Exercise
ABy Admin
Apr 28'23
Answer
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]]