A car dealership sells 0, 1, or 2 luxury cars on any day. When selling a car, the dealer also tries to persuade the customer to buy an extended warranty for the car. Let [math]X[/math] denote the number of luxury cars sold in a given day, and let [math]Y[/math] denote the number of extended warranties sold.

[[math]] \begin{align*} \operatorname{P}[X = 0, Y = 0] &= 1/6 \\ \operatorname{P}[X = 1, Y = 0] &= 1/12 \\ \operatorname{P}[X = 1, Y = 1] &= 1/6 \\ \operatorname{P}[X = 2, Y = 0] &= 1/12 \\ \operatorname{P}[X = 2, Y = 1] &= 1/3 \\ \operatorname{P}[X = 2, Y = 2] &= 1/6 \\ \end{align*} [[/math]]

Calculate the variance of [math]X[/math].

• 0.47
• 0.58
• 0.83
• 1.42
• 2.58

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

Solution: B

[[math]] \begin{align*} \operatorname{P}(X = 0) &= 1/6 \\ \operatorname{P}(X = 1) &= 1/12 + 1/6 = 3/12 \\ \operatorname{P}(X = 2) &= 1/12 + 1/3 + 1/6 = 7/12 \\ \operatorname{E}[X] &= (0)(1/6) + (1)(3/12) + (2)(7/12) = 17/12 \\ \operatorname{E}[X^2] &= (0)2(1/6) + (1)2(3/12) + (2)2(7/12) = 31/12 \\ \operatorname{Var}[X] &= 31/12 – (17/12)2 = 0.58. \\ \end{align*} [[/math]]

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