# SADC Course in Statistics Conditional Probabilities and Independence (Session 03)

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SADC Course in Statistics Conditional Probabilities and Independence (Session 03)

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 explain what is meant by a conditional probability distinguish the concepts of mutual exclusiveness and independence of events. identify events which are independent state and apply Bayes theorem construct a tree diagram for a specific scenario and compute probabilities associated with events along branches of the tree

To put your footer here go to View > Header and Footer 3 Conditional probability There are situations where we want to find the probability of one event, say A, when we know that some related event B has occurred. For example, we may ask what is the probability of rain later today given that it is now very windy what is the probability that a person is HIV positive given that he has Tuberculosis (TB) Such probabilities are called conditional probabilities. In the first example the known condition was that it was windy while in the second example the known condition was that a person is suffering from TB.

To put your footer here go to View > Header and Footer 4 Definition of conditional probability The conditional probability of event A, given event B has occurred, written P(A|B), is defined as It is estimated as the proportion of the event B in which A occurs (Note: A B is a sub-event of B). A B A B provided P(B)>0.

To put your footer here go to View > Header and Footer 5 Practical quiz If the red (lighter) object represents the event that a person has AIDS and the blue (darker) object represents the event that a person has TB, write down a statement that each of the two figures below represents. Which of the two figures below is a better representation of reality, explain. Fig. 1Fig. 2

To put your footer here go to View > Header and Footer 6 Independence of events Events are independent if the occurrence of one does not affect the occurrence of the others. In other words events are independent if knowledge of one does not supply information about the other. Events which are not independent are said to be dependent.

To put your footer here go to View > Header and Footer 7 If events A and B are independent, P(A B) = P(A) x P(B) If P(B) is not equal to zero then P(A|B) = P(A B) / P(B) = P(A) x P(B) / P(B) = P(A) The above says that if A and B are independent, then the occurrence of B does not change the probability of occurrence of A. Conditional Probability and Independence

To put your footer here go to View > Header and Footer 8 Other ways of determining independence Independence cannot always be determined by the formula given above. The most common way is to assume independence on the basis of knowledge of the physical or natural phenomenon. In the following table draw lines joining pairs of events A and B that you believe are dependent. Give an argument for your answers.

To put your footer here go to View > Header and Footer 9 Which pairs of events are dependent? Set ASet B 1.Mortgage rate increase 1.Many people are poor 2.Thunderstorm2.Depreciation of the local currency by 20% 3.A tuberculosis epidemic 3.Interest rates hike by 5% 4.Crime wave4.Increase of HIV prevalence 5.Decline of the number of locals touring abroad 5.Heavy rain

To put your footer here go to View > Header and Footer 10 Which repetitions of each experiment will produce independent outcomes? ExperimentOutcome 1.ChildbearingSex of a child 2.Measuring lifetimes of electric bulbs Lifetime in hours 3.Monitoring air pollution level daily at a city centre Air pollution index 4.Poverty level of a rural household over time Monthly consumption expenditure 5.Rolling of a dieThe number that appears on the upper face

To put your footer here go to View > Header and Footer 11 Law of Total Probability Recall from previous session that P(B) = P(B A) + P(B A c ). A generalisation of this is: P(A) = P(A|E 1 ) P(E 1 ) + P(A|E 2 ) P(E 2 ) + …. ……. + P(A|E k ) P(E k ) where E 1, E 2, …., E k are mutually exclusive events such that E 1 U E 2 U …. U E k = S. i.e. the E i s form a partition of S. Above generalisation is called the Law of Total Probability.

To put your footer here go to View > Header and Footer 12 Bayes Theorem Using above, we can write P(B) = [P(B|A)*P(A) ] + [P(B|A c )*P(A c ) ] Combining this and the definition of conditional probability we obtain This is called Bayes Theorem. This may be generalised to several mutually exclusive events E 1, E 2, …., E k that partition S.

To put your footer here go to View > Header and Footer 13 An application of Bayes rule Consider the following events A is the event that a person is HIV infected. A c is the event that a person is not HIV infected. B is the event that a person tests as HIV positive. Suppose we also know that 15% of the people in the area are HIV infected and research has shown that P(B|A) = 0.98, P(B|A c ) = 0.01. What is the probability that a person who tests positive is actually infected with HIV, that is P(A|B)?

To put your footer here go to View > Header and Footer 14 The multiplication rule One of the major problems in calculating probabilities is to make sure that all logical possibilities are considered. The multiplication rule states that if events A 1, A 2, ….,A k have n 1, n 2, …, n k possible outcomes respectively, then the total number of possible outcomes of the k events is n 1 x n 2 x … x n k.

To put your footer here go to View > Header and Footer 15 The addition rule If, on the other hand, events A 1, A 2, ….,A k are mutually exclusive, we have the addition rule, i.e. that A 1 U A 2 U …. U A k has n 1 + n 2 + … + n k possible outcomes. Note: Independence and mutual exclusive are not the same thing, e.g. if a die and a coin are thrown simultaneously, the event of a six on the die and the event head on the coin, are independent, but are not mutually exclusive.

To put your footer here go to View > Header and Footer 16 Suppose companies X, Y, Z have tendered for a low-cost housing project in a municipality. There are concerns whether the project will be completed in time and whether the cost will be as budgeted or otherwise. The events are: A – company X,Y or Z. B – in time (T) or not in time (N). C – below budget (U), as budgeted (B), over budget (O). These are independent events, so total number of possible outcomes is 3x2x3 = 18. We can represent the possible outcomes using a tree diagram. An Example:

To put your footer here go to View > Header and Footer 17 A tree diagram for housing project UBOUBOUBOUBOUBOUBO T TTN NN X Y Z

To put your footer here go to View > Header and Footer 18 Suppose from previous experience you have the following probabilities: 0.3 that company X will win the bid, 0.45 that Y will win and 0.25 that Z will win. 0.9 that the project will finish in time if X does it, 0.8 it will finish in time if Y does it an 0.7 it will finish in time if Z does it. 0.2 that the project will go over-budget if the project finishes in time and 0.7 that it will go over-budget if it does not finish in time. Calculate the probability that the project will not go over the budgeted cost. Prob. calculations from tree diagram

To put your footer here go to View > Header and Footer 19 UBOUBOUBOUBOUBOUBO T TTN NN X Y Z 0.3 0.45 0.25 0.9 0.80.7 0.2 0.7 0.2 0.3x0.9x0.2 0.3x0.1x0.7 0.45x0.8x0.2 0.45x0.2x0.7 0.25x0.7x0.2 0.25x0.3x0.7

To put your footer here go to View > Header and Footer 20 Answer to question (slide 18) The required probability is 1-P(O), where P(O) = (0.3*0.9*0.2) + (0.3*0.1*0.7) + + (0.45*0.8*0.2) + (0.45*0.2*0.7) + + (0.25*0.7*0.2) + (0.25*0.3*0.7) = 0.2975. Hence 1- P(O) = 0.7025.

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

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