Question

Ed's Tires and Brakes has two locations, one on the northwest side of town and one on the southeast side of town. At both locations are performed routine tire repairs and rotations, as well as expensive brake repairs. This past week,

60%

of the cars serviced at Ed's were serviced at the northwest location, while

40%

of the cars were serviced at the southeast location. (No car was serviced at both locations.) Brake repairs were more typical at the northwest location than at the southeast location:

65%

of the cars at the northwest location required brake repairs, while

40%

of the cars at the southeast location required brake repairs.

Let

N

denote the event that a randomly chosen car (taken to Ed's in the past week) was serviced at the northwest location and

N

denote the event that a randomly chosen car was serviced at the southeast location. Let

B

denote the event that a randomly chosen car required brake repairs and

B

denote the event that a randomly chosen car did not require brake repairs.

Fill in the probabilities to complete the tree diagram below, and then answer the question that follows. Do not round any of your responses.

=PBN =P∩NB =PN 0.6 =P∩NB =PBN 0.35 =PBN =P∩NB =PN =P∩NB =PBN 0.6 What is the probability that a randomly chosen car did not require brake repairs?

Answer 1

Solution

Back-up Theory

If A and B are two events such that probability of B is influenced by occurrence or otherwise of A, then conditional Probability of B given A, denoted by P(B/A) = P(B ∩ A)/P(A)……...............................….(1)

First Part Probability Table:

This past week, at Ed's

60% of the cars serviced were at the northwest location => P(N) = 0.6 …….......….…. (2)

40% of the cars were serviced at the southeast location => P(N^C) = 0.4 …….....…......(3)

65% of the cars at the northwest location required brake repairs, => P(B/N) = 0.65.....(4)

40% of the cars at the southeast location required brake repairs => P(B/N^C) = 0.4......(5)

Now,

P(B ∩ N) = P(B/N) x P(N)   [vide (1)]

                       = 0.65 x 0.6 [vide (4) and (2)]

                       = 0.39  

P(B ∩ N^C) = P(B/N^C) x P(N^C)   [vide (1)]

                       = 0.4 x 0.4 [vide (4) and (2)]

                       = 0.16  

With these, we have the following probability table:

	Northwest	Northeast	Total
Tyre repairs and rotation	0.21*	0.24*	0.45
Brake repairs	0.39	0.16	0.55
Total	0.60	0.40	1.00

* By subtraction

Answer 1

Second Part

Since probability of Brake repairs is 0.55,

probability that a randomly chosen car did not require brake repairs = 1 – 0.55 = 0.45 Answer 2

DONE