At Jaxlyn's Garage, 200 cars were inspected and worked on. There were 92 cars that

needed new brakes, 136 cars that needed a new alternator, and 24 cars that needed

neither repair.

Part A: Determine the probability of a car needing both repairs.

Part B: Determine the probability of a car needing an alternator, but not new brakes.

Question

Coby Mills

Mathematics

27 December, 21:36

At Jaxlyn's Garage, 200 cars were inspected and worked on. There were 92 cars that

needed new brakes, 136 cars that needed a new alternator, and 24 cars that needed

neither repair.

Part A: Determine the probability of a car needing both repairs.

Part B: Determine the probability of a car needing an alternator, but not new brakes.

+4

Answers (1)

Know the Answer?

Hudson Mcclain · Answer 1

A) 0.26

B) 0.42

Step-by-step explanation:

No. of cars which needed both:

92 + 136 - (200 - 24)

52

P (both) = 52/200 = 0.26

No. of cars that need an alternator, but not new brakes:

136 - 52

84

P = 84/200 = 0.42