Question

An eyewitness observes a hit-and-run accident in New York City, where 95% of the cabs are yellow and 5% are blue. A police expert believes the witness is 80% reliable. That is, the witness will correctly identify the color of a cab 80% of the time. If the cab is actually yellow, there is an 80% chance the witness will assert it is yellow, and if the cab is actually blue, there is an 80% chance the witness will assert it is blue. If a witness asserts the cab was blue, what is the probability that the cab actually was blue

Answer 1

0.1739 = 17.39% probability that the cab actually was blue

Explanation:

Conditional Probability

We use the conditional probability formula to solve this question. It is

`P(B|A) = (P(A \cap B))/(P(A))`

In which

P(B|A) is the probability of event B happening, given that A happened.

`P(A \cap B)` is the probability of both A and B happening.

P(A) is the probability of A happening.

In this question:

Event A: Witness asserts the cab is blue.

Event B: The cab is blue.

Probability of a witness assessing that a cab is blue.

20% of 95%(yellow cab, witness assesses it is blue).

80% of 5%(blue cab, witness assesses it is blue). So

`P(A) = 0.2*0.95 + 0.8*0.05 = 0.23`

Probability of being blue and the witness assessing that it is blue.

80% of 5%. So

`P(A \cap B) = 0.8*0.05 = 0.04`

What is the probability that the cab actually was blue?

`P(B|A) = (P(A \cap B))/(P(A)) = (0.04)/(0.23) = 0.1739`

0.1739 = 17.39% probability that the cab actually was blue