Question

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?

Answer 1

Answer: The required probability is 0.414.

Step-by-step explanation:

Since we have given that

Probability of taxis in a certain city by Blue Cab P(B)= 15%

Probability of taxis by Green Cab P(G) = 85%

Let A be the event that eyewitness said that vehicle was blue.

P(A|B)=0.80

P(A|G)=0.80

P(A'|B)=0.20=P(A'|G)

Using the "Bayes theorem":

Probability that the taxi at fault was blue is given by

$=\dfrac{P(B).P(A|B)}{P(B).P(A|B)+P(G).P(A'|G)}\\\\=\dfrac{0.15\times 0.8}{0.15\times 0.8+0.85\times 0.2}\\\\=\dfrac{0.12}{0.12+0.17}\\\\=\dfrac{0.12}{0.29}\\\\=0.414$

Hence, the required probability is 0.414.