An actuary compiles the following information from a portfolio of 1000 homeowners insurance policies:

1. 130 policies insure three-bedroom homes.
2. 280 policies insure one-story homes.
3. 150 policies insure two-bath homes.
4. 30 policies insure three-bedroom, two-bath homes.
5. 50 policies insure one-story, two-bath homes.
6. 40 policies insure three-bedroom, one-story homes.
7. 10 policies insure three-bedroom, one-story, two-bath homes.

Calculate the number of homeowners policies in the portfolio that insure neither one-story nor two-bath nor three-bedroom homes.

• 310
• 450
• 530
• 550
• 570

Copyright 2023 . The Society of Actuaries, Schaumburg, Illinois. Reproduced with permission.

Solution: D

Let A, B, and C be the sets of policies in the portfolio on three-bedroom homes, one-story homes, and two-bath homes, respectively. We are asked to calculate 1000 − n( A ∪ B ∪ C ) , where n(D) denotes the number of elements of the set D. Then,

[[math]] \begin{align*} n( A ∪ B ∪ C ) &= n( A) + n( B ) + n(C ) − n( A ∩ B ) − n( A ∩ C ) − n( B ∩ C ) + n( A ∩ B ∩ C ) \\ &= 130 + 280 + 150 − 40 − 30 − 50 + 10 = 450. \end{align*} [[/math]]

The answer is 1000 – 450 = 550.

Copyright 2023 . The Society of Actuaries, Schaumburg, Illinois. Reproduced with permission.