v × IG ABox In A Certain Supply R : + 49 lahniel . Strongest Sol | yt Chapter 2

Question

v × IG ABox In A Certain Supply R : + 49 lahniel . Strongest Sol | yt Chapter 2 18266985 06 ho irsurance companyoftes fuer litrent eductible levels-none, low, medium, and high-for its homeowner's policyholders and high -for its automobile policyholders. The accompanying table gives the proportion of individuals with both low homeowner's proportions for the various categories of pollcyholders whoh deductible and low auto deductible is 0.06 (6% of all such individuals), Homeowner's 0.04 0.07 0.06 0.08 0.05 0.20 0.03 0.12 Suppose an Individual having both types of policies is randomly selected. (a) What is the probability that the Individual has a medium auto deductible and a high homeowner's deductible? (b) What is the probabilty that the individual has a low auto deductible? What is the probability that the individual has a low homeowner's deductible? c) What is the probability that the individual is in the same category for both auto and homeowner's deductibles? (d) Based on your answer in part (c), what is the probability that the two categories are different? (e) what is the probability that the individual has at least one low deductible level? the answer in part (e), what is the probability that neither deductible level is low? Type here to search 9

Explanation / Answer

Solution:-

a) The probability that individual has a medium auto deductible and a high homeowner's deductible is 0.12.

b) The probability that individual has a low auto deductible is 0.18.

P(Low Auto) = 0.04 + 0.06 + 0.05 + 0.03

P(Low Auto) = 0.18

The probability that individual has a low homeowner deductible is 0.17.

P(Low homeowner) = 0.06 + 0.08 + 0.03

P(Low homeowner) = 0.17

c) The probability that individual is in the same category for both auto and homeowners deductables is 0.41.

P(Same) = 0.06 + 0.20 + 0.15

P(Same) = 0.41

d) The probability that two categories are different is 0.59.

P(Different) = 1 - P(Same)

P(Different) = 1 - P(Same)

P(Different) = 1 - 0.41

P(Different) = 0.59

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