## Related questions with answers

An actuary studying the insurance preferences of automobile owners makes the following conclusions: (i) An automobile owner is twice as likely to purchase collision coverage as disability coverage. (ii) The event that an automobile owner purchases collision coverage is independent of the event that he or she purchases disability coverage. (iii) The probability that an automobile owner purchases both collision and disability coverages is 0.15. What is the probability that an automobile owner purchases neither collision nor disability coverage? Choose one of the following. (a) 0.18 (b) 0.33 (c) 0.48 (d) 0.67 (e) 0.82

Solution

VerifiedDefine events

E = " owner purchases collision coverage",

F = " owner purchases disability coverage",

(i) gives us $P(E)=2P(F)$,

(ii) gives us $P(E\cap F)=P(E)\cdot P(F)$

(iii) gives us $P(E\cap F)=0.15.$

## Create an account to view solutions

## Create an account to view solutions

## Recommended textbook solutions

#### Thomas' Calculus

14th Edition•ISBN: 9780134438986 (2 more)Christopher E Heil, Joel R. Hass, Maurice D. Weir#### Finite Mathematics

11th Edition•ISBN: 9780321979438 (1 more)Margaret L. Lial, Nathan P. Ritchey, Raymond N. Greenwell#### Calculus: Early Transcendentals

8th Edition•ISBN: 9781285741550 (1 more)James Stewart#### Calculus: Early Transcendentals

9th Edition•ISBN: 9781337613927 (2 more)Daniel K. Clegg, James Stewart, Saleem Watson## More related questions

- college algebra
- college algebra

1/4

- college algebra
- college algebra

1/7