For the development of this problem we will use the basic concept of Probability and relation of sets.
Our values are:
[tex]P(Grill)= P_g(x)= 0.29[/tex]
[tex]P(fans)=P_f(x)= 0.6[/tex]
[tex]P(both)=P_b(x)= 0.13[/tex]
PART A ) What is the probability that the household has a ceiling fan or an outdoor grill?
[tex]P_f(x) \cup P_g(x)= P_f(x) +P_g(x)-P_b(x)[/tex]
[tex]P_f(x) \cup P_g(x)= 0.6+0.29-0.13[/tex]
[tex]P_f(x) \cup P_g(x)= 0.76[/tex]
PART B) What is the probability that the household has neither a ceiling fan nor an outdoor grill?
[tex]P_f(x)' \not\cup P_g(x)'=1-P_f(x) \cup P_g(x)\\P_f(x)' \not\cup P_g(x)'=1-0.76\\P_f(x)' \not\cup P_g(x)'= 0.24[/tex]
PART c) What is the probability that the household does not have a ceiling fan and does have an outdoor grill?
[tex]P_f(x)' \cap P_g(x)=P_g(x)-P_f(x) \cap P_g(x)\\P_f(x)' \cap P_g(x)= 0.29-0.13\\P_f(x)' \cap P_g(x)=0.16[/tex]
PART D) What is the probability that the household does have a ceiling fan and does not have an outdoor grill?
[tex]P_f(x) \cap P_g(x)' = P_f(x)-P_f(x) \cap P_g(x)\\P_f(x) \cap P_g(x)' =0.6-0.13\\P_f(x) \cap P_g(x)' =0.47[/tex]
