[2] The members of a consulting firm rent cars from three rental agencies: \( 60 \% \) from agency 1 , \( 30 \% \) from agency 2 , and \( 10 \% \) from agency 3 . If \( 9 \% \) of the cars from agency 1 need a tune-up, \( 20 \% \) of the cars from agency 2 need a tune-up, and \( 6 \% \) of the cars from agency 3 need a tune-up, what is the probability that a rental car delivered to the firm will need a tune-up?

To find the overall probability that a rental car delivered to the firm will need a tune-up, we can use the law of total probability. Let \( A_1 \), \( A_2 \), and \( A_3 \) be the events that the car is rented from agency 1, agency 2, and agency 3, respectively. The probabilities of renting from each agency are: - \( P(A_1) = 0.60 \) - \( P(A_2) = 0.30 \) - \( P(A_3) = 0.10 \) Let \( B \) be the event that a car needs a tune-up. Given the probabilities of needing a tune-up for each agency: - \( P(B | A_1) = 0.09 \) - \( P(B | A_2) = 0.20 \) - \( P(B | A_3) = 0.06 \) Now, we can use the law of total probability to find \( P(B) \): \[ P(B) = P(B | A_1) P(A_1) + P(B | A_2) P(A_2) + P(B | A_3) P(A_3) \] Substituting in the probabilities: \[ P(B) = (0.09)(0.60) + (0.20)(0.30) + (0.06)(0.10) \] Calculating each term: - \( (0.09)(0.60) = 0.054 \) - \( (0.20)(0.30) = 0.060 \) - \( (0.06)(0.10) = 0.006 \) Now, adding them together: \[ P(B) = 0.054 + 0.060 + 0.006 = 0.120 \] Therefore, the probability that a rental car delivered to the firm will need a tune-up is: \[ \boxed{0.12} \] or \( 12\% \).