An actuary studying the insurance preferences of automobile owners makes the following conclusions:

1. An automobile owner is twice as likely to purchase collision coverage as disability coverage.
2. The event that an automobile owner purchases collision coverage is independent of the event that he or she purchases disability coverage.
3. The probability that an automobile owner purchases both collision and disability coverages is 0.15.

Calculate the probability that an automobile owner purchases neither collision nor disability coverage.

• 0.18
• 0.33
• 0.48
• 0.67
• 0.82

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

Solution: B

Let

[math]C[/math] = Event that a policyholder buys collision coverage

[math]D[/math] = Event that a policyholder buys disability coverage

Then we are given that [math]\operatorname{P}[C] = 2\operatorname{P}[D][/math] and [math]\operatorname{P}[C ∩ D] = 0.15[/math]. By the independence of [math]C[/math] and [math]D[/math], it therefore follows that

[[math]] 0.15 = \operatorname{P}[C ∩ D] = \operatorname{P}[C] \operatorname{P}[D] = 2\operatorname{P}[D] \operatorname{P}[D] = 2(\operatorname{P}[D])2 (\operatorname{P}[D])2 = 0.15/2 = 0.075 \operatorname{P}[D] = 0.075 [[/math]]

and [math]\operatorname{P}[C] = 2\operatorname{P}[D] = 2*0.075[/math]. Now the independence of C and D also implies the independence of [math]C^c[/math] and [math]D^c[/math] . As a result, we see that

[[math]] \operatorname{P}[C^c ∩ D^c] = \operatorname{P}[C^c] \operatorname{P}[D^c] = (1 – \operatorname{P}[C]) (1 – \operatorname{P}[D]) = (1 – 2 \sqrt{0.075} ) (1 – \sqrt{0.075} ) = 0.33 . [[/math]]

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