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SADC Course in Statistics Laws of Probability (Session 02)
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SADC Course in Statistics Laws of Probability (Session 02)

Mar 28, 2015

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Page 1: SADC Course in Statistics Laws of Probability (Session 02)

SADC Course in Statistics

Laws of Probability

(Session 02)

Page 2: SADC Course in Statistics Laws of Probability (Session 02)

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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

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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

Page 4: SADC Course in Statistics Laws of Probability (Session 02)

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An Example

A household survey in a certain districtproduced 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 was3400.

Page 5: SADC Course in Statistics Laws of Probability (Session 02)

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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.

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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

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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.

Page 8: SADC Course in Statistics Laws of Probability (Session 02)

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17501200

60050

700

a

bc

d

Birth Certificate

School Attendance

Support grant

e = outside of the three circles

2450

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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 usingthe above equations.

Finding a, b, c, d, e

Page 10: SADC Course in Statistics Laws of Probability (Session 02)

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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

Answers to Questions:

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• 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

Conclusions:

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The language of probabilityThe 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.

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• 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.

Further definitions

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• 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.

Definitions continued…

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• Complement – of an event A, denoted by Ac, is the set of outcomes in S which are not in A.

A

S

Ac

The complement of event A is represented by the sky-blue (darker shaded) area.

Definitions continued…

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• 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.

Definitions continued…

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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).

Page 18: SADC Course in Statistics Laws of Probability (Session 02)

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Consequences of probability laws

i. P(Ac) = 1 – P(A).

This follows from the fact that S = A U Ac, and because A and its complement are mutually exclusive.

Law 3 implies P(S) = P(A) + P(Ac). Now apply Law 2.

ii. P(Ø) = 0.

This easily follows from (i) since Sc = Ø. There is nothing outside the universe S.

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Consequences (continued)

iii. P(A) = P(A B) + P(A Bc).

This also easily follow from Law 3 because events A B and A Bc are disjoint and together they make up the event A.

iv. P(A U B) = P(A) + P(B) – P(A B)This follows from noting that B and ABc aremutually exclusive, and that their union is AUB.

Hence P(A U B) = P(B) + P(A Bc).

Substituting for P(A Bc) from (iii) abovegives the desired result.

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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.

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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!

Sub-events: an example

Page 22: SADC Course in Statistics Laws of Probability (Session 02)

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Answers to questions in slide 9

Values of a, b, c, d and e are:

a = 200, b = 250, c = 300

d = 1200, e = 100

Page 23: SADC Course in Statistics Laws of Probability (Session 02)

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