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Example

Example: An insurance company that believes that people can be divided inot two distinct classes -- those who are accident prone and those who are not. During any given year an accident-prone person will have an accident with probability .4, whereas the corresponding figure for a non-accident-prone person is .2. (a) If we assume that 30 percent of the population is accident phone, what is the probability that a new policyholder whill have an accident within a year of purchasing a policy?
(b) What is the conditional probability that a new policyholder will have an accident in his or her second year of policy ownership, given that the policyholder has had an accident in the first year?

[Solution:] (a)
We shall obtain the desired probability by first conditioning upon whether or not the policyholder is accident prone. Let A₁ denote the event that the policyholder will have an accident within a year of purchase; and let A denote the event that the policyholder is accident prone. Hence the desired probability, P(A₁), is given by

$\begin{array}{rcl} P(A_1)&=&P(A_1\vert A)P(A)+P(A_1\vert A^c)P(A^c) \\ &=&(.4)(.3)+(.2)(.7)=.26 \\ \end{array}$

(b)
If we let A be the event that the policyholder is accident prone and we let A_i, i=1,2, be the event that he or she has had an accident in the ith year, then the desired probability P(A₂|A₁) may be obtained by conditioning on whether or not the policyholder is accident prone, as follows: P(A₂|A₁)=P(A₂|AA₁)P(A|A₁)+P(A₂|A^cA₁)P(A^c|A₁)
Now, $\displaystyle P(A\vert A_1)=\frac{P(A_1A)}{P(A_1)}=\frac{P(A_1A)P(A)}{P(A_1)}$
However, P(A) is assumed to equal $\displaystyle\frac{3}{10}$ , and it was shown in (a) that P(A₁)=.26. Hence

$\displaystyle P(A\vert A_1)=\frac{(.4)(.3)}{.26}=\frac{6}{13}$

and thus

$\displaystyle P(A^c\vert A_1)=1-P(A\vert A_1)=\frac{7}{13}$

Since P(A₂|AA₁)=.4 and P(A₂|A^cA₁)=.2, we see that

$\displaystyle P(A_2\vert A_1)=(.4)\frac{6}{13}+(.2)\frac{7}{13}\approx .29\qquad\rule[0.02em]{1.0mm}{1.5mm}$