Conditional Probabilities and Independence

1. Conditional Probabilities and Independence (Session 03)

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

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.

4. Definition of conditional probability The conditional probability of event A, given event B has occurred, written P(A|B), is defined as provided P(B)>0. It is estimated as the proportion of the event B in which A occurs (Note: A  B is a sub-event of B). B A A  B

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. 1 Fig. 2

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.

7. Conditional Probability and Independence • 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.

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.

9. Which pairs of events are dependent?

10. Which repetitions of each experiment will produce independent outcomes?

11. Law of Total Probability Recall from previous session that P(B) = P(B A) + P(B Ac). A generalisation of this is: P(A) = P(A|E1) P(E1) + P(A|E2) P(E2) + …. ……. + P(A|Ek) P(Ek) where E1, E2, …., Ek are mutually exclusive events such that E1 U E2 U …. U Ek = S. i.e. the Ei’s form a partition of S. Above generalisation is called the Law of Total Probability.

12. Bayes’ Theorem Using above, we can write P(B) = [P(B|A)*P(A) ] + [P(B|Ac)*P(Ac) ] Combining this and the definition of conditional probability we obtain This is called Bayes’ Theorem. This may be generalised to several mutually exclusive events E1, E2, …., Ek that partition S.

13. An application of Bayes’ rule Consider the following events • A is the event that a person is HIV infected. • Ac 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|Ac) = 0.01. What is the probability that a person who tests positive is actually infected with HIV, that is P(A|B)?

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 A1, A2, ….,Ak have n1, n2, …, nk possible outcomes respectively, then the total number of possible outcomes of the k events is n1x n2x … x nk .

15. The addition rule If, on the other hand, events A1, A2, ….,Ak are mutually exclusive, we have the addition rule, i.e. that A1 U A2 U …. U Ak has n1 + n2 + … + nk 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.

16. An Example: 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.

17. U U U U U U B B B B B B O O O O O O A tree diagram for housing project Z Y X T N T T N N

18. Prob. calculations from tree diagram 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.

19. 0.25 0.3 0.45 Z Y X 0.8 0.7 0.9 T N T T N N 0.2 0.7 0.2 0.7 0.7 0.2 U U U U U U B B B B B B O O O O O O 0.3x0.9x0.2 0.45x0.8x0.2 0.25x0.7x0.2 0.3x0.1x0.7 0.45x0.2x0.7 0.25x0.3x0.7

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.