Question

In Rebecca's neighborhood, 64% of the houses have garages and 49% have a garage and a pool. What is the probability (in percent) that a house in her neighborhood has a pool, given that it has a garage? Round your answer to 1 decimal place.​

Answer 1

Let P(A) be the probability of a house having a garage and P(B) be the probability of a house having a pool. We are given that P(A) = 64% and P(A ∩ B) = 49%. We want to find P(B|A), the probability of a house having a pool given that it has a garage.

Using Bayes' theorem, we have:

P(B|A) = P(A ∩ B) / P(A)

Substituting the values we have:

P(B|A) = 49% / 64% = 0.7656

Rounding to 1 decimal place, we get:

P(B|A) = 76.6% (rounded to 1 decimal place)

Answer 2

To solve this problem, we can use conditional probability. Let's denote the event that a house has a garage by G, and the event that a house has a pool by P. We are given that P(G) = 0.64 (i.e., 64% of the houses have garages) and P(G and P) = 0.49 (i.e., 49% of the houses have both a garage and a pool).

The conditional probability that a house has a pool, given that it has a garage, can be calculated using the formula:

P(P|G) = P(G and P) / P(G)

Substituting the given probabilities, we get:

P(P|G) = 0.49 / 0.64 = 0.7656

Multiplying by 100 to convert to a percentage and rounding to 1 decimal place, we get:

P(P|G) = 76.6%

So the probability (in percent) that a house in Rebecca's neighborhood has a pool, given that it has a garage, is 76.6%.