laws of probability

SADC Course in Statistics

Laws of Probability(Session 02)

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Learning ObjectivesAt 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 sessionsIn 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

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An ExampleA 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.

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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.

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17501200

60050

700

ab

c

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

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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 = 0.49while 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 boyConsidering 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 probabilityThe 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, thenP(A U B) = P(A) + P(B).

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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

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Answers to questions in slide 9Values of a, b, c, d and e are:a = 200, b = 250, c = 300d = 1200, e = 100

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