Chapter 13, Problem 26RE

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 $0.02$.

a. What is the probability that a car requires either a tune-up or a brake job?

b. What is the probability that a car requires a tune-up but not a brake job?

c. What is the probability that a car requires neither a tune-up nor a brake job?

To determine

To find:

a. What is the probability that a car requires either a tune-up or a brake job?

Answer to Problem 26RE

a. $0.68$

Explanation of Solution

Given:

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 $0.02$.

Formula used:

$P\left(AUB\right) = P\left(A\right) + P\left(B\right) - P\left(A{{\displaystyle \cap }}^{\text{}}B\right)$

$P\left(A but not B\right) = P\left(A\right) - P\left(A{{\displaystyle \cap }}^{\text{}}B\right)$

$P\overline{\left(A\right)} = 1 - P\left(A\right)$

Calculation:

Let T be a car that needs a tune-up.

$P\left(T\right) = 0.6$

Let B be a car that needs a brake job.

$P\left(B\right) = 0.1$

$P\left(T{{\displaystyle \cap }}^{\text{}}B\right) = 0.02$

a. The probability that a car requires either a tune-up or a brake job is given by

$P\left(T or B\right) = P\left(T{{\displaystyle \cup }}^{\text{}}B\right)$

$P\left(AUB\right) = P\left(A\right) + P\left(B\right) - P\left(A{{\displaystyle \cap }}^{\text{}}B\right)$

$P\left(T{{\displaystyle \cup }}^{\text{}}B\right) = P\left(T\right) + P\left(B\right) - P\left(T{{\displaystyle \cap }}^{\text{}}B\right)$

$P\left(T or B\right) = 0.6 + 0.1 - 0.02 = 0.68$

$P\left(tune - up or brakejob\right) = 0.68$

To determine

To find:

b. What is the probability that a car requires a tune-up but not a brake job?

Answer to Problem 26RE

b. $0.58$

Explanation of Solution

Given:

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 $0.02$.

Formula used:

$P\left(AUB\right) = P\left(A\right) + P\left(B\right) - P\left(A{{\displaystyle \cap }}^{\text{}}B\right)$

$P\left(A but not B\right) = P\left(A\right) - P\left(A{{\displaystyle \cap }}^{\text{}}B\right)$

$P\overline{\left(A\right)} = 1 - P\left(A\right)$

Calculation:

Let T be a car that needs a tune-up.

$P\left(T\right) = 0.6$

Let B be a car that needs a brake job.

$P\left(B\right) = 0.1$

$P\left(T{{\displaystyle \cap }}^{\text{}}B\right) = 0.02$

b. The probability that a car requires a tune-up but not a brake job.

$P\left(A but not B\right) = P\left(A\right) - P\left(A{{\displaystyle \cap }}^{\text{}}B\right)$

$P\left(Tune - up but not a brake job\right) = P\left(T\right) - P\left(T{{\displaystyle \cap }}^{\text{}}B\right)$

$= 0.6 - 0.02 = 0.58$

To determine

To find:

c. What is the probability that a car requires neither a tune-up nor a brake job?

Answer to Problem 26RE

c. $0.32$

Explanation of Solution

Given:

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 $0.02$.

Formula used:

$P\left(AUB\right) = P\left(A\right) + P\left(B\right) - P\left(A{{\displaystyle \cap }}^{\text{}}B\right)$

$P\left(A but not B\right) = P\left(A\right) - P\left(A{{\displaystyle \cap }}^{\text{}}B\right)$

$P\overline{\left(A\right)} = 1 - P\left(A\right)$

Calculation:

Let T be a car that needs a tune-up.

$P\left(T\right) = 0.6$

Let B be a car that needs a brake job.

$P\left(B\right) = 0.1$

$P\left(T{{\displaystyle \cap }}^{\text{}}B\right) = 0.02$

c. The probability that a car requires neither a tune-up nor a brake job.

$P\overline{\left(A\right)} = 1 - P\left(A\right)$

$P\left(neither Tune - up nor a brake job\right) = 1 - P\left(tune - up or a brake job\right)$

$P\left(neither Tune - up nor a brake job\right) = 1 - 0.68 = 0.32$