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SADC Course in Statistics Laws of Probability (Session 02)

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Presentation on theme: "SADC Course in Statistics Laws of Probability (Session 02)"— Presentation transcript:

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

2 To put your footer here go to View > Header and Footer 2 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

3 To put your footer here go to View > Header and Footer 3 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

4 To put your footer here go to View > Header and Footer 4 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.

5 To put your footer here go to View > Header and Footer 5 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.

6 To put your footer here go to View > Header and Footer 6 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

7 To put your footer here go to View > Header and Footer 7 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.

8 To put your footer here go to View > Header and Footer a b c d Birth Certificate School Attendance Support grant e = outside of the three circles 2450

9 To put your footer here go to View > Header and Footer 9 From survey results (slide 6), we have (i) a b = 1750 (ii) b = 850 (iii) c + b = 1200 (iv) b + c + d = 2450 (v) c + d + e = 3400 Class exercise: Determine values for a, b, c, d and e using the above equations. Finding a, b, c, d, e

10 To put your footer here go to View > Header and Footer 10 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:

11 To put your footer here go to View > Header and Footer 11 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:

12 To put your footer here go to View > Header and Footer 12 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.

13 To put your footer here go to View > Header and Footer 13 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

14 To put your footer here go to View > Header and Footer 14 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…

15 To put your footer here go to View > Header and Footer 15 Complement – of an event A, denoted by A c, is the set of outcomes in S which are not in A. A S AcAc The complement of event A is represented by the sky-blue (darker shaded) area. Definitions continued…

16 To put your footer here go to View > Header and Footer 16 Mutually exclusive – also called disjoint events, are events which do not have any outcomes in common. No overlap. A baby girlA 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…

17 To put your footer here go to View > Header and Footer 17 Fundamental laws of probability The probability of an event A is a number P(A) which satisfies the following three conditions: 1.0P(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).

18 To put your footer here go to View > Header and Footer 18 Consequences of probability laws i.P(A c ) = 1 – P(A). This follows from the fact that S = A U A c, and because A and its complement are mutually exclusive. Law 3 implies P(S) = P(A) + P(A c ). Now apply Law 2. ii.P(Ø) = 0. This easily follows from (i) since S c = Ø. There is nothing outside the universe S.

19 To put your footer here go to View > Header and Footer 19 Consequences (continued) iii.P(A) = P(A B) + P(A B c ). This also easily follow from Law 3 because events A B and A B c 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 A B c are mutually exclusive, and that their union is AUB. Hence P(A U B) = P(B) + P(A B c ). Substituting for P(A B c ) from (iii) above gives the desired result.

20 To put your footer here go to View > Header and Footer 20 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.

21 To put your footer here go to View > Header and Footer 21 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

22 To put your footer here go to View > Header and Footer 22 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

23 To put your footer here go to View > Header and Footer 23


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