Each time a hurricane arrives, a new home has a 0.4 probability of experiencing damage. The occurrences of damage in different hurricanes are mutually independent.

Calculate the mode of the number of hurricanes it takes for the home to experience damage from two hurricanes.

• 2
• 3
• 4
• 5
• 6

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

Solution: B

Let [math]X[/math] equal the number of hurricanes it takes for two losses to occur. Then [math]X[/math] is negative binomial with “success” probability [math]p = 0.4[/math] and [math]r = 2[/math] “successes” needed.

[[math]] \operatorname{P}[X = n] = \frac{n-1}{r-1} p^{r}(1-p)^{n-r} = \binom{n-1}{2-1} (0.4)^2(1-0.4)^{n-2} = (n-1)(0.4)^2(0.6)^{n-2}, \, n \geq 2. [[/math]]

We need to maximize [math]\operatorname{P}[X = n][/math]. Note that the ratio

[[math]] \frac{\operatorname{P}[X=n+1]}{\operatorname{P}[X=n]} = \frac{n(0.4)^2(0.6)^{n-1}}{(n-1)(0.4)^2(0.6)^{n-2}} = \frac{n}{n-1} (0.6). [[/math]]

This ratio of “consecutive” probabilities is greater than 1 when [math]n = 2[/math] and less than 1 when [math]n ≥ 3.[/math] Thus, [math]\operatorname{P}[X = n][/math] is maximized at [math]n = 3[/math]; the mode is 3. Alternatively, the first few probabilities could be calculated.

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