An insurance company looks at its auto insurance customers and finds that (a) all insure at least on

Jameson Powers

Jameson Powers

Answered question

2022-02-13

An insurance company looks at its auto insurance customers and finds that (a) all insure at least one car, (b) 85% insure more than one car, (c) 23% insure a sports car, and (d) 17% insure more than one car, including a sports car. Find the probability that a customer selected at random insures exactly one car and it is not a sports car.

Answer & Explanation

Dzikowiec5wa

Dzikowiec5wa

Beginner2022-02-14Added 13 answers

If 100% insure at least one car and 85% insure more than one car then 15% insure one car.
23% insure a sports car, and 17% insure more than one car including a sports car so 6% insure only a sports car.
If we go back to the 15% that insure one care and subtract the 6% who have one car that is a sports car then u get 9% which is the probability that the person selected randomly is insuring one non-sports car.
star233

star233

Skilled2023-06-15Added 403 answers

Answer: 0.83
Explanation:
Let's denote the following events:
- A: A customer insures at least one car.
- B: A customer insures more than one car.
- C: A customer insures a sports car.
We are given the following probabilities:
- P(A)=1 (since all customers insure at least one car).
- P(B)=0.85 (85% insure more than one car).
- P(C)=0.23 (23% insure a sports car).
- P(BC)=0.17 (17% insure more than one car, including a sports car).
We want to find the probability that a customer selected at random insures exactly one car and it is not a sports car. Let's denote this event as D.
To find P(D), we can use the formula for conditional probability:
P(D)=P(A¬C)=P(A¬C)P(A).
To calculate P(A¬C), we can use the inclusion-exclusion principle:
P(A¬C)=P(A)P(AC).
Now we can substitute the given probabilities into the formula:
P(D)=P(A¬C)P(A)=P(A)P(AC)P(A).
Substituting the known probabilities:
P(D)=1P(AC)1.
Since P(AC)=P(BC) (as all customers who insure more than one car, including a sports car, also insure a sports car), we can write:
P(D)=1P(BC)1.
Substituting P(BC)=0.17:
P(D)=10.171=0.83.
Therefore, the probability that a customer selected at random insures exactly one car and it is not a sports car is 0.83.
karton

karton

Expert2023-06-15Added 613 answers

Let's denote the following events:
A: Insures at least one car.
B: Insures more than one car.
C: Insures a sports car.
We are given the following probabilities:
P(A)=1 (All insure at least one car).
P(B)=0.85 (85% insure more than one car).
P(C)=0.23 (23% insure a sports car).
P(BC)=0.17 (17% insure more than one car, including a sports car).
We are required to find the probability that a customer selected at random insures exactly one car and it is not a sports car. This can be expressed as P(A¬C), where ¬C represents the complement of event C (not insuring a sports car).
Using the formula for conditional probability, we have:
P(A¬C)=P(A)P(AC)
P(A¬C)=1P(C)
Substituting the given value of P(C), we get:
P(A¬C)=10.23
P(A¬C)=0.77
Therefore, the probability that a customer selected at random insures exactly one car and it is not a sports car is 0.77.

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