Answer to Question #202341 in Statistics and Probability for Aasia Tariq

Question

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

Let $C$  represent that US household has ceiling fan: $P(C)=0.60.$

Let $G$  represent that US household has outdoor grill: $P(G)=0.29.$

The probability that both ceiling fan and outdoor grill is $P(C\cap G)=0.13.$

a.

$$P(C\cup G)=P(C)+P(G)-P(C\cap G)$$

$$=0.60+0.29-0.13=0.76$$

The probability that the household has a ceiling fan or an outdoor grill is 0.76.

b.

$$P(C'\cap G')=1-P(C\cup G)=1-0.76$$

$$=0.24$$

The probability that the household has neither a ceiling fan nor an outdoor grill IS 0.24.

c.

$$P(C'\cap G)=P(G)-P(C\cap G)=0.29-0.13$$

$$=0.16$$

The probability that the household does not have a ceiling fan and does have an outdoor grill is 0.16.

d.

$$P(C\cap G')=P(C)-P(C\cap G)=0.60-0.13$$

$$=0.47$$

The probability that the household does have a ceiling fan and does not have an outdoor grill iis 0.47.