Fill in the probabilities to complete the tree diagram below, and then answer th

Question

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

P(N)=

P()= 0.3

P(B N)=

P(N?B)=

What is the probability that a randomly chosen car required brake repairs?

___________

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, of the cars serviced at Ed's were serviced at the northwest location, while 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: of the cars at the northwest location required brake repairs, while of the cars at the southeast location required brake repairs.

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

P(N)=

P()= 0.3

P(B N)=

P( N)=

P(B )= 0.3

P(  ) =

P(N?B)=

P(N?)=

P( ? B)=

P(?)=

What is the probability that a randomly chosen car required brake repairs?

___________

Explanation / Answer

Can you repost the question or write the question in words as the images of the question is not clear. i.e. it is not visible whether a union is required or an intersection.

However, I am answering as many as I can understand.

P(N) = 0.7

P(N-bar) = 0.3

P( B N) = P(B) * P(N) = (0.7*0.5) = 0.35 =35%

P(B-bar N) = P(B-bar) * P(N) = 0.5 * 0.7 = 0.35 = 35%

P(B N-bar) = P(B) P(N-bar) = 0.3 * 0.3 = 0.09 = 9%

P(B-bar N-bar) = 0.3 * 0.7 = 0.21 = 21%

Now,

Probability that a randomly chosen car requires a brake repairs :

( 0.7 * 0.5) + (0.3 * 0.3)

= 0.35 + 0.09

= 0.44

= 44%

Hope this helps.

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