A cab was involved in a hit-and-run accident. Two cab companies serve the city: the Green, which operates 85% of the cabs, and the Blue, which operates the remaining 15%. A witness identifies the hit-and-run cab as Blue. When the court tests the reliability of the witness under circumstances similar to those on the night of the accident, he correctly identifies the color of the cab 80% of the time and misidentifies it the other 20%.
What's the probability that the cab involved in the accident was Blue, as the witness stated?

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ajeigbeibraheem ajeigbeibraheem · Answer 1 · 2020-02-01T09:43:28+01:00

Answer:

0.4138

Step-by-step explanation:

Probability of an event X given event Y

P (X ║ Y) = [tex]\frac{P ( X and Y)}{P(Y)}[/tex]

The Green, which operates 85% of the cabs, and the Blue, which operates the remaining 15%.

⇒ P ( Cab = Green ) = 0.85

P ( Cab = Blue ) = 0.15

The witness correctly identified the color of a cab 80% of the time and misidentified it the other 20%.

∴ we can say;

⇒ P ( identify = Blue ║ cab = Blue ) = 0.80

P ( identify = Blue ║ cab = Green ) = 0.20

P ( identify = Blue ║ cab = Green ) = 0.80

P ( identify = Blue ║ cab = Blue ) = 0.20

The witness identifies the hit-and-run cab as Blue.

Probability that the cab involved in the accident was Blue will be;

=[tex]\frac{(P(Identify=Blue|cab=Blue)*P(cab=Blue)}{P(identify=Blue|cab=Blue)*P(cab=Blue)+P(Identify=Blue|cab=Green)*P(cab=Green)}[/tex]

=[tex]\frac{0.80*0.15}{(0.8*0.15)+(0.2*0.85)}[/tex]

=[tex]\frac{0.12}{(0.12+0.17)}[/tex]

=[tex]\frac{0.12}{(0.29)}[/tex]

= 0.4138