in a subdivision, 67 perecent of the homes have detached garages, 32 perecent have a patio, and 13 percent have both. What is the probability that a house has a patio if you know that it has a detached garage?

Good guess. Only 13% have BOTH, so it is actually 13/67.

If 13% (.13) have both detached garages and patios, isn't that the probability?

Please only post your questions once. Repeating posts will not get a quicker response. In addition, it wastes our time looking over reposts that have already been answered in another post. Thank you.

No, the probability of a house having a patio given that it has a detached garage is not simply 32/67.

To find the probability, we need to use conditional probability. In this case, we want to find the probability of a house having a patio, given that it has a detached garage.

Conditional probability can be calculated using the formula:

P(A|B) = P(A ∩ B) / P(B)

Where:
P(A|B) represents the probability of event A occurring given that event B has occurred.
P(A ∩ B) represents the probability of both events A and B occurring.
P(B) represents the probability of event B occurring.

In this case, event A represents having a patio, and event B represents having a detached garage.

Given that 67% of homes have detached garages, the probability of having a detached garage (P(B)) is 67/100 or 0.67.

Given that 13% of homes have both a detached garage and a patio (P(A ∩ B) = 0.13), we need to find the probability of having a patio (P(A)).

To find P(A), we can use the principle of inclusion-exclusion:

P(A) = P(A ∩ B) + P(A ∩ ~B)

Where:
P(A) represents the probability of having a patio.
P(A ∩ ~B) represents the probability of having a patio but not having a detached garage.

Since P(A ∩ B) = 0.13, we need to find P(A ∩ ~B).

P(A ∩ ~B) = P(A) - P(A ∩ B)
P(A ∩ ~B) = P(A) - 0.13

To find P(A), we subtract the probability of not having a patio from 1:

P(A) = 1 - P(~A)

Since 32% of homes have a patio (P(~A) = 0.32), we have:

P(A ∩ ~B) = 1 - 0.32 - 0.13
P(A ∩ ~B) = 0.55

Now, we can calculate the probability of having a patio given that a house has a detached garage:

P(A|B) = P(A ∩ B) / P(B)
P(A|B) = 0.13 / 0.67

Therefore, the probability of a house having a patio, given that it has a detached garage, is approximately 0.194 or 19.4%.