Answer to Question #202134 in Statistics and Probability for Aasia Tariq

Question

iii) The U.S. Energy Department states that 60% of all U.S. households have ceiling fans. In addition, 29% of all U.S. households have an outdoor grill. Suppose 13% of all U.S. households have both a ceiling fan and an outdoor grill. A U.S. household is randomly selected.

a. What is the probability that the household has a ceiling fan or an outdoor grill?

b. What is the probability that the household has neither a ceiling fan nor an outdoor grill?

c. What is the probability that the household does not have a ceiling fan and does have an outdoor grill?

d. What is the probability that the household does have a ceiling fan and does not have an outdoor grill?

Answer 1

A= ceiling fan

P(A)=0.60

B= outdoor grill

P(B) = 0.29

$P(A \cap B) = 0.13$

a. P(ceiling fan or an outdoor grill)$=P(A) + P(B) - P(A \cap B)$

= 0.60+0.29-0.13

=0.76

b. P(neither a ceiling fan nor an outdoor grill) $= 1 - (P(A) + P(B) - P(A \cap B))$

= 1 - (0.60+0.29-0.13)

= 1 - 0.76

= 0.24

c. P(does not have a ceiling fan and does have an outdoor grill) $= P(B) - P( A \cap B)$

= 0.29 - 0.13

= 0.16

d. P(does have a ceiling fan and does not have an outdoor grill) $= P(A) - P(A \cap B)$

= 0.60 - 0.13

= 0.47