Question

Sixty percent of the families in a certain community own their own car, thirty percent own their own home, and twenty percent own both their own car and their own home. If a family is randomly chosen, what is the probability that this family owns a car or a house but not both?

Short answer

The probability of a randomly chosen family owning either a car or a house but not both is 50%. We found this by using the Inclusion-Exclusion principle: P(A or B but not both) = P(A or B) - P(A and B) = (60% + 30% - 20%) - 20% = 50%.

Step-by-step solution

Find the probability of A or B#

Using the Inclusion-Exclusion principle, the probability of a family owning a car or a house is the sum of the individual probabilities minus the probability of both events occurring together: P(A or B) = P(A) + P(B) - P(A and B) We have the values of P(A), P(B), and P(A and B) from the problem, so we can plug those in to find P(A or B): P(A or B) = 60% + 30% - 20% = 70%.

Find the probability of A or B but not both#

Now, we need to find the probability of A or B happening but not both. Since we already found P(A or B), we only need to subtract the probability of both events occurring together: P(A or B but not both) = P(A or B) - P(A and B) This gives us: P(A or B but not both) = 70% - 20% = 50%. So, the probability of a family owning a car or a house but not both is 50%.