The question is incomplete! Complete question along with answer and step by step explanation is provided below.
Question:
An insurance company examines its pool of auto insurance customers and gathers the following information: (i) All customers insure at least one car. (ii) 70% of the customers insure more than one car. (iii) 20% of the customers insure a sports car. (iv) 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 car is not a sports car?
Answer:
P( X' ∩ Y' ) = 0.205
Step-by-step explanation:
Let X is the event that the customer insures more than one car.
Let X' is the event that the customer insures exactly one car.
Let Y is the event that customer insures a sport car.
Let Y' is the event that customer insures not a sport car.
From the given information we have
70% of customers insure more than one car.
P(X) = 0.70
20% of customers insure a sports car.
P(Y) = 0.20
Of those customers who insure more than one car, 15% insure a sports car.
P(Y | X) = 0.15
We want to find out the probability that a randomly selected customer insures exactly one car, and that car is not a sports car.
P( X' ∩ Y' ) = ?
Which can be found by
P( X' ∩ Y' ) = 1 - P( X ∪ Y )
From the rules of probability we know that,
P( X ∪ Y ) = P(X) + P(Y) - P( X ∩ Y ) (Additive Law)
First, we have to find out P( X ∩ Y )
From the rules of probability we know that,
P( X ∩ Y ) = P(Y | X) × P(X) (Multiplicative law)
P( X ∩ Y ) = 0.15 × 0.70
P( X ∩ Y ) = 0.105
So,
P( X ∪ Y ) = P(X) + P(Y) - P( X ∩ Y )
P( X ∪ Y ) = 0.70 + 0.20 - 0.105
P( X ∪ Y ) = 0.795
Finally,
P( X' ∩ Y' ) = 1 - P( X ∪ Y )
P( X' ∩ Y' ) = 1 - 0.795
P( X' ∩ Y' ) = 0.205
Therefore, there is 0.205 probability that a randomly selected customer insures exactly one car, and that car is not a sports car.
