Blue Cab operates 15% of the taxis in a certain city, and Green Cab operates the other 85%. After a nighttime hit-and-run accident involving a taxi, an eyewitness said the vehicle was blue. Suppose, though, that under night vision conditions, only 80% of individuals can correctly distinguish between a blue and a green vehicle. What is the (posterior) probability that the taxi at fault was blue? In answering, be sure to indicate which probability rules you are using. [Hint: A tree diagram might help. Note: This is based on an actual incident.]

Answer Step 1 of 3 P(blue taxi)=P(B)=15%=0.15 P(green taxi)=P(G)=85%=0.85 P(witness saying blue when the cab is really blue)= P(SB|B) =0.80 P(witness saying green when the cab is really green)= P(SG|G) =0.80 the probabilities of misidentifications are: P(SG|B) =0.2 and P(SB|G) =0.2 Step 2 of 3 Now we want to know the probability that the cab really was blue, given that witness said it was blue, i.e., we want to know P(B|SB). According to Bayes’ rule, this probability is given by: P(B|SB) =P(B)×P(SB|B) / P(SB) . We have the values for the two expressions in the numerator: P(B) = 0.15 and P(SB|B) =0.8 Then P(saying blue)=P(SB) = P(B)×P(SB|B)+P(G)×P(SB|G) = (.15×.80)+(.85×.20) =0.12+0.17 =0.29