The members of a consulting firm rent cars from three rental agencies :
50% from agency X, 30% from agency Y and 20% from agency Z. From past experience, it is known that 9% of the cars from agency X need a service and tunning before renting, 12% of the cars from agency Y need a service and tunning before renting and 10% of the cars from agency Z need a service and tunning before renting. If the renting car delivered to the firm neeeds service and tunning, find the probability that agency Z is not to be blamed.

Answer

To solve the problem, we will use Bayes' theorem and the law of total probability. We need to find the probability that a car needing service is not from agency Z.

### Step 1: Define the events
Let:
- \( A_X \): the event that the car is from agency X
- \( A_Y \): the event that the car is from agency Y
- \( A_Z \): the event that the car is from agency Z
- \( B \): the event that the car needs service

### Step 2: Determine the probabilities
From the problem statement, we have:
- \( P(A_X) = 0.5 \)
- \( P(A_Y) = 0.3 \)
- \( P(A_Z) = 0.2 \)

The probabilities that a car from each agency needs service are:
- \( P(B | A_X) = 0.09 \)
- \( P(B | A_Y) = 0.12 \)
- \( P(B | A_Z) = 0.10 \)

### Step 3: Calculate the total probability of needing service
Using the law of total probability, we can find \( P(B) \):
\[ P(B) = P(B | A_X)P(A_X) + P(B | A_Y)P(A_Y) + P(B | A_Z)P(A_Z) \]
Substituting the values:
\[ P(B) = (0.09)(0.5) + (0.12)(0.3) + (0.10)(0.2) \]
Calculating each term:
\[ P(B) = 0.045 + 0.036 + 0.02 = 0.101 \]

### Step 4: Calculate the probability that the car is from agency Z given that it needs service
Using Bayes' theorem:
\[ P(A_Z | B) = \frac{P(B | A_Z)P(A_Z)}{P(B)} \]
Substituting the values:
\[ P(A_Z | B) = \frac{(0.10)(0.2)}{0.101} = \frac{0.02}{0.101} \]
Calculating this gives:
\[ P(A_Z | B) \approx 0.198 \]

### Step 5: Find the probability that the car is not from agency Z given that it needs service
We need to find \( P(A_Z^c | B) \):
\[ P(A_Z^c | B) = 1 - P(A_Z | B) \]
Substituting the value we found:
\[ P(A_Z^c | B) = 1 - 0.198 \approx 0.802 \]

### Final Answer
Thus, the probability that agency Z is not to be blamed when a rented car needs service is approximately \( 0.802 \) or \( \frac{81}{101} \).

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