Chapter PIV, Problem 1RE

(a)

To determine

To find: the probability that if it's a new car, it's got some type of defect.

Answer to Problem 1RE

0.34

Explanation of Solution

Given:

  $P\left(C\right)=0.29$

  $P\left(F\right)=0.07$

  $P\left(C\cap F\right)=0.02$

Formula used:

  $P\left(C\cap F\right)=P\left(C\right)+P\left(F\right)-P\left(C\cap F\right)$

Calculation:

Event C represents a cosmetic defect in the new cars.

Event F represents a functional defect in the car.

Required probability is

  $\begin{array}{c}P\left(C\cap F\right)=P\left(C\right)+P\left(F\right)-P\left(C\cap F\right)\\ =0.29+0.07-0.02\\ =0.36-0.02\\ =0.34\end{array}$

Therefore, the required probability is 0.34.

(b)

To determine

To find: the possibility that it has a cosmetic defect but no functional defect.

Answer to Problem 1RE

0.27

Explanation of Solution

Given:

  $P\left(C\right)=0.29$

  $P\left(F\right)=0.07$

  $P\left(C\cap F\right)=0.02$

Formula used:

  $P\left(C\cap F^{\text{'}}\right)=P\left(C\right)-P\left(C\cap F\right)$

Calculation:

  $\begin{array}{c}P\left(C\cap F^{\text{'}}\right)=P\left(C\right)-P\left(C\cap F\right)\\ =0.29-0.02\\ =0.27\end{array}$

Therefore, the required probability is 0.27

(c)

To determine

To find: the probability that if a crack on a new car is found, it has a functional defect.

Answer to Problem 1RE

0.069

Explanation of Solution

Given:

  $P\left(C\right)=0.29$

  $P\left(F\right)=0.07$

  $P\left(C\cap F\right)=0.02$

Formula used:

  $P\left(F|C\right)=\frac{P\left(C\cap F\right)}{P\left(C\right)}$

Calculation:

  $\begin{array}{c}P\left(F|C\right)=\frac{P\left(C\cap F\right)}{P\left(C\right)}\\ =\frac{0.02}{0.29}\\ =0.0689\\ =0.069\end{array}$

Therefore, the required probability is 0.069

(d)

To determine

To explain: the events are disjointed by the two forms of defects.

Explanation of Solution

Given:

  $P\left(C\right)=0.29$

  $P\left(F\right)=0.07$

  $P\left(C\cap F\right)=0.02$

The two types of events are not cases that are disjointed. Therefore, the chance of new vehicles getting all kinds of problems is 2%.

(e)

To determine

To explain: the two forms of defects are independent cases.

Answer to Problem 1RE

Independent

Explanation of Solution

Given:

  $P\left(C\right)=0.29$

  $P\left(F\right)=0.07$

  $P\left(C\cap F\right)=0.02$

Calculation:

If the events should say to be independent then,

  $P\left(C\cap F\right)=P\left(C\right)\times P\left(F\right)$

Therefore,

  $\begin{array}{c}P\left(C\cap F\right)=P\left(C\right)\times P\left(F\right)\\ 0.02=0.29\times 0.07\\ 0.02=0.02\end{array}$

Therefore, both events are independent.