Probability - the Australian Mathematical Sciences Institute

A guide for teachers – Years 11 and 12 • {33}Then, using an obvious notation:• A = {HHT,HTH,THH,HHH}• B = {HHH,THH,HTT,TTT}• C = {HHH,HHT,HTT,TTT}.Thus A ∩ B ∩C = {HHH}. Assuming independence of the tosses, there are eight elementaryoutcomes (two for each of the three tosses), all equally probable, so the probabilityof each of them equals 1 8 . HencePr(A ∩ B ∩C ) = 1 8 = ( 12) 3 = Pr(A)Pr(B)Pr(C ).So equation (∗) is satisfied. However, B ∩C = {HHH,HTT,TTT} and hencePr(B ∩C ) = 3 8 ≠ 1 4 = ( 12) 2 = Pr(B)Pr(C ).It follows that B and C are not independent, and hence the events A, B and C are notmutually independent.Examples like the previous one are not really of practical importance. Of much greaterimportance is the fact that, if events are physically independent, then they are mutuallyindependent, and so then equation (∗) is true. This is the result that is useful in practice.Tree diagramsProbability problems sometimes involve more than one step or stage, and at the differentstages various outcomes may occur. It can be tricky to keep track of the stages, outcomesand probabilities, and a ‘tree diagram’ is often helpful. Such a diagram is usually labelledin a sensible way with the outcomes that may occur at each stage and their probabilities.The probabilities of the outcomes at the ends of the branches of the tree may be foundusing the multiplication rule.Example: Traffic lightsRose rides a bicycle to work. There is a sequence of two sets of traffic lights on her route,not far apart. For simplicity, we ignore amber and assume the lights are only red or green.Suppose the probability that the first set is green when she arrives at it is 0.3. If she getsthrough on green at the first set, the probability that the second set is green when shearrives is 0.6; if she has to stop on red at the first set, the probability that the second setis green when she arrives is 0.4. What is the probability that she gets through both setson green?