Probability - the Australian Mathematical Sciences Institute

A guide for teachers – Years 11 and 12 • {39}This is a natural context for a tree diagram, which is shown below. This is because thereare two stages. The first stage concerns whether or not the woman has cancer, and thesecond stage is the result of the test, positive (indicating cancer) or not. We find, forexample, that Pr(T + ∩C ) = Pr(C )Pr(T + |C ) = 0.01 × 0.85 = 0.0085.Pr(T+⋂C) = 0.0085Pr(T+|C) = 0.85Pr(C) = 0.01Pr(T−|C) = 0.15Pr(T−⋂C) = 0.0015Pr(C’) = 0.99Pr(T+|C’) = 0.07Pr(T+⋂C’) = 0.0693Pr(T−|C’) = 0.93Pr(T−⋂C’) = 0.9207Tree diagram for the diagnostic-testing example.We can now apply the usual rule for conditional probability to find Pr(C|T + ):Pr(C|T + ) = Pr(C ∩ T +)Pr(T + )= 0.0085Pr(T + ) .To find Pr(T + ), we use the law of total probability:HencePr(T + ) = Pr(C ∩ T + ) + Pr(C ′ ∩ T + )= Pr(C )Pr(T + |C ) + Pr(C ′ )Pr(T + |C ′ )= (0.01 × 0.85) + (0.99 × 0.07)= 0.0085 + 0.0693= 0.0778.Pr(C|T + ) = Pr(C ∩ T +)Pr(T + )= 0.00850.0778 ≈ 0.1093.