On a block of ten houses, [math]k[/math] are not insured. A tornado randomly damages three houses on the block. The probability that none of the damaged houses are insured is 1/120.

Calculate the probability that at most one of the damaged houses is insured.

• 1/5
• 7/40
• 11/60
• 49/60
• 119/120

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

Solution: C

The probability that none of the damaged houses are insured is

[[math]] \frac{1}{120} = \frac{\binom{10-k}{0}\binom{k}{3}}{\binom{10}{3}} = \frac{k(k-1)(k-2)}{720} \Rightarrow k (k − 1)(k − 2) = 6. [[/math]]

This cubic equation could be solved by expanding, subtracting 6, and refactoring. However, because k must be an integer, the three factors must be integers and thus must be 3(2)(1) for k =3.

The probability that at most one of the damaged houses is insured equals

[[math]] \frac{1}{120} + \frac{\binom{10-3}{1}\binom{3}{2}}{\binom{10}{3}} = \frac{1}{120} + \frac{7(3)}{120} = \frac{22}{120} = \frac{11}{60}. [[/math]]

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