Question

14) World Wide Insurance Company found that 53% of the residents of a city had homeowner’s insurance (H) with the company. Of these clients, 27% also had automobile insurance (A) with the company. If a resident is selected at random, find the probabil- ity that the resident has both homeowner’s and automobile insurance with World Wide Insurance Company.

Answer 1

The probability that the resident has both homeowner’s and automobile insurance with World Wide Insurance Company is 0.143 OR 14.3%.

Explanation:

We are given that:

53% of the residents had homeowner's insurance i.e. P(H) = 0.53

Of these people, 27% also had automobile insurance i.e. they had automobile insurance given that they had homeowner's insurance. So, P(A|H) = 0.27.

We need to find out the probability that the resident has both homeowner's and automobile insurance i.e. P(A and H) or P(A∩H). To compute P(A∩H) we will use the conditional probability formula:a

P(A|B) = P(A∩B)/P(B)

Here we have:

P(A|H) = P(A∩H)/P(H)

P(A∩H) = P(A|H) * P(H)

= 0.27 * 0.53

P(A∩H) = 0.143

The probability that the resident has both homeowner’s and automobile insurance with World Wide Insurance Company is 0.143 OR 14.3%.

Answer 2

The probability that the resident has both homeowner’s and automobile insurance with World Wide Insurance Company is 14% (P=0.1431).

Explanation:

The probability of having homeowner’s insurance (H) with the company is P(H)=0.53.

The probability that, given that the resident has homeowner insurance with the company, also had a automobile insurance is P(A|H)=0.27.

We have to calculate P(A&H): probability of having automovile and homeowner:

`P(A\&H)=P(H)*P(A|H)=0.53*0.27=0.1431`