A client spends [math]X[/math] minutes in an insurance agent’s waiting room and [math]Y[/math] minutes meeting with the agent. The joint density function of [math]X[/math] and [math]Y[/math] can be modeled by

[[math]] f(x,y) = \begin{cases} \frac{1}{800} e^{-x/40}e^{-y/20}, \,\, x \gt 0, y \gt 0\\ 0, \, \textrm{Otherwise.} \end{cases} [[/math]]

Determine which of the following expressions represents the probability that a client spends less than 60 minutes at the agent’s office.

• [math]\frac{1}{800}\int_0^{40}\int_0^{20}e^{-x/40}e^{-y/20} dy dx[/math]
• [math]\frac{1}{800} \int_0^{40}\int_0^{20-x}e^{-x/40}e^{-y/20} dy dx[/math]
• [math]\frac{1}{800} \int_0^{20}\int_0^{40-x}e^{-x/40}e^{-y/20} dy dx[/math]
• [math] \frac{1}{800} \int_0^{60}\int_0^{60}e^{-x/40}e^{-y/20} dy dx[/math]
• [math]\frac{1}{800} \int_0^{60}\int_0^{60-x}e^{-x/40}e^{-y/20} dy dx [/math]

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

Solution: E

The total time is less than 60 minutes, so if x minutes are spent in the waiting room, less than 60 − x minutes are spent in the meeting itself.

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