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
WHAT IS PROBABILITY?
The word probability
derives
from the Latin probare
(to prove, or to test).
WHAT IS PROBABILITY?
PROBABLE
is almost synonym of
• likely
• hazardous
• risky
• uncertain
• doubtful
WHAT IS PROBABILITY?
theory of probability
attempts to quantify the
notion of probable.
HOW PROBABLE /LIKELY SOMETHING IS?
To answer, we need a number!!!!!
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 Pascal
Vocabulary
An experiment
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.
EXPERIMENT
Experiments!





Rolling a single 6-sided die
Running an horse race
Driving a car race
Picking a card from a deck
Tossing a coin.
OUTCOME
EVENT
EXPERIMENT
OUTCOME
Outcomes!
EVENT
Rolling a single 6-sided die:
“a number six was drawn”
Possible outcomes
in the experiment
(to be drawn = uscire);
“a number three was drawn” ;
“a number eight can’t be drawn”
Impossible outcome
in the experiment
EXPERIMENT
Outcomes!
OUTCOME
EVENT
Experiment: Driving a car race
Outcome: “Schumacher wins”
EXPERIMENT
Outcomes!
OUTCOME
EVENT
Experiment: Picking a card from a deck
Outcome: “A king is drawn”
EXPERIMENT
Outcomes!
OUTCOME
EVENT
Experiment: Tossing a coin
Outcome: “a tail has been tossed”
EXPERIMENT
Outcomes!
OUTCOME
EVENT
THE SET OF ALL THE POSSIBLE
OUTCOMES IS CALLED SAMPLE
SPACE and is denoted by S.
EXPERIMENT
Outcomes! Examples
OUTCOME
EVENT

Experiment: Rolling a die once:

Experiment: Tossing a coin:

Experiment: Measuring the height (cms)
of a girl on her first day at school:
• Sample space S = {1,2,3,4,5,6}
• Sample space S = {Heads,Tails}
• Sample space S = the set of all (?) possible
real numbers
EXPERIMENT
Event!
OUTCOME
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
This is OUR event!
are picked up from a deck of card”?
“Rolling a die once, how possible is it that
This is OUR event!
the score is < 4?”
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?
LET’S REPEAT!!
ANY POSSIBLE SITUATION THAT OCCURS WHEN I CARRY OUT
THE EXPERIMENT IS AN
OUTCOME
EXAMPLES?
LET’S REPEAT!!
ALL THE POSSIBLE OUTCOMES THAT CAN OCCUR WHEN I
EXECUTE THE EXPERIMENT, FORM THE
SAMPLE SPACE
EXAMPLES?
LET’S REPEAT!!
THE PARTICOLAR OUTCOME or SET OF OUTCOMES WE’RE
INTERESTED IN IS AN
EVENT
EXAMPLES?
LET’S REPEAT!!
THE MEASURE OF HOW LIKELY AN EVENT IS, IS CALLED
PROBABILITY
WHAT IS PROBABILITY?
CLIL project
Class II C
LESSON 2
CLIL project
Class II C
HOW TO EVALUATE PROBABILITY?
Probability is a number!
We need a formula or a procedure to find it!
HOW TO EVALUATE PROBABILITY?
THERE ARE THREE POSSIBLE
WAYS TO FIND THIS VALUE



CLASSICAL DEFINITION
SUBJECTIVE PROBABILITY
FREQUENTIST DEFINITION
HOW TO EVALUATE PROBABILITY?
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
HOW TO EVALUATE PROBABILITY?
WE’LL SEE ONLY THE
CLASSICAL DEFINITION OF
PROBABILITY.
by
SIMON DE LAPLACE (1749-1827)
CLASSICAL PROBABILITY
SIMON DE LAPLACE (17491827) 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 FORMULA FOR THE CLASSICAL
PROBABILITY
The probability of event A is the
number of ways event A can occur
divided by the total number of
possible outcomes.
THE FORMULA FOR THE CLASSICAL
PROBABILITY
The probability of event A is the
number of favorable cases
(outcomes) divided by the total
number of possible cases
(outcomes).
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?
Outcomes: The possible outcomes of this
experiment are 1, 2, 3, 4, 5 and 6.
P(1) = number of ways to roll a 1
total number of sides
P(2) = number of ways to roll a 2
total number of sides
P(3) = number of ways to roll a 3
total number of sides
P(4) = number of ways to roll a 4
total number of sides
P(5) = number of ways to roll a 5
total number of sides
P(6) = number of ways to roll a 6
total number of sides
= 1
6
= 1
6
= 1
6
= 1
6
= 1
6
= 1
6
All the values
are the
same!!! The
outcomes are
equally likely
.
EQUALLY LIKELY EVENTS
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  3  1
6 2
# 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
3
 0.5 
6
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”.
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”
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?
In this case SURE!
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!
Is it possible that event A certainly will occur?
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) =1
Is it possible that event A certainly will occur?
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!
WHAT IS PROBABILITY?
CLIL project
Class II C
LESSON 3
CLIL project
Class II C
LET’S REPEAT!!
WHEN I DO SOMETHING I SAY THAT I EXECUTE AN
EXPERIMENT
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!!
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 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 FORMULA FOR THE CLASSICAL
PROBABILITY
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
Rolling a die, which is the probability of rolling any
number except 2?
Let’s calculate the
probability of Ā
PROBLEM
Rolling a die, which is the probability of
rolling any number except 2?
What do we need?
Favorable cases.
All the elements of subset
1,3,4,5,6 
Possible cases.
1,2,3,4,5,6 
PROBLEM
Rolling a die, which is the probability of
rolling any number except 2?
The probability of this
event is
5
6
PROBLEM
Rolling a die, which is the probability of
rolling any number except 2?
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
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
PROBABILITY OF THE COMPLEMENT
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

n A  n A nE 

nE 
nE 
and than, with an obvious (?) passage,

n A n A

1
n  E  n E 

P  A  P A  1
PROBABILITY OF THE COMPLEMENT
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)
PROBABILITY OF THE COMPLEMENT
In general, if we know the
probability of an event, we can
immediately calculate the
probability of its complement !
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)
4
48 12
 1


52 52 13
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!!
We have proved that the following
formula for finding the value of the
complement of an event:
P(A) + P (Ā) = 1
WHAT IS PROBABILITY?
CLIL project
Class II C
LESSON 4
CLIL project
Class II C
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)
4
48 12
 1


52 52 13
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.
1. We roll a die…
AND
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
COMPOUND EVENT
This is a typical example of
COMPOUND EVENT
COMPOUND EVENT
In the COMPOUND
EVENT we want that
both the events occur
COMPOUND EVENT
In the COMPOUND
EVENT we
want that one AND the other
event occur
The KEY CONJUNCTION is
AND
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”
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)
45 1


90 2
EVENT B = “rolling a multiple of 3”
P(3,6)
2 1
 
6 3
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 can use the multiplication rule because the two events are
evidently independent!
1 1 1
P(odd AND die) = P(odd) * P(die) 
 
2 3 6
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(3 on die)
1

2
1

6
P(head AND 3)
1 1 1
  
2 6 12
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)
4
1


52 13
4
1
P(eigth) 

52 13
P(jack AND eight)
1 1
1
  
13 13 169
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)
4
1


52 13
P(eigth)
4

51
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)
4
1


52 13
P(jack AND eight)
P(eigth)
4

51
1 4
4
 

13 51 663
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
6
3
P(yellow) 

16 8
P(yellow AND green)
with replacing
P(green)
3 5
15
 

8 16 128
5

16
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 = 16
6
3
P(yellow) 

16 8
without replacing
P(green)
P(yellow AND green)
5 1


15 3
3 1 1
  
8 3 8
ANOTHER EXPERIMENT
3 red, 5 green, 2 blue and 6 yellow
with replacing
P(yellow AND green)
with no replacing
P(yellow AND green)
3 5
15
 

8 16 128
3 1 1
  
8 3 8
PROBABILITY HAS CHANGED AGAIN!
COMPOUND EVENT

If the events are independent, multiplication rule is valid
and probability is just
P(A and B) = P(A) · P(B)

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)
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).
WHAT IS PROBABILITY?
CLIL project
Class II C
LESSON 5
CLIL project
Class II C
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.
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
LET’S REPEAT!!
In the COMPOUND
EVENT we
want that one AND the other
event occur
The KEY CONJUNCTION is
AND
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!!
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
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)
4
1


52 13
4
1
P(eigth) 

52 13
P(jack AND eight)
1 1
1
  
13 13 169
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)
MUTUALLY EXCLUSIVE 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
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 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)
6
3


16 8
first favorable event
P(yellow OR green)
P(green)
5

16
second favorable event
3 5 11
 

8 16 16
ONE MORE EXERCISE
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)
1 1 1
  
6 6 3
MUTUALLY EXCLUSIVE EVENTS
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
Outcomes which satisfy
satisfy event A and event B elementary event B
A  B =  → Disjoint sets
A  B =  … In a class:


Event A : randomly, choosing a greeneyes-student;
Event B : randomly, choosing a browneyes-student;
A  B =  … In a class:
A : green-eyes-student
B : brown-eyes-student
A
B
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
LET’S REPEAT :
the addition rule
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)
WHAT IS PROBABILITY?
CLIL project
Class II C
LESSON 6
CLIL project
Class II C
LET’S REPEAT :
the addition rule
When two events can’t occur
contemporarily, they are said
- MUTUALLY INDEPENDENT EVENTS
- DISJOINT EVENTS
LET’S REPEAT :
the addition rule
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)
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 OR
NOT MUTUALLY EXCLUSIVE EVENTS
In this case the event A (“choosing a
jack”) and the event B (“choosing a
club ”) are not disjoint.
Outcomes
which satisfy
In this case the event
A (“choosing
a jack”) and
elementary event A: 4 jacks
the event B (“choosing a club”) are not disjoint.
sample space : 52 cards
Outcomes which satisfy both the
events : the jack of clubs
7
J
J
J
J
10
1
2
3
Q
K
6
8
5
4
9
HOW
TO
EVALUATE
THEHAVE
PROBABILITY
IN THIS
CASE
Outcomes
which
satisfy
THEY
THE
THEY
ARE
EVENTS
ARE
NOT
NOT
MUTUALLY
DISJOINT
AN INTERSECTION
EXCLUSIVE
elementary
event B: 13 clubs.
OF NOT MUTUALLY EXCLUSIVE
EVENTS?
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)
How to evaluate the number of favorable cases
when events are not mutually exclusive?
How many outcomes are favorable?
4
(number of jacks)
+
13
(number of clubs  )
Is it
NO!
correct?
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
7
2
J
5
J
7
4
J
10
J
1
10
1
Q
9
7
6
3
9
3
2
1
3
2
Q
8
5
K
6
4
3
10
4
9
13 clubs 
The real number of favorable cases is 16, and not 17:
we have counted the jack of clubs twice!!!!!!!
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)
=
4
+ 13
 n(jacks  clubs)
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
4  13  1 16 4


P( A  B) 
52 13
52
But let’s write the first fraction in another way
4  13  1
4
13
1
P( jack  clubs ) 



52
52
52 52
P(jack)
OR
P(clubs)
P(jackclubs)
AND
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
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)
Playing tombola, what is the probability that the first extracted
number is
MULTIPLE OF 10 (A) OR GREATER THAN 70 (B)?
1
P(A)=   0,10
90 10
10,20,30,40…,90
20 2
P(B)=   0,22
90 9
71,72,73,…89,90
9
2
 0,02
P(AB)=
90
80,90
1 2 2 27 3
P(mult   70)   


 0,30
10 9 90 90 10
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)