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Laws of Probability

Laws of Probability. (Session 02). 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

nigel-holcomb
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Laws of Probability

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  1. Laws of Probability (Session 02)

  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. 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. 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. 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. 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. 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. Support grant Birth Certificate 1200 1750 50 600 a b c 700 d 2450 e = outside of the three circles School Attendance

  9. Finding a, b, c, d, e From survey results (slide 6), we have • a + 600 + b + 700 = 1750 • b + 600 = 850 • 600 + 50 + c + b = 1200 • 700 + b + c + d = 2450 • 1750 + 50 + c + d + e = 3400 Class exercise: Determine values for a, b, c, d and e using the above equations.

  10. Answers to Questions: Let X, Y, be events that a child gets a grant, given that • he/she has a birth certificate • 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

  11. 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

  12. The language ofprobability 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. 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.

  14. 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.

  15. Definitions continued… • Complement – of an event A, denoted by Ac, is the set of outcomes in S which are not in A. S A Ac The complement of event A is represented by the sky-blue (darker shaded) area.

  16. 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.

  17. Fundamental laws of probability The probability of an event A is a number P(A) which satisfies the following three conditions: • 0≤P(A)≤1, i.e. probability is a measure that is restricted between 0 and 1. • 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. Consequences of probability laws • P(Ac) = 1 – P(A). This follows from the fact that S = A UAc, and because A and its complement are mutually exclusive. Law 3 implies P(S) = P(A) + P(Ac). Now apply Law 2. • P(Ø) = 0. This easily follows from (i) since Sc = Ø. There is nothing outside the universe S.

  19. Consequences (continued) • 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. • P(A U B) = P(A) + P(B) – P(A  B) This follows from noting that B and ABc are mutually exclusive, and that their union is AUB. Hence P(A U B) = P(B) + P(A  Bc). Substituting for P(A  Bc) from (iii) above gives the desired result.

  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. 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!

  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

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