Thread: Random Probability

Hi, I am not able to figure out the following question. Any help is appreciated.

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.

Originally Posted by delishus

Hi, I am not able to figure out the following question. Any help is appreciated.

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.

Here is what I read from this problem:

Conditions a) and b) suggest that 15% insure EXACTLY one car.
A = exactly one car
P(A) = 0.15
P(A_) = 0.85
B = insure a sports car
P(B) = 0.23
P(B_) = 0.77

68% insure more than one car BUT no sports car --> P(A_ and B_) = 0.68

Want: P(A and B_) = P(A|B_)P(A) = P(B_ |A) P(B_)

still thinking.

Originally Posted by delishus

Hi, I am not able to figure out the following question. Any help is appreciated.

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.

Draw a Karnaugh Table:




It's not difficult to see that a = 0.06, e = 0.15, b = 0.09, d = 0.77 and c = 0.68.

Use this table to answer the question.