Help with Probability question.

[Kiziaru]

So I've been trying to do probability problems to take the P/1 Exam, and I've come across a problem that I don't quite understand.

Problem:An insurance company estimates that 40% of policyholders who have only an auto policy
will renew next year and 60% of policyholders who have only a homeowners policy will
renew next year. The company estimates that 80% of policyholders who have both an
auto and a homeowners policy will renew at least one of those policies next year.
Company records show that 65% of policyholders have an auto policy, 50% of
policyholders have a homeowners policy, and 15% of policyholders have both an auto
and a homeowners policy. Using the company’s estimates, calculate the percentage of
policyholders that will renew at least one policy next year.Attempt at solution:

Pr(A ∩ H(compliment)) = Pr(A - H) = Pr(A) - Pr(H)

Pr(A) = .65
Pr(H) = .50

.65 - .50 = .15

However, the solution claims that Pr(A - H) = Pr(A-(A ∩ H)) = .50. How is that possible? How did they derive that?

Edit: Disregard, I understand it now. I had to apply the Pr(A U B) = Pr(A) + Pr(B) - Pr(A ∩ B) rule.

 

[mmwave]

So the solution is actually .65 + .50 - .15 = 1.00. This means that 100% of policyholders will renew at least one policy next year. This makes sense because the problem states that 80% of policyholders with both policies will renew at least one, and since 15% have both policies, that accounts for 80% of the total policyholders. The remaining 20% would be the policyholders with only one policy, and the problem states that 40% of those with only an auto policy will renew, and 60% of those with only a homeowners policy will renew. So .40(.35) + .60(.50) = .14 + .30 = .44. .44 + .80 = 1.00.

Great job on figuring out the solution! The key to understanding this problem is to remember that the probabilities of A and H are not mutually exclusive. In other words, there are policyholders who have both an auto and homeowners policy, so they would be counted in both A and H. This is why we have to subtract the probability of A ∩ H from the sum of Pr(A) and Pr(H). Keep up the good work with your probability problems!