A car is new at the beginning of a calendar year. The time, in years, before the car experiences its first failure is exponentially distributed with mean 2.

Calculate the probability that the car experiences its first failure in the last quarter of some calendar year.

• 0.081
• 0.088
• 0.102
• 0.205
• 0.250

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

Solution: D

Using the law of total probability, the requested probability is

[[math]] \sum_{k=0}^{\infty} \operatorname{P}(k + 0.75 \lt X ≤ k + 1| k \lt X ≤ k + 1)\operatorname{P}(k \lt X ≤ k + 1) . [[/math]]

The first probability is

[[math]] \begin{align*} \operatorname{P}(k + 0.75 \lt X ≤ k + 1 | k \lt X ≤ k + 1) &= \frac{\operatorname{P}(k + 0.75 \lt X ≤ k + 1)}{\operatorname{P}(k \lt X ≤ k + 1)} \\ &= \frac{F(k + 1) − F (k + 0.75)}{F (k + 1) − F (k)} \\ &= \frac{1 − e^{-( k +1)/2} − 1 + e^{-(k + 0.75)/2}}{1 − e^{−( k +1)/2} − 1 + e^{− k /2}} \\ &= \frac{e^{−0.375} − e^{−0.5}}{1-e^{-0.5}} \\ &= 0.205 \end{align*} [[/math]]

This probability factors out of the sum and the remaining probabilities sum to 1 so the requested probability is 0.205.

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