An automobile manufacturer is concerned about a possible recall of its best selling four-door sedan. If there was a recall there is a probability that a defect is in the brake system, in the transmission, in the fuel system and in some other area. Part 1 What is the probability that the defect is in the brakes or the transmission system if the probability of defect in both the systems simultaneously is

Complete question is;

An automobile manufacturer is concerned about a possible recall of its best - selling four-door sedan. If there were a recall, there is a probability of 0.25 of a defect in the brake system, 0.18 of a defect in the transmission, 0.17 of a defect in the fuel system, and 0.40 of a defect in some other area.

(a) What is the probability that the defect is the brakes or the fueling system if the probability of defects in both systems simultaneously is 0.15?

(b) What is the probability that there are no defects in either the brakes or the fueling system?

A) 0.27

B) 0.73

Step-by-step explanation:

Let B denote that there is a defect in the break system

Let T demote that there is a defect in transmission

Let F denote that there is a defect in the fuel system.

Let P denote that there is a defect in some other area.

Now, we are given;

P(B) = 0.25

P(T) = 0.18

P(F) = 0.17

P(O) = 0.40

A) We are given that probability of defects in both brakes and the fueling system simultaneously is 0.15.

Thus, it means; P(B ⋂ F) = 0.15

Now, we want to find the probability that the defect is the brakes or the fueling system. This is expressed as;

P(B ⋃ F) = P(B) + P(F) - P(B ⋂ F)

Plugging in the relevant values to give;

P(B ⋃ F) = 0.25 + 0.17 - 0.15

P(B ⋃ F) = 0.27

B) We want to find the probability that there are no defects in either the brakes or the fueling system.

This is expressed as;

P(B' ⋃ F') = 1 - P(B ⋃ F)

Plugging in relevant value;

P(B' ⋃ F') = 1 - 0.27

P(B' ⋃ F') = 0.73

look a the sides they are perpendicular

$4034.40

the perimeter of the field is the sum of the lengths of the four sides:

47 yd + 158 yd + 47 yd + 158 yd = 410 yd

the cost of 3 ft (1 yd) of fence is 3×$3.28 = $9.84.

so the cost of jake's wood perimeter fance is

(410 yd) × ($9.84 /yd) = $4034.40