(king)(king)13

12

52

48

52

41

Let’s see now a more Let’s see now a more complicated situation, complicated situation, involving involving more actions.more actions.

COMPOUND EVENT

Experiment.

Let’s suppose to propose an experiment in which we do two actions.

1. We roll a die…

2. We take a number, playing tombola

AND

and …

Experiment.

…we set up an event which includes both actions

For instance: For instance:

What is the probability of drawing an odd What is the probability of drawing an odd number from the sack of tombola number from the sack of tombola AND AND rolling a multiple of 3 on the die?

INDEPENDENT EVENTS

First of all, we can notice that the rolling of the die and the drawing of the number are

INDEPENDENT EVENTS.

Two events, A and B, are Two events, A and B, are independentindependent if the fact that A if the fact that A occurs occurs does not affectdoes not affect the the probability of B occurring.probability of B occurring.

INDEPENDENT EVENTS

The rolling of the die and the drawing of the number playing tombola are

INDEPENDENT EVENTS …

…because the tombola number “doesn’t see” the outcome of the die and it isn’t

influenced by it !!!!!

Experiment.

What is the probability of drawing an odd What is the probability of drawing an odd number from the sack of tombola number from the sack of tombola AND AND rolling a multiple of 3 on the die?

We are looking for the probability of a We are looking for the probability of a more complicated event which involves more complicated event which involves two simpler INDEPENDENT eventstwo simpler INDEPENDENT events

COMPOUND EVENT

TWO ACTIONS

ONE REQUIREMENT

CONNECTOR CONNECTOR ANDAND

COMPOUND EVENT

This is a typical example of

COMPOUND EVENT

COMPOUND EVENT

In the COMPOUND EVENT we want that bothboth the events occur

COMPOUND EVENT

In the COMPOUND EVENT we want that oneone AND AND the other the other

event occurevent occur

The KEY CONJUNCTION is ANDAND

COMPOUND EVENT

We could prove that

the probability of the compound event is always the product of the single probabilities of two independent events which compose the compound event..

Multiplication rule

When two events, A and B, are independent, the probability of both occurring is:

P(A and B) = P(A) · P(B)

This is called the “multiplication rule”This is called the “multiplication rule”

Experiment.

2

1

90

45

What is the probability of choosing an odd number from the What is the probability of choosing an odd number from the sack of tombola sack of tombola AND AND rolling a multiple of 3 on the die?

EVENT EVENT AA = “drawing an odd tombola number” = “drawing an odd tombola number”

P(odd)P(odd)

EVENT EVENT BB = “rolling a multiple of 3” = “rolling a multiple of 3”

P(3,6)P(3,6)3

1

6

2

Experiment.

What is the probability of drawing an odd number from the What is the probability of drawing an odd number from the sack of tombola sack of tombola AND AND rolling a multiple of 3 on the die.

P(odd AND die) = P(odd) * P(die)P(odd AND die) = P(odd) * P(die)6

1

3

1

2

1

We can use the multiplication rule because the two events are evidently independent!

OTHER EXPERIMENTS

2

1

A coin is tossed and a single 6-sided die is rolled. Find the probability of tossing heads AND rolling a 3 on the die.

P(head)P(head)

6

1P(3 on die)P(3 on die) 12

1

6

1

2

1P(head AND 3)P(head AND 3)

OTHER EXPERIMENTS

A card is chosen at random from a deck of 52 cards. It is then put back and a second card is chosen. What is the probability of choosing a jack AND an eight, replacing the chosen card?

OTHER EXPERIMENTS. A card is chosen at random from a deck of 52 cards. It is then put back and a second card is chosen. What is the probability

of getting a jack AND an eight, replacing the chosen card?

P(jack)P(jack)13

1

52

4

P(eigth)P(eigth)13

1

52

4

169

1

13

1

13

1P(jack AND eight)P(jack AND eight)

DEPENDENT EVENTS

What happens if we decide not to put back the first card in the deck?

In this case, the second draw would be conditioned by the first one. In fact, in the second draw there would be only 51 cards in the deck! So, in the second draw, the possible cases would be 51 (one card has been removed!), while the favorable cases of picking an eight would remain the same (4).

P(jack)P(jack)13

1

52

4 P(eigth)P(eigth)

51

4

DEPENDENT EVENTS

What happens if we decide not to put back the first card in the deck?

In this case, the second draw would be conditioned by the first one. In fact, in the second draw there would be only 51 cards in the deck! So, in the second draw, the possible cases would be 51 (one card has been removed!), while the favorable cases of picking an eight would remain the same (4).

THE EVENTS AREN’T INDEPENDENT ANYMORE. THEY ARE DEPENDENT.

DEPENDENT EVENTS

P(jack)P(jack)13

1

52

4 P(eigth)P(eigth)

51

4

P(jack AND eight)P(jack AND eight)663

4

51

4

13

1

PROBABILITY HAS CHANGED!

AN OTHER EXPERIMENT

A jar contains 3 red, 5 green, 2 blue and 6 yellow marbles. A marble is picked at random from the jar. After putting it back, a second marble is picked. What is the probability of getting a green and a yellow marble?

AN OTHER EXPERIMENT

8

3

16

6

Possible cases == 16 16 with replacingwith replacing

P(yellow)16

5P(green)

128

15

16

5

8

3P(yellow AND green)

DEPENDENT EVENTS

But what happens if we decide not to put back the first marble in the jar ?

ANOTHER EXPERIMENT 3 red, 5 green, 2 blue and 6 yellow

8

3

16

6

8

1

3

1

8

3

Possible cases == 16 16 without replacingwithout replacing

P(yellow) P(green)

P(yellow AND green)

3

1

15

5

ANOTHER EXPERIMENT 3 red, 5 green, 2 blue and 6 yellow

8

1

3

1

8

3

with no replacingwith no replacing

P(yellow AND green)

PROBABILITY HAS CHANGED AGAIN!

with replacingwith replacing

128

15

16

5

8

3P(yellow AND green)

COMPOUND EVENT

If the events are independent, multiplication rule is valid and probability is just

If the events are dependent, multiplication rule is still valid, but the second factor depends on the first


P(A and B) = P(A) · P(B | A)B | A)

Read: B occurs given that B occurs given that event A has occurredevent A has occurred

LET’S REPEAT : the multiplication rule



When two events, A and B, are dependent, the probability of both occurring is

P(A and B) = P(A) · P(B | AB | A)

The usual notation for "event B occurs given that event A has The usual notation for "event B occurs given that event A has occurred" is “B | A" (B given A).occurred" is “B | A" (B given A).



LESSON 5


LET’S REPEAT!! Independent events

Two events, A and B, are Two events, A and B, are independentindependent if the fact that A if the fact that A occurs occurs does not affectdoes not affect the the probability of B occurring.probability of B occurring.

LET’S REPEAT!!

The rolling of a die and the drawing of The rolling of a die and the drawing of a tombola number are a tombola number are

INDEPENDENT events!INDEPENDENT events!

LET’S REPEAT!!

When we have two events and we want that bothboth of those occur, we are considering a

COMPOUND EVENT

LET’S REPEAT!!

In the COMPOUND EVENT we want that oneone AND AND the other the other

event occurevent occur

The KEY CONJUNCTION is ANDAND

LET’S REPEAT!!

We could prove that

the probability of the compound event is always the product of the single probabilities of two independent events which form the compound event..

LET’S REPEAT!!



This is called the “multiplication rule”This is called the “multiplication rule”

LET’S REPEAT!!

A card is chosen at random from a deck of 52 cards. It is then put back and a second card is chosen. What is the probability of drawing a jack AND an eight, putting back the chosen card?

DEPENDENT EVENTS

In this case we say that we are valuating

A COMPOUND EVENT

of TWO DEPENDENT EVENTS

LET’S REPEAT!!! A card is chosen at random from a deck of 52 cards. It is then replaced and a second card is chosen. What is the probability of

choosing a jack AND an eight, replacing the chosen card?

P(jack)P(jack)13

1

52

4

P(eigth)P(eigth)13

1

52

4

169

1

13

1

13

1P(jack AND eight)P(jack AND eight)

LET’S REPEAT : the multiplication rule



When two events, A and B, are dependent, the probability of both occurring is

P(A and B) = P(A) · P(B | AB | A)

The usual notation for "event B occurs given that event A has The usual notation for "event B occurs given that event A has occurred" is “B | A" (B given A).occurred" is “B | A" (B given A).

Let’s see know a new situation, involving only one experiment and one action (one rolling of the die, one choosing of a card, and so on…), but where we accept

MORE FAVORABLE OUTCOMES

MORE FAVORABLE OUTCOMES

Experiment. MORE FAVORABLE OUTCOMES

We choose a card from an ordinary deck, and we accept

either a King a numbered card.

first favorable eventsecond favorable event

OR

MUTUALLY EXCLUSIVE EVENTS

First of all, we can notice that the choosing of a King and the choosing of a

numbered card CAN’T OCCURE at the same time. We say that:

They are two

MUTUALLY EXCLUSIVE EVENTS(disjoint events)


Two events are mutually exclusive Two events are mutually exclusive (or disjoint) (or disjoint) if it is impossible for if it is impossible for them to occur together.them to occur together.

Experiment.

We are looking for the probability of a We are looking for the probability of a more complicated event which involves more complicated event which involves two simpler MUTUALLY EXCLUSIVE two simpler MUTUALLY EXCLUSIVE events.events.

ONE ACTION

TWO REQUESTS

CONNECTOR CONNECTOR OROR

COMPOUND EVENT

We could prove that

the probability of the occurring of one of two MUTUALLY EXCLUSIVE events is the

sum of the probabilities of each event..

Addition rule

When two events, A and B, are mutually exclusive, the probability of just one occurring is:

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

This is called the addition ruleThis is called the addition rule

ANOTHER EXPERIMENT

A jar contains 3 red, 5 green, 2 blue and 6 yellow marbles. A marble is chosen at random from the jar. What is the probability of choosing a green OR a yellow marble?

ANOTHER EXPERIMENT

8

3

16

6

16

11

16

5

8

3

Possible cases == 16 16

P(yellow)16

5P(green)

P(yellow OR green)

first favorable eventsecond favorable event

ONE MORE EXERCISE

What is the probability of rolling a 2 OR a 5 on a single 6-sided die?


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

P(A or B)6

1

6

1

3

1

Read: OROR


Formally, if two events A and B are mutually exclusive we can write:

A B =

WHY? WHAT DOES IT MEAN?

A B =

sample sample

spacespace AB

A B = → Disjoint sets

Outcomes which satisfy elementary event A

Outcomes which satisfy elementary event B

There aren’t outcomes which contemporarily satisfy event A and event B

A B = … In a class:

Event A : randomly, choosing a green-eyes-student;

Event B : randomly, choosing a brown-eyes-student;

Obviously, no student has Obviously, no student has contemporarily green and brown eyescontemporarily green and brown eyes

AB


A : green-eyes-student B : brown-eyes-student


Obviously, no student has Obviously, no student has contemporarily contemporarily greengreen AND AND brownbrown eyes eyes

Therefore, the probability of choosing an element of A and an element of B is = 0

The sets are disjoint

The events are disjoint

When two events can’t occurcan’t occur contemporarily, contemporarily, they are said

LET’S REPEAT : the addition rule

- MUTUALLY INDEPENDENT EVENTS

- DISJOINT EVENTS

When two events can’t occurcan’t occur contemporarily, there’s no element in the contemporarily, there’s no element in the

intersection between A and B intersection between A and B


A B =

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



LESSON 6


When two events can’t occurcan’t occur contemporarily, contemporarily, they are said


- MUTUALLY INDEPENDENT EVENTS

- DISJOINT EVENTS

When two events can’t occurcan’t occur contemporarily, there’s no element in the contemporarily, there’s no element in the



A B =

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

Addition rule for NOT MUTUALLY EXCLUSIVE EVENTS

BUT WHAT ABOUT WHEN TWO EVENTS CAN OCCUR CONTEMPORARILY?

i.e. WHAT IS THE PROBABILITY OF TWO NOT MUTUALLY EXCLUSIVE EVENTS?

LET’S SEE AN EXAMPLE!

OTHER EXPERIMENTS

A card is chosen at random from a deck of 52 cards.

What is the probability of choosing a jack OR a club ?

ONE ACTION

(one picking of a card)

TWO REQUESTS

(club; jack)

CONNECTOR CONNECTOR OROR

NOT MUTUALLY EXCLUSIVE EVENTS

In this case the event A (“choosing a jack”) and the event B (“choosing a club ”) are not disjoint.

sample space : 52 cardssample space : 52 cards

Outcomes which satisfy elementary event A: 4 jacks

Outcomes which satisfy elementary event B: 13 clubs.

J

9

JJ 5Q

3

4

2

1

7

J8

6

10 K

Outcomes which satisfy both the events : the jack of clubs

In this case the event A (“choosing a jack”) and the event B (“choosing a club”) are not disjoint.

THE EVENTS HAVE AN INTERSECTIONTHEY ARE NOT DISJOINTTHEY ARE NOT MUTUALLY EXCLUSIVEHOW TO EVALUATE THE PROBABILITY IN THIS CASE OF NOT MUTUALLY EXCLUSIVE EVENTS?

How many outcomes are possible (= form the sample space)?

How to evaluate the probability in this case of not mutually exclusive events?

52 (number of cards)

How many outcomes are favorable?

How to evaluate the number of favorable cases when events are not mutually exclusive?

4 (number of jacks)

+13

(number of clubs )

17 (number of jacks OR clubs)

Is it correct?NO!


sample space : 52 cardssample space : 52 cards

4 jacks

13 clubs

J

9

JJ 5Q

3

4

2

1

7

J8

6

10 K

Outcome which satisfies both the events : the jack of clubs

The real number of favorable cases is 16, and not 17: we have counted the jack of clubs twice!!!!!!!

57

2

9

4

97

10

10

K

1

Q 6

3

7

4

3 21

3


The correct procedure to find the number of favorable cases is:

n(fav.cases) = n(jacks) + n(clubs) n(jacks clubs)

= 4 + 13 1 = 16

We must subtract the number of the elements of the intersection not to count them twice!

So, the probability of choosing a jack OR a club is

52

1134)(

BAP

13

4

52

16

But let’s write the first fraction in another way

52

1134)clubsjack(

P

52

4

52

13

52

1

P(jack)

P(jackclubs)

P(clubs)OR

AND

GENERAL FORMULA

We can generalize what we have just found:

P(jackclubs) = P(jacks) +P(clubs) P(jackclubs)

P(AB) = P(A) +P(B) P(AB)

AN OTHER EXPERIMENT

Playing tombola, what is the probability that the first extracted number is

MULTIPLE OF 10 (A) OR GREATER THAN 70 (B)?

The events A and B are not disjont!We will use formula for not mutually

exclusive events

P(AB) = P(A) +P(B) P(AB)

Playing tombola, what is the probability that the first extracted number is MULTIPLE OF 10 (A) OR GREATER THAN 70 (B)?

P(A)= 10,010

1

90 10,20,30,40…,90 9

P(B)= 22,09

2

90 71,72,73,…89,90 20

P(AB)= 02,090

80,90 2

30,010

3

90

27

90

2

9

2

10

1)70mult ( P

When two events can’t occurcan’t occur contemporarily, there’s no element at the contemporarily, there’s no element at the


LET’S REVIEW : the addition rule

A B =

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

When two events can can occuroccur contemporarily, contemporarily, there’s some element at the intersection there’s some element at the intersection

between A and B between A and B

LET’S REVIEW : the addition rule

A B

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

Documents

WHAT IS PROBABILITY? prof. Renzo Nicolini I.M.”G.CARDUCCI” - Trieste