ABy Admin
Apr 28'23
Exercise
Insurance company examines its pool of auto insurance customers and gathers the following information:
- All customers insure at least one car.
- 64% of the customers insure more than one car.
- 20% of the customers insure a sports car.
- Of those customers who insure more than one car, 15% insure a sports car.
Calculate the probability that a randomly selected customer insures exactly one car, and that the car is not a sports car.
- 0.16
- 0.19
- 0.26
- 0.29
- 0.31
ABy Admin
Apr 28'23
Solution: C
Consider the following events about a randomly selected auto insurance customer:
A = customer insures more than one car
B = customer insures a sports car
We want to find the probability of the complement of A intersecting the complement of B (exactly one car, non-sports). We have [math]\operatorname{P}(A^c \cap B) = 1 - \operatorname{P}(A \cup B). [/math] By the additive Law,
[[math]]
\operatorname{P}( A ∪ B )= \operatorname{P}( A) + \operatorname{P}( B) − \operatorname{P}( A ∩ B)
[[/math]]
By the Multiplicative Law,
[[math]]
\operatorname{P}( A ∩ B ) \operatorname{P}( B | A) \operatorname{P}( A) = (0.15) (0.64) = 0.096.
= 0.096 .
[[/math]]
Then,
[[math]]
\operatorname{P}( A ∪ B ) = 0.64 + 0.20 − 0.096 = 0.744 .
[[/math]]
Finally,
[[math]]
\operatorname{P}( A^c ∩ B^c ) = 1 − 0.744 = 0.256.
[[/math]]