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Question
| CASE-BASED/DATA-BASED |
 |
| An insurance company believes that people can be divided into two classes: those who are accident prone and those who are not. The company’s statistics show that an accident-prone person will have an accident at some time within a fixed one-year period with a probability 0.6, whereas this probability is 0.2 for a person who is not accident prone. The company knows that 20 percent of the population is accident prone. |
Based on the given information, answer the following questions.
- What is the probability that a new policyholder will have an accident within a year of purchasing a policy?
- Suppose that a new policyholder has an accident within a year of purchasing a policy. What is the probability that he or she is accident prone?
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Solution
Let E1 = The policyholder is accident prone.
E2 = The policyholder is not accident prone.
E = The new policyholder has an accident within a year of purchasing a policy.
i. `"P"("E") = "P"("E"_1) × "P"("E"/"E"_1) + "P"("E"_2) × "P"("E"/"E"_2)`
= `20/100 xx 6/10 + 80/100 xx 2/10`
= `7/25`
ii. By Bayes’ Theorem, `"P"("E"_1/"E") = ("P"("E"_1) xx "P"("E"/"E"_1))/("P"("E"))`
= `(20/100 xx 6/10)/(280/1000)`
= `3/7`
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Solution: Let A, C and T be the events that Mr. X goes to office by Auto, Car and Train respectively. Let L be event that he is late.
Given that P(A) = `square`, P(C) = `square`
P(T) = `square`
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P(L) = P(A ∩ L) + P(C ∩ L) + P(T ∩ L)
`="P"("A")*"P"("L"//"A") + "P"("C")*"P"("L"//"C") + "P"("T")*"P"("L"//"T")`
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`"P"("C"//"L") = ("P"("L" ∩ "C"))/("P"("L"))`
= `("P"("C") * "P"("L"//"C"))/("P"("L"))`
`= (square * square)/square`
`= square`
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