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? [Hint: A tree diagram might help. Note: This is based on an actual incident.] (Round your answer to four decimal places.)
Indicate the probability rule you used.
Bayes' Theorem
The Product Rule for Ordered Pairs    
The Division Rule
The Product Rule for k-Tuples
The Rule of Permutations

This problem can be approached using **Bayes' Theorem**, which is well-suited for updating probabilities based on new evidence. Let's define the events involved:

- **B:** The taxi is blue.
- **G:** The taxi is green.
- **E:** The eyewitness says the taxi is blue.

We are asked to find the **posterior probability** \( P(B|E) \), the probability that the taxi was blue given that the eyewitness said it was blue. Bayes' Theorem states:

\[
P(B|E) = \frac{P(E|B)P(B)}{P(E)}
\]

### Step 1: Define the Probabilities
- \( P(B) = 0.15 \) (The probability that the taxi is blue).
- \( P(G) = 0.85 \) (The probability that the taxi is green).
- \( P(E|B) = 0.80 \) (The probability that the eyewitness correctly identifies a blue taxi).
- \( P(E|G) = 0.20 \) (The probability that the eyewitness incorrectly identifies a green taxi as blue).

### Step 2: Calculate \( P(E) \) (Total Probability of the Eyewitness Saying Blue)
The total probability that the eyewitness says the taxi is blue, \( P(E) \), is found using the law of total probability:

\[
P(E) = P(E|B)P(B) + P(E|G)P(G)
\]

Substituting the known values:

\[
P(E) = (0.80 \times 0.15) + (0.20 \times 0.85)
\]

\[
P(E) = 0.12 + 0.17 = 0.29
\]

### Step 3: Apply Bayes' Theorem
Now we can compute the posterior probability \( P(B|E) \):

\[
P(B|E) = \frac{P(E|B)P(B)}{P(E)} = \frac{0.80 \times 0.15}{0.29} = \frac{0.12}{0.29} \approx 0.4138
\]

### Final Answer
The probability that the taxi was blue, given that the eyewitness said it was blue, is approximately **0.4138** or **41.38%**.

### Rule Used: Bayes' Theorem

Would you like further clarification, or do you have any questions on how the solution was derived?

Here are 5 related questions for further exploration:
1. How would the result change if Green Cab operated 90% of the taxis instead of 85%?
2. What if the eyewitness's accuracy was 90% instead of 80%?
3. How would the probability change if the witness had seen the taxi during daylight instead of at night?
4. How can the total probability rule be applied to other eyewitness testimony scenarios?
5. What is the significance of Bayes' Theorem in real-world decision-making under uncertainty?

**Tip:** When dealing with conditional probabilities, drawing a tree diagram can simplify visualization and computation.

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? [Hint: A tree diagram might help. Note: This is based on an actual incident.] (Round your answer to four decimal places.) Indicate the probability rule you used.

Solve a real-world Bayesian probability problem involving a hit-and-run accident. The question asks for the posterior probability that a blue cab was involved, based on an eyewitness report and conditional probabilities. This detailed solution uses Bayes' Theorem and the law of total probability.

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