Question

In Example$3\mathrm{a}$, what is the probability that someone has an accident in the second year given that he or she had no accidents in the first year?

Short answer

The probability that someone has an accident in the second year given that he or she had no accidents in the first year is$0.24864$.

Step-by-step solution

Given information

In Example $3a$someone has an accident in the second year given that he or she had no accidents in the first year.

Solution

Let $A$be the possibility that the policyholder is accident prone and $A_1$be the event that the policyholder will have an accident within a year of purchasing the policy.

The tree diagram is given below,

Thus,

$P\left(A\mid A_1^c\right)=\frac{P\left(A\cap A_1^c\right)}{P\left(A_1^c\right)}$

$=\frac{0.3\times 0.6}{0.3\times 0.6+0.7\times 0.8}$

$=\frac{0.18}{0.74}$

$=0.2432$

Then,

$P\left(A^c\mid A_1^c\right)=\frac{P\left(A^c\cap A_1^c\right)}{P\left(A_1^c\right)}$

$=\frac{0.7\times 0.8}{0.3\times 0.6+0.7\times 0.8}$

$=\frac{0.56}{0.74}$

$=0.7568$

solution 

Let $A_2$be the event that the policy holder will have an accident in the second year.

The tree diagram will be,

The probability that a person has a casualty in the second year given that he or she kept no accident in the first year is,

$P\left(A_2\mid A_1^c\right)=P\left(A_2\mid A\cap A_1^c\right)+P\left(A_2\mid A^c\cap A_1^c\right)$

$=0.4\times P\left(A\mid A_1^c\right)+0.2\times P\left(A^c\mid A_1^c\right)$

$=0.4\times 0.2432+0.2\times 0.7568$

$=0.24864$

Final answer

The probability that someone has an accident in the second year given that he or she had no accidents in the first year is$0.24864$.