I am studying for the Exam P for actuarial credentialing. I came across this problem, which is in reality, likely quite easy but for some reason I got wrong. I will give the relevant information below and the question statement and my quick work. Any help in seeing where I went wrong and helping to develop an intuition for problems of this nature would be greatly appreciated.
An insurance company has data that indicates that $65$% of policyholders has an auto insurance policy, $50$% have a homeowners policy and $15$% have both. Analysis predicts that $40$% with only an auto policy will renew next year, $60$% with only a homeowners policy will renew next year and $80$% with both will renew next year. What percentage of policyholders will renew at least one policy next year?
For my attempt I basically just took a linear combination of the percentage of each policy with the percentage of those policyholders that will renew next year to find that: $$(65 \cdot 0.4) + (50 \cdot 0.6) + (15 \cdot 0.8) = 26 +30 +12 = 68$$ Hence $68$% of policyholders will renew at least one policy next year. However, this is incorrect as the solution guide states the correct answer is $53$%. Could it be an application of the inclusion exclusion principle? Considering we are dealing with orders of two sets and their intersection?
The only thing I would add to the above comments is that the "linear combination" you refer to is due to conditional probability. If the events $A_1, A_2, \ldots A_n$ form a partition of your sample space -- that is, all $A_i$ are disjoint and their union is the entire sample space -- then for some other event $B$,
$$ P(B) = \sum_i P(B \, | \, A_i) P (A_i). $$
You can find a proof in any study manual, I imagine.
In your example, let $A$ be the event that the policyholder has only auto insurance; $H$, only home insurance; and $B$, both; and let $R$ be the event that the policy is renewed. Then $A$, $H$ and $B$ form a partition -- after all, the policyholder can't hold no policies -- and so
$$P(R) = P(R \, | \, A) P (A) + P(R \, | \, H) P (H) + P(R \, | \, B) P (B), $$
an equation to which you've already assigned values.