Answer to Question #139082 in Statistics and Probability for qwerty

Question

5. 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 and the transmission 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 and the fueling system?

Answer 1

given : p(B)= 0.25 P(T)= 0.18 P(F)=0.17 P(O)=0.40

Solution :

i)from given data P(B n F)=0.15

We know P(B U F)= P(B)+P(F)-P(B n F)

=0.25+0.17-0.15=0.27

ii)To find :$\overline{P(BUF)}= 1- P(BUF)$

= 1- 0.27

=0.73