INTRODUCTION
The word probability derives from the Latin probare (to prove, or to test).
The word PROBABLE is almost synonym of likely hazardous, risky, uncertain, doubtful and the theory of probability attempts to quantify the notion of probable.
The question is:
HOW PROBABLE/LIKELY SOMETHING IS?
To answer, we need a number. Thetheory of probability is a mathematical theory to find this number.
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.
When?
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).
VOCABULARY
To start, we need some terms:
- 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.
Examples of:
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” (to be drawn = uscire); “a number three was drawn” ; “a number eight can’t be drawn”
Driving a car race:“Schumacher wins”
Picking a card from a deck : “A king is drawn”
Tossing a coin:“a tail has been tossed”
The set of all the possible outcomes is called sample space and is denoted by S.
Rolling a die once: Sample space S = {1,2,3,4,5,6}
Tossing a coin: Sample space S = {Heads,Tails}
Measuring the height (cms) of a girl on her first day at school: Sample space S = the set of all possible real numbers.
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 “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?”
Probability
(It’s the value we estimate for a single event)
“Which is the probability that a Queen is picked up from a deck of card”?
“Which is the probability that a Jack OR a King is picked up from a deck of card”?
LET’S CALCULATE THE 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.
During this course we’ll see the classical definition of probability, proposed by Simon De Laplace (1749-1827) .
In order to measure probabilities, he has proposed the following formula for finding the probability of an event.
Probability Of An Event P(A) = The Number Of Ways an Event A Can Occur
The Total Number Of Possible Outcomes
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 favourable cases (outcomes) divided by the total number of possible cases (outcomes).
Examples:
A single 6-sided die is rolled. What is the probability of each outcome?
The possible outcomes of this experiment are 1, 2, 3, 4, 5 and 6. The outcomes are equally likely
Equally likely events have the same probability to occur.
What is the probability of rolling an even number ?
The probability of rolling an even number is one half = 0,5
Of rolling an odd number?
The probability of rolling an odd number is one half = 0,5
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 must know everything about the experiment BEFOREresults (=outcomes)come out. That’s why we say that classical probability is a “ a priori” (Latin: “da prima”)
IMPOSSIBLE EVENT. CERTAIN EVENT.
Some more about the formula for probability:TThe “impossible” event, and Tthe “certain event”
Is it possible that there are no ways event A can occur?SURE!
In this case the formula for probability
has numerator equal 0!
P(A) =0.THE EVENT IS IMPOSSIBLE!
It has no probability to happen!
Which is the probability of rolling number 7 on a 6 sided die?
The nNumber oOf wWays eEvent AcCan oOccuris 0. P(A) =0
Is it possible that event A certainly will occur?SURE!
In this case the formula for probability 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?
WHICH IS THE PROBABILITY THAT, ROLLING A DIE, A NUMBER BETWEEN 0 AND 7 COMES OUT ? P(A) = 6/6 = 1 It certainly will happen!
P(A) =1. THE EVENT IS CERTAIN!
P(A) =1. It certainly will happen!
LET’S REPEAT!!
PROBABILITY is a positive real number, between 0 and 1; Zero for the impossible event; One for the certain event.
TO FIND THE CLASSICAL PROBABILITY (Laplace) we need the following definition:
Probability Of An Event P(A) = The Number Of Ways an Event A Can Occur
The Total Number Of Possible Outcomes
(at the denominator there’s the number of elements of the sample space)
WHEN TWO EVENTS HAVE THE SAME PROBABILITY, WE SAY THAT THEY ARE EQUALLY LIKELY. Heads and tails are equally likely!!!
THE COMPLEMENT OF THE EVENT
Which is the probability that next time you’ll appreciate our CLIL lesson? Classical probability doesn’t give any answer to this question, because it’s not a problem solving “a priori”. 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!!
I HOPE THIS PROBABILITY IS NOT ZERO!
IImagine to be asked to solve the following exercise:
Rolling a die, which is the probability of rolling any number except 2?
[WHICHEVER number BUT 2 means: any, BUT NOT number 2]
The statement “any number except 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”
We say that the EVENT “any number except 2” is the complement of the event A COMPLEMENT OF THE EVENT A “rolling a number 2”
It’s the opposite statement of the EVENT A; Wwe use to indicate it with Ā. Ā (A bar) is the complement of A
Let’s calculate the probability of Ā .
What do we need?
- Favourable 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 then 5/6.
Can we solve this problem in another way? Yes… but how?
Let’s start considering the problem of the EVENT A: “Probability of rolling a 2”. It’s obviously
IT’S OBVIOUSLY: 1/6
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. OBSERVE THAT ONE-SIXTH PLUS FIVE-SIXTH EQUAL TO 1.
The total probability we found is 1. In other words, the sum between P(A) and P (Ā) is 1.
P(A) + P (Ā) = 1
Is it a fortuitous caseCOMBINAZIONE)? NO! It’s a general rule!!!! Let’s prove it!
PROOF
- nLet n(A) be the number of favourable cases for the event A
- Let n(Ā) be the number of favourable 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)
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)
PE \* MERGEFORMAT
WE GOT THE RESULT: P(A) + P (Ā) = ; OR, IN OTHER TERMS, The probability P (Ā) of the complement of an event A is given by the subtraction between 1 and the probability P (A) of the event A.
P(A)+P (Ā) = 1 - P(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 !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
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?
is not a King? P (not a king) = 1 – P(king)
COMPOUND EVENT FORMED BY INDEPENDENT EVENTS
LET’S REPEAT!! When an event B has the opposite request of an event A, we say that the event B is the
COMPLEMENT OF EVENT A. We can also indicate it as Ā
We have proved that the following formula for the complement values: P(A) + P (Ā) = 1
COMPOUND EVENT
Let’s see now a more complicated situation involving more actions.
Let’s suppose to propose an experiment in which we do two actions: We 1.(roll a die)…AND…2. (Playing Tombola, Wwe take a number), playing tombola
Aand we set up an event which includes both actions.
This is a typical example of compound event. In a compound event we want that both the events occur
In other words, in the compound event we want that one AND the other event occur
Example:
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?
First of all, we can notice that the rolling of the die and the drawing of the number are twoIindependent eventsNDEPENDENT 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 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 !!!!!
We could prove that
Let’s solve the exercise:
What is the probability of choosing 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
TWO ACTIONS
ONE REQUEST
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 AND.
We could prove that
the probability of the compound event is always the product between the single probabilities of two independent events which form the compound event.
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.
What is the probability of drawing an odd number from the sack of tombola AND rolling a multiple of 3 on the die?
- EVENT A = “drawing an odd number in tombola”
P(odd)
- EVENT B = “rolling a multiple of 3”
P(3,6)
Which is the probability of choosing 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)
Example
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) = ½1/2 P(3 on die)= 1/6P(head AND 3)
Example
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
Let’s consider the last exercise. But wWhat happens in that case, if we don’t decide 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 card in the deck! So, in the second draw, the possible cases would be 51 (one card has been removed!), while the favourable cases of picking an eight would remain the same (4).
P(jack) =4/52 = 1/13P(eight)=4/51
THE EVENTS AREN’T INDEPENDENT ANYMORE. THEY ARE DEPENDENT.
P(jack) P(eigth) P(jack AND eight)
PROBABILITY HAS CHANGED!
AN OTHER EXPERIMENTExamples
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?
Possible cases = 16 w with replacinreplacing_ g
P(yellow)P(green) P(yellow AND green)
But what should have happenedhappens if we hadn’t decideddecide for the putting back of the first marble in the jar ?
Possible cases = 16 without replacingwithout replacing
P(yellow)P(green) P(yellow AND green)
PROBABILITY HAS CHANGED AGAIN!
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. In symbols, we indicate this fact with the following expression
P(A and B) = P(A) · P(B | A)
MORE FAVOURABLE OUTCOMES in MUTUALLY EXCLUSIVE EVENTS
MORE FAVORABLE OUTCOMES
Let’s see know a new situation, involving only one experiment and one action (one rolling of the die, one draw of a card, and so on…), but where we acceptMORE FAVOURABLE OUTCOMES
Example
We draw a card from an ordinary deck, and we accept either a King OR a numbered card.
First of all, we can notice that the drawing of a King and the drawing of a numbered card CANNOT OCCURE at the same time.
We say that: they are twomutually exclusive events (disjoint events)
Two events are mutually exclusive (or disjoint), if it is impossible for them to occur together.
We are looking for the probability of a more complicated event which involves two simpler mutually exclusive events.
Example
A jar contains 3 red, 5 green, 2 blue and 6 yellow marbles. A marble is drawn at random from the jar. What is the probability of picking a green OR a yellow marble?
P(yellow)P(green)P(yellow OR green)
Example
What is the probability of rolling a 2 OR a 5 on a single 6-sided die?
P(A or B)= 1/6 +1/6 = 2/6 = 1/3
Formally, if two events A and B are mutually exclusive we can then write: A B =
Why? What does it mean?
Example
A B = … In a class:
- Event A : randomly, choose a green-eyes-student;
- Event B : randomly, choose a brown-eyes-student;
Let’s use Venn’s diagrams to explain the situation and the symbolism.
Obviously, no student has contemporarily green and brown eyes
Therefore, the probability of choosing an element of A and an element of B is zero.
There isn’t intersection. There aren’t outcomes which contemporarily satisfy event A and event B
A B = → Disjoint sets→ Disjoint events
When two events can’t occur contemporarily, they are said to be
MUTUALLY INDEPENDENT EVENTS
DISJOINT EVENTS
MORE FAVOURABLE OUTCOMES in NOT MUTUALLY EXCLUSIVE EVENTS.
“TOTAL PROBABILTY”
What about, when two events can occur contemporarily?
i.e. What is the probability of two not mutually exclusive events?.
Let’s an example.
Example
A card is chosen at random from a deck of 52 cards. What is the probability of drawig a jack OR a club ?
ONE ACTIONTWO REQUESTSCONNECTOR OR
(one draw of a card) (club; jack)
In this case the event A (“drawing a jack”) and the event B (“drawing a club ”) are not disjoint.
Example
A B = … In a class:
- Event A : randomly, choose a green-eyes-student;
- Event B : randomly, choose a brown-eyes-student;
Let’s use Venn’s diagrams to explain the situation and the symbolism.
The events have intersection. They are not mutually exclusive. They are not disjoint.
Let’s return to our
Example
A card is chosen at random from a deck of 52 cards. What is the probability of drawing a jack OR a club ?
How can we 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 many outcomes are favourable?
4 (number of jacks) + 13 (number of clubs ) = 17 (number of jacks OR clubs)?
The real number of favourable cases is 16, and not 17: we have counted the jack of clubs twice!
The correct procedure to find the number of favourable cases is:
4 (number of jacks) + 13 (number of clubs ) 1 (jack of clubs) = 16 (number of jacks OR clubs)?
n (fav.cas) = n (jacks) + n (clubs) n(jacks clubs)
We must subtract the number of the elements of the intersection in order not to count them twice!
n (favourable cases) = n (jacks) + n(clubs) n (jacks clubs)
So, the probability of drawing a jack OR a club is
We can write the first fraction in another way:
P(jackclubs) = P(jacks) +P(clubs) P(jackclubs)
We can generalize what we have just found.
When we accept
P(AB) = P(A) +P(B) P(AB)
This the most general expression for more favourable outcomes.
This is known as the formula for the TOTAL PROBABILITY.
Example.
Playing tombola, what is the probability that the first drawn number is a multiple of 10 OR is greater than 70?
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)
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