91Ó°ÊÓ

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

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,

$P\left(Aâˆ\pounds A_1^c\right)=\frac{P\left(Aâˆ\copyright A_1^c\right)}{P\left(A_1^c\right)}$

$=\frac{0.3×0.6}{0.3×0.6+0.7×0.8}$

$=\frac{0.18}{0.74}$

$=0.2432$

Then,

$P\left(A^câˆ\pounds A_1^c\right)=\frac{P\left(A^câˆ\copyright A_1^c\right)}{P\left(A_1^c\right)}$

$=\frac{0.7×0.8}{0.3×0.6+0.7×0.8}$

$=\frac{0.56}{0.74}$

$=0.7568$

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,

$P\left(A_2âˆ\pounds A_1^c\right)=P\left(A_2âˆ\pounds Aâˆ\copyright A_1^c\right)+P\left(A_2âˆ\pounds A^câˆ\copyright A_1^c\right)$

$=0.4×P\left(Aâˆ\pounds A_1^c\right)+0.2×P\left(A^câˆ\pounds A_1^c\right)$

$=0.4×0.2432+0.2×0.7568$

$=0.24864$

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