Q.3.39 - A First Course in Probability

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?

Step-by-Step Solution

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 1: 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.

Step 2: 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,

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

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

$=\frac{0.56}{0.74}$

$=0.7568$

Step 3: 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,

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

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

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

$=0.24864$

Step 4: 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 $0.24864$.