Question

11) The probability of an Emerald Gates subdivision resident having a privacy fence is 60%. The probability of a resident owning a dog is 35%, and the probability they own a dog and have a privacy fence is 25%. (8 points) a) Given a resident has a dog, what is the probability they have a privacy fence? b) Are having a dog and having a privacy fence independent events? Explain using probabilities. c) Are having a dog and having a privacy fence disjoint events? Explain using probabilities. 12) It is believed that 47% of American households own no American made vehicles, 27% own 1 American made vehicle, 19% own 2 American made vehicles and the rest own 3 or more American made vehicles (7 points) a) Find the probability that a randomly selected American household owns 3 or more American made vehicles. b) If 3 American households are considered, what is the probability that none of the 3 own 1 American made vehicle? c) If 3 American households are considered, what is the probability at least one owns 3 or more American made vehicles? 13) Fill in the blank: (1 point) The states that, as the number of observations drawn at random from a population with finite mean I increases, the mean of the observed values gets closer and closer to the population mean .

Answer 1

Solution:

11) Let us define some events as follows:

A : A resident has a privacy fence

B : A resident has a dog

a) We have to obtain P(A|B).

Conditional probability of A given B is given as follows:

P(AB) = P(ANB) P(B)

P(A ∩ B) = Probability that a resident own a dog and also he has privacy fence

P(B) = Probability that a resident own a dog

We have, P(A ∩ B) = 25% = 0.25 and P(B) = 35% = 0.35

Hence, P(A|B) = 0.25/0.35 = 5/7

Given that a resident own a dog, the probability that he has a privacy fence is 5/7.

b) If two events A and B are independent then, P(A ∩ B) = P(A).P(B)

We have, P(A) = 0.60, P(B) = 0.35 and P(A ∩ B) = 0.25

Now, P(A).P(B) = 0.60×0.35 = 0.21 which is not equal to P(A ∩ B).

Hence, having a dog and having a fence are not independent events.

c) If two events A and B are disjoint events then P(A ∩ B) = 0

We are given that, P(A ∩ B) = 0.25 which is not equal to zero.

Therefore, having a dog and having a privacy fence are not disjoint events.