The time until the next car accident for a particular driver is exponentially distributed with a mean of 200 days. Calculate the probability that the driver has no accidents in the next 365 days, but then has at least one accident in the 365-day period that follows this initial 365-day period.

• 0.026
• 0.135
• 0.161
• 0.704
• 0.839

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

Solution: B

The desired event is equivalent to the time of the next accident being between 365 and 730 days from now. The probability is

[[math]] F(730) - F(365) = 1-e^{-730/200} - (1-e^{-365/200}) = e^{-1.825} - e^{-3.65} = 0.1352. [[/math]]

Note that the problem provides no information about the distribution of the time to subsequent accidents, but that information is not needed. With nothing given, anything can be assumed. If the time to subsequent accidents has the same exponential distribution and the times are independent, then the number of accidents in each 365 day period is Poisson with mean 1.825. Then the required probability is

[[math]] e^{-1.825}(1-e^{-1.825}) = 0.1352. [[/math]]

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