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?
Answer:
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
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