Of households in the United States, 18 million, or 16%, have three or more vehicles, as stated in USA Today (June 12, 2002), quoting the Census Bureau as the source.
(a) If two U.S. households are randomly selected, find the probability that both will have three or more vehicles. (Give your answer correct to four decimal places.)
(b) If two U.S. households are randomly selected, find the probability that neither of the two has three or more vehicles. (Give your answer correct to four decimal places.)
(c) If four households are selected, what is the probability that all four have three or more vehicles? (Give your answer correct to four decimal places.)
To solve these probability questions, we need to consider the information given and use certain probability concepts.
(a) To find the probability that both households will have three or more vehicles, we need to calculate the probability of the first household having three or more vehicles and then multiply it by the probability of the second household having three or more vehicles.
The probability of the first household having three or more vehicles is given as 16% or 0.16. Since we are assuming a random selection, this probability will remain the same for both households.
So, P(both households have three or more vehicles) = P(first household has three or more vehicles) * P(second household has three or more vehicles) = 0.16 * 0.16 = 0.0256.
Therefore, the probability that both households will have three or more vehicles is 0.0256, or 2.56% (rounded to four decimal places).
(b) To find the probability that neither of the two households has three or more vehicles, we need to calculate the probability of the first household not having three or more vehicles and then multiply it by the probability of the second household also not having three or more vehicles.
The probability of a household not having three or more vehicles is the complement of the probability of it having three or more vehicles, which is 1 - 0.16 = 0.84.
So, P(neither household has three or more vehicles) = P(first household does not have three or more vehicles) * P(second household does not have three or more vehicles) = 0.84 * 0.84 = 0.7056.
Therefore, the probability that neither of the two households will have three or more vehicles is 0.7056, or 70.56% (rounded to four decimal places).
(c) To find the probability that all four households have three or more vehicles, we use the same logic. The probability of each household having three or more vehicles is 0.16.
Using the multiplication rule again, we can calculate the probability as follows:
P(all four households have three or more vehicles) = P(first household has three or more vehicles) * P(second household has three or more vehicles) * P(third household has three or more vehicles) * P(fourth household has three or more vehicles) = 0.16 * 0.16 * 0.16 * 0.16 = 0.0016.
Therefore, the probability that all four households will have three or more vehicles is 0.0016, or 0.16% (rounded to four decimal places).