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

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

WHAT IS PROBABILITY? CLIL project Class II C

LESSON 1 CLIL project Class II C

derives from the Latin probare (to prove, or to test).
WHAT IS PROBABILITY? The word probability derives from the Latin probare (to prove, or to test).

PROBABLE is almost synonym of likely hazardous risky uncertain
WHAT IS PROBABILITY? PROBABLE is almost synonym of likely hazardous risky uncertain doubtful

theory of probability attempts to quantify the notion of probable.
WHAT IS PROBABILITY? theory of probability attempts to quantify the notion of probable. HOW PROBABLE SOMETHING IS? /LIKELY To answer, we need a number!!!!!

The scientific study of probability is a modern development. Gambling
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.

Christian Huygens(1657) Daniel Bernoulli (1713)
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 Pascal

Vocabulary An experiment An outcome An event Probability
is a situation involving chance or probability that leads to results called outcomes. An outcome is the result of a single trial of an experiment. An event is one or more outcomes of an experiment. Probability is the measure of how likely an event is.

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

Outcomes! Rolling a single 6-sided die: “a number six was drawn”
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” Possible outcomes in the experiment Impossible outcome in the experiment

Outcomes! Experiment: Driving a car race Outcome: “Schumacher wins”
EVENT Outcomes! Experiment: Driving a car race Outcome: “Schumacher wins”

Outcomes! Experiment: Picking a card from a deck
EVENT Outcomes! Experiment: Picking a card from a deck Outcome: “A king is drawn”

Outcomes! Experiment: Tossing a coin Outcome: “a tail has been tossed”
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.

Outcomes! Examples Experiment: Rolling a die once:
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?” This is OUR event! This is OUR event!

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”?

WHEN I DO SOMETHING I SAY THAT I CARRY OUT AN
LET’S REPEAT!! WHEN I DO SOMETHING I SAY THAT I CARRY OUT AN EXPERIMENT EXAMPLES?

LET’S REPEAT!! OUTCOME EXAMPLES?
ANY POSSIBLE SITUATION THAT OCCURS WHEN I CARRY OUT THE EXPERIMENT IS AN OUTCOME EXAMPLES?

LET’S REPEAT!! SAMPLE SPACE EXAMPLES?
ALL THE POSSIBLE OUTCOMES THAT CAN OCCUR WHEN I EXECUTE THE EXPERIMENT, FORM THE SAMPLE SPACE EXAMPLES?

THE PARTICOLAR OUTCOME or SET OF OUTCOMES WE’RE INTERESTED IN IS AN
LET’S REPEAT!! THE PARTICOLAR OUTCOME or SET OF OUTCOMES WE’RE INTERESTED IN IS AN EVENT EXAMPLES?

THE MEASURE OF HOW LIKELY AN EVENT IS, IS CALLED
LET’S REPEAT!! THE MEASURE OF HOW LIKELY AN EVENT IS, IS CALLED PROBABILITY

HOW TO EVALUATE PROBABILITY?
Probability is a number! We need a formula or a procedure to find it!

THERE ARE THREE POSSIBLE WAYS TO FIND THIS VALUE CLASSICAL DEFINITION SUBJECTIVE PROBABILITY FREQUENTIST DEFINITION

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 ( )

CLASSICAL PROBABILITY
SIMON DE LAPLACE ( ) 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).

A single 6-sided die is rolled.
EXAMPLE/EXERCISE A single 6-sided die is rolled. What is the probability of each outcome? What is the probability of rolling an even number? Of rolling an odd number?

All the values are the same!!! The outcomes are equally likely
Outcomes: The possible outcomes of this experiment are 1, 2, 3, 4, 5 and 6. P(1) = number of ways to roll a 1 = 1 total number of sides P(2) = number of ways to roll a 2 = 1 total number of sides P(3) = number of ways to roll a 3 = 1 total number of sides P(4) = number of ways to roll a 4 = 1 total number of sides P(5) = number of ways to roll a 5 = 1 P(6) = number of ways to roll a 6 = 1 total number of sides All the values are the same!!! The outcomes are equally likely .

EQUALLY LIKELY EVENTS HAVE THE SAME PROBABILITY TO OCCUR

What is the probability of rolling an even number?
P(even) = # ways to roll an even number # total number of sides probability of rolling an even number is one half = 0,5

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

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”.

Some more about the formula for probability
Probability Of An Event P(A) = The Number Of Ways Event A Can Occur The Total Number Of Possible Outcomes Some more about the formula for probability The “impossible” event The “certain event”

THE EVENT IS IMPOSSIBLE!
Is it possible that there are no ways event A can occur? SURE! In this case the formula for probability The Number Of Ways Event A Can Occur The Total Number Of Possible Outcomes has numerator equal to 0! P(A) =0 THE EVENT IS IMPOSSIBLE!

Is it possible that there are no ways event A can occur?
THE EVENT IS IMPOSSIBLE? P(A) =0 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 The Total Number Of Possible Outcomes The Number Of Ways Event A Can Occur is 0 because number 7 doesn’t exist in such a die!!! P(A) =0

Is it possible that event A certainly will occur?
SURE! In this case the formula for probability The Number Of Ways Event A Can Occur The Total Number Of Possible Outcomes has numerator equal to the denominator. The fraction values 1 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 P(A) =1

THE EVENT IS CERTAIN! P(A) =1 It will certainly happen!

LET’S REPEAT!! PROBABILITY is a positive real number, between 0 and 1
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 The number of elements of the sample space

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!

WHEN I DO SOMETHING I SAY THAT I EXECUTE AN
LET’S REPEAT!! WHEN I DO SOMETHING I SAY THAT I EXECUTE AN EXPERIMENT

ANY POSSIBLE SITUATION THAT OCCURS WHEN I EXECUTE THE EXPERIMENT IS AN
LET’S REPEAT!! ANY POSSIBLE SITUATION THAT OCCURS WHEN I EXECUTE THE EXPERIMENT IS AN OUTCOME ALL THE POSSIBLE OUTCOMES THAT CAN OCCUR WHEN I EXECUTE THE EXPERIMENT, FORM THE SAMPLE SPACE

LET’S REPEAT!! EVENT PROBABILITY
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

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 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? Let’s calculate the probability of Ā

What do we need? Favorable cases. All the elements of subset 1,3,4,5,6  Possible cases. 1,2,3,4,5,6 

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 OBVIOUSLY 1 6

Probability one-sixth for EVENT A Probability five-sixth for EVENT Ā
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 FORTUITOUS 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) In fact, the number of cases for A and the number of cases for Ā exhaust all the possibilities, i.e. all the possible cases. Let’s divide the equality by n(E)

P(A) + P (Ā) = 1 and than, with an obvious (?) passage,

WE GOT THE RESULT: P(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? P (not a king) = 1 – P(king)

We can also indicate it as Ā
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

P(A) + P (Ā) = 1 LET’S REPEAT!!
We have proved that the following formula for finding the value of the complement of an event: P(A) + P (Ā) = 1

We have proved the following formula for the complement :
LET’S REPEAT!! We have proved the following formula for the complement : P(A) + P (Ā) = 1

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? P (not a king) = 1 – P(king)

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

Experiment. Let’s suppose to propose an experiment in which we do two actions. AND 1. We roll a die… 2. We take a number, playing tombola and …

Experiment. …we set up an event which includes both actions
For instance: What is the probability of drawing an odd number from the sack of tombola 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 independent if the fact that A occurs does not affect the 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 number from the sack of tombola AND rolling a multiple of 3 on the die? We are looking for the probability of a more complicated event which involves two simpler INDEPENDENT events

COMPOUND EVENT TWO ACTIONS ONE REQUIREMENT CONNECTOR AND

This is a typical example of COMPOUND EVENT

In the COMPOUND EVENT we want that both the events occur

In the COMPOUND EVENT we want that one AND the other event occur
The KEY CONJUNCTION is

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.

This is called the “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”

Experiment. What is the probability of choosing an odd number from the sack of tombola AND rolling a multiple of 3 on the die? EVENT A = “drawing an odd tombola number” P(odd) EVENT B = “rolling a multiple of 3” P(3,6)

Experiment. P(odd AND die) = P(odd) * P(die)
What is the probability of drawing an odd number from the sack of tombola AND rolling a multiple of 3 on the die. We can use the multiplication rule because the two events are evidently independent! P(odd AND die) = P(odd) * P(die)

OTHER EXPERIMENTS 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 AND 3) P(3 on die)

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?

P(jack) P(eigth) P(jack AND eight)
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(eigth) 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(eigth) P(jack)

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.

PROBABILITY HAS CHANGED!
DEPENDENT EVENTS P(jack) P(eigth) P(jack AND eight) 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 Possible cases = 16 with replacing P(yellow)
P(green) P(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
Possible cases = without replacing P(yellow) P(green) P(yellow AND green)

ANOTHER EXPERIMENT 3 red, 5 green, 2 blue and 6 yellow
with replacing P(yellow AND green) with no replacing P(yellow AND green) PROBABILITY HAS CHANGED AGAIN!

P(A and B) = P(A) · P(B | A)
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) P(A and B) = P(A) · P(B | A) Read: B occurs given that event A has occurred

LET’S REPEAT : the multiplication rule
When two events, A and B, are independent, the probability of both occurring is: P(A and B) = P(A) · P(B) When two events, A and B, are dependent, the probability of both occurring is P(A and B) = P(A) · P(B | A) The usual notation for "event B occurs given that event A has occurred" is “B | A" (B given A).

LET’S REPEAT!! Independent events
Two events, A and B, are independent if the fact that A occurs does not affect the probability of B occurring.

The rolling of a die and the drawing of a tombola number are
LET’S REPEAT!! The rolling of a die and the drawing of a tombola number are INDEPENDENT events!

LET’S REPEAT!! When we have two events and we want that both of those occur, we are considering a COMPOUND EVENT

In the COMPOUND EVENT we want that one AND the other event occur
LET’S REPEAT!! In the COMPOUND EVENT we want that one AND the other event occur AND The KEY CONJUNCTION is

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.

This is called the “multiplication rule”
LET’S REPEAT!! 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”

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

P(jack) P(eigth) P(jack AND eight)
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(eigth) P(jack AND eight)

LET’S REPEAT : the multiplication rule
When two events, A and B, are independent, the probability of both occurring is: P(A and B) = P(A) · P(B) When two events, A and B, are dependent, the probability of both occurring is P(A and B) = P(A) · P(B | A) The usual notation for "event B occurs given that event A has occurred" is “B | A" (B given A).

MORE FAVORABLE OUTCOMES
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

Experiment. MORE FAVORABLE OUTCOMES
We choose a card from an ordinary deck, and we accept either a King OR a numbered card. first favorable event second favorable event

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 (or disjoint) if it is impossible for them to occur together.

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

ONE ACTION TWO REQUESTS CONNECTOR OR

sum of the probabilities of each event.
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.

This is called the 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 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 Possible cases = 16 P(yellow) P(green)
second favorable event first favorable event P(yellow OR green)

P(A or B) ONE MORE EXERCISE MUTUALLY EXCLUSIVE EVENTS
What is the probability of rolling a 2 OR a 5 on a single 6-sided die? MUTUALLY EXCLUSIVE EVENTS P(A  B) = P(A) + P(B) Read: OR P(A or B)

Formally, if two events A and B are mutually exclusive we can write: A  B =  WHY? WHAT DOES IT MEAN?

A  B =  Outcomes which satisfy elementary event A sample space A B There aren’t outcomes which contemporarily satisfy event A and event B Outcomes which satisfy elementary event B A  B =  → Disjoint sets

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 contemporarily green and brown eyes
A  B =  … In a class: A : green-eyes-student B : brown-eyes-student A B Obviously, no student has contemporarily green and brown eyes

Obviously, no student has contemporarily green AND brown eyes
A  B =  … In a class: Obviously, no student has contemporarily green AND brown 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

LET’S REPEAT : the addition rule
When two events can’t occur contemporarily, they are said - MUTUALLY INDEPENDENT EVENTS - DISJOINT EVENTS

When two events can’t occur contemporarily, there’s no element in the intersection between A and B A  B =  P(A  B) = P(A) + P(B)

When two events can’t occur contemporarily, they are said - MUTUALLY INDEPENDENT EVENTS - DISJOINT EVENTS

When two events can’t occur contemporarily, there’s no element in the intersection between A and B A  B =  P(A  B) = P(A) + P(B)

LET’S SEE AN EXAMPLE! 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 TWO REQUESTS CONNECTOR OR (one picking of a card)
(club; jack) CONNECTOR OR

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

THEY ARE NOT MUTUALLY EXCLUSIVE THE EVENTS HAVE AN INTERSECTION
In this case the event A (“choosing a jack”) and the event B (“choosing a club”) are not disjoint. Outcomes which satisfy elementary event A: 4 jacks sample space : 52 cards Outcomes which satisfy both the events : the jack of clubs 7 2 3 J J J Q 8 5 10 J K 1 4 9 6 HOW TO EVALUATE THE PROBABILITY IN THIS CASE OF NOT MUTUALLY EXCLUSIVE EVENTS? THEY ARE NOT MUTUALLY EXCLUSIVE THE EVENTS HAVE AN INTERSECTION THEY ARE NOT DISJOINT Outcomes which satisfy elementary event B: 13 clubs.

How many outcomes are possible (= form the sample space)?
How to evaluate the probability in this case of not mutually exclusive events? How many outcomes are possible (= form the sample space)? 52 (number of cards)

(number of jacks OR clubs)
How to evaluate the number of favorable cases when events are not mutually exclusive? How many outcomes are favorable? 4 (number of jacks) + Is it correct? NO! 13 (number of clubs  ) 17 (number of jacks OR clubs)

How to evaluate the number of favorable cases when events are not mutually exclusive?
sample space : 52 cards Outcome which satisfies both the events : the jack of clubs 4 jacks 7 K 9 7 7 2 2 4 3 J J J Q 8 5 5 7 10 J K 4 10 1 4 10 9 6 1 13 clubs  Q 6 1 3 3 2 3 9 The real number of favorable cases is 16, and not 17: we have counted the jack of clubs twice!!!!!!!

The correct procedure to find the number of favorable cases is:
How to evaluate the number of favorable cases when events are not mutually exclusive? 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!

But let’s write the first fraction in another way
So, the probability of choosing a jack OR a club  is But let’s write the first fraction in another way

P(jack) OR P(clubs) P(jackclubs) AND

We can generalize what we have just found:
GENERAL FORMULA We can generalize what we have just found: P(jackclubs) = P(jacks) +P(clubs)  P(jackclubs) P(AB) = P(A) +P(B)  P(AB)

AN OTHER EXPERIMENT The events A and B are not disjont!
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(AB) = P(A) +P(B)  P(AB)

9 10,20,30,40…,90 P(A)= 20 71,72,73,…89,90 P(B)= 2 80,90 P(AB)=
Playing tombola, what is the probability that the first extracted number is MULTIPLE OF 10 (A) OR GREATER THAN 70 (B)? 9 10,20,30,40…,90 P(A)= 20 71,72,73,…89,90 P(B)= 2 80,90 P(AB)=

LET’S REVIEW : the addition rule
When two events can’t occur contemporarily, there’s no element at the intersection between A and B A  B =  P(A  B) = P(A) + P(B)

LET’S REVIEW : the addition rule
When two events can occur contemporarily, there’s some element at the intersection between A and B A  B   P(A  B) = P(A) + P(B) - P(A  B)

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

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