From 27 pieces of luggage, an airline luggage handler damages a random sample of four. The probability that exactly one of the damaged pieces of luggage is insured is twice the probability that none of the damaged pieces are insured.

Calculate the probability that exactly two of the four damaged pieces are insured.

• 0.06
• 0.13
• 0.27
• 0.30
• 0.31

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

Solution: C

The ratio of the probability that one of the damaged pieces is insured to the probability that none of the damaged pieces are insured is

[[math]] \frac{\binom{r}{1}\binom{27-r}{3}}{\binom{27}{4}} \Big / \frac{\binom{r}{0}\binom{27-r}{4}}{\binom{27}{4}} = \frac{4r}{24-r} [[/math]]

where [math]r[/math] is the total number of pieces insured. Setting this ratio equal to 2 and solving yields [math]r[/math] = 8.

The probability that two of the damaged pieces are insured is

[[math]] \begin{align*} \frac{\binom{r}{2}\binom{27-r}{2}}{\binom{27}{4}} &= \frac{\binom{8}{2}\binom{19}{2}}{\binom{27}{4}} \\ &= \frac{(8)(7)(19)(18)(4)(3)(2)(1)}{(27)(26)(25)(24)(2)(1)(2)(1)} \\ &= \frac{266}{975} \\ &= 0.27. \end{align*} [[/math]]

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