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Nat2105 [25]
3 years ago
15

Blue Cab operates 15% of the taxis in a certain city, and Green Cab operates the other 85%. After a nighttime hit-and-run accide

nt 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?
Mathematics
1 answer:
NikAS [45]3 years ago
8 0

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.

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