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? In answering, be sure to indicate which probability rules you are using. [Hint: A tree diagram might help. Note: This is based on an actual incident.]

Answer 1

Answer:

0.4138 or 41.38%

Step-by-step explanation:

Prob(a) = 0.15

Prob(b|a) = 0.80

A = event that taxi is blue

B = witness says it is blue

Prob(taxi is not blue) = 1 - prob(a)

= 1- 0.15

= 0.85

Prob(b|a') = 1 - prob(b|a)

= 1 - 0.8

= 0.20

Using addition rule

P(b) = p(a)p(b|a) + p(a')p(b|a')

= 0.15 * 0.8 + 0.85 * 0.20

= 0.29

Now we are to find probability that the taxi that is at fault is blue

According to bayes theorem

P(a|b) = P(a)*p(b|a) / p(b)

= 0.15 x 0.8 / 0.29

= 0.4138

Answer 2

In this exercise we have to use the knowledge of probability to calculate the chances of the taxi being blue in this way we can describe as:

$41.38\%$

So we find that some information found in the text is from ;

• Prob(a) = 0.15

• Prob(b|a) = 0.80

• A = event that taxi is blue

• B = witness says it is blue

So the probability that the taxi is not blue is:

$Prob(taxi \ is \ not \ blue) = 1 - prob(a)\\ = 1- 0.15\\ = 0.85$

So the probability that the taxi is blue is:

$Prob(ba') = 1 - prob(ba)\\ = 1 - 0.8\\ = 0.20$

Using addition rule

$P(b) = p(a)p(b|a) + p(a')p(b|a')\\ = 0.15 * 0.8 + 0.85 * 0.20\\ = 0.29$

Now we are to find probability that the taxi that is at fault is blue, so according to bayes theorem:

$P(a|b) = P(a)*p(b|a) / p(b)\\ = 0.15 * 0.8 / 0.29\\ = 0.4138$

See more about probability at brainly.com/question/795909