91Ó°ÊÓ

At the Milex tune-up and brake repair shop, the manager has found that a car 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.

Let A be the event that the car requires a tune-up and B be the event that the car requires a brake job.

To find the probability that a car requires either a tune-up or a brake job we will use the equation $P\left(A\text{or}B\right)=P\left(A\right)+P\left(B\right)-P\left(A\text{and}B\right)$

Substitute $P\left(A\right)=0.6,P\left(B\right)=0.1,P\left(AandB\right)=0.02$in the equation and evaluate

$$\begin{gathered} P\left(AorB\right)=P\left(A\right)+P\left(B\right)-P\left(AandB\right) \\ P\left(AorB\right)=0.6+0.1-0.02 \\ P\left(AorB\right)=0.68 \end{gathered}$$

To find the probability that a car requires a tune-up but not a brake job can be calculated using the equation $P(Aâˆ\copyright \stackrel{¯}{B})=P\left(A\right)-P(Aâˆ\copyright B)$

Substitute $P\left(A\right)=0.6,P\left(AandB\right)=0.02$and evaluate

$$\begin{gathered} P(Aâˆ\copyright \stackrel{¯}{B})=P\left(A\right)-P(Aâˆ\copyright B) \\ P(Aâˆ\copyright \stackrel{¯}{B})=0.6-0.02 \\ P(Aâˆ\copyright \stackrel{¯}{B})=0.58 \end{gathered}$$

From the previous part, the probability that a car requires either a tune-up or a brake job is 0.68

To find the probability that a car requires neither a tune-up nor a brake job can be calculated by using the equation $P(\stackrel{¯}{\text{tune up}}or\stackrel{¯}{\text{brake job}})=1-P\left(\text{tune uporbrake job}\right)$

Substitute $P\left(\text{tune uporbrake job}\right)=0.68$in the equation and evaluate\

$P(\stackrel{¯}{\text{tune up}}∪\stackrel{¯}{\text{brake job}})=1-0.68=0.32$