Probability - the Australian Mathematical Sciences Institute

A guide for teachers – Years 11 and 12 • {37}H H ′A 1 Pr(A 1 )A 2 Pr(A 2 )A 3 Pr(A 3 ∩H) = Pr(A 3 )Pr(H|A 3 ) Pr(A 3 )A 4 Pr(A 4 )A 5 Pr(A 5 )A 6 Pr(A 6 )Pr(H) Pr(H ′ )Example: Traffic lights, continuedWe have seen an illustration of the law of total probability, in the traffic-lights examplefrom the previous section. We obtained the probability of the rider meeting green at thesecond set of traffic lights by using the possible events at the first set of traffic lights, G 1and R 1 . These two events are a partition: note that R 1 = G1 ′ . Hence,Pr(G 2 ) = Pr(G 1 ∩G 2 ) + Pr(R 1 ∩G 2 )= Pr(G 1 )Pr(G 2 |G 1 ) + Pr(R 1 )Pr(G 2 |R 1 )= (0.3 × 0.6) + (0.7 × 0.4)= 0.18 + 0.28= 0.46.In contexts with sequences of events, for which we use tree diagrams, we may be interestedin the conditional probability of an event at the first stage, given what happenedat the second stage. This is a reversal of what seems to be the natural order, but it issometimes quite important. In the traffic-lights example, it would mean considering aconditional probability such as Pr(G 1 |G 2 ): the probability that the rider encounters greenat the first set of traffic lights, given that she encounters green at the second set.Again, consider the general situation involving a partition A 1 , A 2 ,..., A k and an event H.The law of total probability enables us to find Pr(H) from the probabilities Pr(A i ) andPr(H|A i ). However, we may be interested in Pr(A i |H). This is found using the standardrule for conditional probability:Pr(A i |H) = Pr(A i ∩ H).Pr(H)