SOLUTION: I dont know what to do! At the Stop 'n Go tune-up and brake shop, the manager has found that an SUV will require a tune-up with a probability of 0.6, a brake job with a probabil

Question 536184: I dont know what to do!
At the Stop 'n Go tune-up and brake shop, the manager has found that an SUV will require a tune-up with a probability of 0.6, a brake job with a probability of 0.1 and both with a probability of 0.02. What is the probability that an SUV requires neither type of repair?
Answer by Earlsdon(6294) (Show Source): You can put this solution on YOUR website!
Let P(tu) = the probability of a tune-up and P(b) = the probability of a brake job.
P(tu) = 0.6 and P(b) = 0.1
P(tu and b) = 0.2
The probability of neither is just 1-P(tu and b) = 1-0.2 = 0.8
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This answer is incorrect!
First, the P(tu and b) is 0.02 not 0.2.
Second:The probability of tu or b is:
P(tu or b) = P(tu)+P(b)-P(tu and b) if the events are not mutually exclusive.
P(tu or b) = 0.6+0.1-0.02 = 0.7-0.02 = 0.68
The probability of neither Tu nor b is 1-P(tu or b) = 1-0.68 = 0.32
If they are mutually exclusive then:
P(tu or b) = P(tu)+P(b) = 0.6+0.1 = 0.7 and neither would be 1-0.7 = 0.3 which is not on your list of answers.

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