Download Laws of Probability - University of Reading

Laws of Probability
(Session 02)
SADC Course in Statistics
Learning Objectives
At the end of this session you will be able to
• state and explain the fundamental laws of
probability
• apply Venn diagrams and the laws of
probability to solve basic problems
• explain what is meant by the universal
event, union and intersection of events,
complement of an event and mutually
exclusive events
To put your footer here go to View > Header and Footer
2
Aims of probability sessions
In sessions 3-10, the aim is to:
• build a firm mathematical foundation for
the theory of probability
• introduce the laws of probability as a
unifying framework for modelling and
solving statistical problems
• develop problem solving skills for basic
probability type questions
To put your footer here go to View > Header and Footer
3
An Example
A household survey in a certain district
produced the following information:
• Access to child support grants (yes/no)
• Possession of a birth certificate (yes/no)
• School attendance (yes/no)
The total number of children surveyed was
3400.
To put your footer here go to View > Header and Footer
4
Two questions of interest…
• Is a child more likely to get a grant if
he/she attends school, or if she/he has a
birth certificate?
• What is the probability that a child chosen
at random from the surveyed children will
attend school, given that he/she does not
possess a birth certificate
We will aim to answer these questions below.
To put your footer here go to View > Header and Footer
5
Some survey results
• 1750 children have a birth certificate
• 850 children have a birth certificate and receive
a child support grant
• 1200 children receive a child support grant
• 600 children have a birth certificate and receive
a child support grant, but do not attend school
• 700 attend school and have a birth certificate
but do not receive a child support grant
• 50 children neither go to school nor have a birth
certificate but receive a child support grant
• 2450 children attend school
To put your footer here go to View > Header and Footer
6
Answering the questions…
To answer the questions posed in slide 5, it
is necessary to determine values for a, b, c,
d and e in the graphical representation
below.
This diagram is called a Venn diagram.
It is a valuable tool for use in computing
probabilities associated with specific events.
To put your footer here go to View > Header and Footer
7
Support grant
Birth Certificate
1200
1750
50
600
a
b
700
c
d
2450
e = outside of
the three circles
School Attendance
To put your footer here go to View > Header and Footer
8
Finding a, b, c, d, e
From survey results (slide 6), we have
(i)
a + 600 + b + 700 = 1750
(ii) b + 600 = 850
(iii) 600 + 50 + c + b = 1200
(iv) 700 + b + c + d = 2450
(v) 1750 + 50 + c + d + e = 3400
Class exercise:
Determine values for a, b, c, d and e using
the above equations.
To put your footer here go to View > Header and Footer
9
Answers to Questions:
Let X, Y, be events that a child gets a grant,
given that
(i) he/she has a birth certificate
(ii) he/she attends school.
Let Z be the event that a child attends
school, given he/she has no birth certificate.
Then, P(X) = (600 + b)/1750 = 0.49
while P(Y) = (b + c)/(700+b+c+d) = 0.22
Further, P(Z) = (c+d)/(3400–1750) = 0.91
To put your footer here go to View > Header and Footer
10
Conclusions:
• The results suggest that access to child
support is based more on possession of
birth certificate than on school attendance.
• There is a high likelihood that a child will
attend school even if he/she does not
possess a birth certificate
To put your footer here go to View > Header and Footer
11
The language of probability
The first step towards a good understanding
of a culture is to learn the language. In the
probability culture, the following terms are
commonly used:
• Experiment – any action that can produce
an outcome. Try the following experiments
and record the outcome: smile to your
neighbour, count the number of colleagues
with cellphones.
• Sample space – the set of all possible
outcomes of an experiment. Denoted by S.
To put your footer here go to View > Header and Footer
12
Further definitions
• Event – any set of outcomes. Thus S is
also an event called the Universal event.
In the children example, we can define an
event E = selecting a child who attends
school and receives child support.
• Union – the union of events A and B,
written A U B (also A or B), is the event
that contains all outcomes in A and
outcomes in B.
The shaded area represent the union.
To put your footer here go to View > Header and Footer
13
Definitions continued…
• Intersection – the intersection of events
A and B, written A  B (also A and B), is the
set of outcomes that belong to both A and
B, i.e. it is the overlap of A and B.
The shaded area represents the
intersection of the two events A and B.
• Null – or empty set is the event with no
outcomes in it. Denoted by Ø.
To put your footer here go to View > Header and Footer
14
Definitions continued…
• Complement – of an event A, denoted by
c
A , is the set of outcomes in S which are
not in A.
S
A
c
A
The complement of event A is represented
by the sky-blue (darker shaded) area.
To put your footer here go to View > Header and Footer
15
Definitions continued…
• Mutually exclusive – also called disjoint
events, are events which do not have any
outcomes in common. No overlap.
A baby girl
A baby boy
Considering the experiment of giving birth,
there are two mutually exclusive possible
outcomes, either a girl or a boy. Of course
we exclude rare events of abnormality.
To put your footer here go to View > Header and Footer
16
Fundamental laws of probability
The probability of an event A is a number P(A)
which satisfies the following three conditions:
1. 0≤P(A)≤1, i.e. probability is a measure
that is restricted between 0 and 1.
2. P(S) = 1, where S is the sample space.
That is, the universal set is the sure event.
3. If events A and B are disjoint events, then
P(A U B) = P(A) + P(B).
To put your footer here go to View > Header and Footer
17
Consequences of probability laws
c
i. P(A ) = 1 – P(A).
c
This follows from the fact that S = A U A ,
and because A and its complement are
mutually exclusive.
c
Law 3 implies P(S) = P(A) + P(A ).
Now apply Law 2.
ii. P(Ø) = 0.
c
This easily follows from (i) since S = Ø.
There is nothing outside the universe S.
To put your footer here go to View > Header and Footer
18
Consequences (continued)
c
iii. P(A) = P(A  B) + P(A  B ).
This also easily follow from Law 3 because
c
events A  B and A  B are disjoint and
together they make up the event A.
iv. P(A U B) = P(A) + P(B) – P(A  B)
c
This follows from noting that B and AB are
mutually exclusive, and that their union is AUB.
Hence
c
P(A U B) = P(B) + P(A  B ).
c
Substituting for P(A  B ) from (iii) above
gives the desired result.
To put your footer here go to View > Header and Footer
19
Sub-events: definition
A is said to be a sub-event of the event B,
if P(A) ≤ P(B), i.e.
If every outcome in A is also an outcome in B.
To put your footer here go to View > Header and Footer
20
Sub-events: an example
Let A be the event that
a baby girl is born
and B the event that a baby is born.
Hence if A happens we know that B has also
happened. However, if B happens we cannot
be sure that A has happened.
Thus, the probability of getting a baby girl, in
the sample space of all potential mothers, is
smaller than the probability of getting a
baby!
To put your footer here go to View > Header and Footer
21
Answers to questions in slide 9
Values of a, b, c, d and e are:
a = 200,
b = 250,
d = 1200,
e = 100
c = 300
To put your footer here go to View > Header and Footer
22
To put your footer here go to View > Header and Footer
23