Stats 151 homework question Please help I am struggling

Question

Recent MLS listings of houses for sale in Edmonton show that 45% have attached garages, 15% have detached garages, and 16% have a home theatre. Additionally, 60% of houses with a home theatre have an attached garage, and 80% of houses with a detached garage do not have a home theatre. (Round your answers to 3 decimal places, if needed.)

(a) What is the probability that a randomly chosen house has a garage?

Answer: 

(b) What is the probability that a house has an attached garage and a home theatre?

Answer: 

(c) What is the probability that a house has an attached garage, but no home theatre?

Answer: 

(d) What is the probability that a house has a detached garage, but no home theatre?

Answer: 

(e) What is the probability that a house has a garage, but no home theatre?

Answer: 

(f) What is the probability that a house has a home theatre given it has an attached garage?

Answer: 

Answer 1

You might consider drawing a Venn Diagram using the verbiage to assist you. It might look like this:

I used 1000 as my number of houses just so all the results have whole numbers.

Answer 2

Let A represent the event that a house has an attached garage, D represent the event that a house has a detached garage, and T represent the event that a house has a home theatre. We will calculate probabilities accordingly.

(a) Probability that a randomly chosen house has a garage

This is the probability that a house has either an attached or a detached garage.

P(Garage) = P(A) + P(D) = 0.45 + 0.15 = 0.60

(b) Probability that a house has an attached garage and a home theatre

We know that 60% of houses with a home theatre have an attached garage.

P(A ∩ T) = P(T) × P(A∣T) = 0.16 × 0.60 = 0.096

(c) Probability that a house has an attached garage, but no home theatre

This is the probability of a house having an attached garage minus the probability of having both an attached garage and a home theatre.

P(A ∩ T^c) = P(A) − P(A ∩ T) = 0.45 − 0.096 = 0.354

(d) Probability that a house has a detached garage, but no home theatre

We know that 80% of houses with a detached garage do not have a home theatre.

P(D ∩ T^c) = P(D) × P(T^c∣D) = 0.15 × 0.80 = 0.12

(e) Probability that a house has a garage, but no home theatre

This is the probability that a house has either an attached or detached garage without a home theatre.

P(Garage ∩ T^c) = P(A ∩ T^c) + P(D ∩ T^c) = 0.354 + 0.12 = 0.474

(f) Probability that a house has a home theatre given it has an attached garage

This is the conditional probability of having a home theatre given that the house has an attached garage.

P(T∣A) = P(A∩T) / P(A) = 0.096 / 0.45 = 0.213

Hope this was helpful.