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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

Sagot :

Answer:

0.1739 = 17.39% probability that the cab actually was blue

Step-by-step explanation:

Conditional Probability

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

[tex]P(B|A) = \frac{P(A \cap B)}{P(A)}[/tex]

In which

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

[tex]P(A \cap B)[/tex] 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

[tex]P(A) = 0.2*0.95 + 0.8*0.05 = 0.23[/tex]

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

80% of 5%. So

[tex]P(A \cap B) = 0.8*0.05 = 0.04[/tex]

What is the probability that the cab actually was blue?

[tex]P(B|A) = \frac{P(A \cap B)}{P(A)} = \frac{0.04}{0.23} = 0.1739[/tex]

0.1739 = 17.39% probability that the cab actually was blue

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