Question

Automobile policies are separated into two groups: low-risk and high-risk. Actuary Rahul examines low-risk policies, continuing until a policy with a claim is found and then stopping. Actuary Toby follows the same procedure with high-risk policies. Each low-risk policy has a 10% probability of having a claim. Each high-risk policy has a 20% probability of having a claim. The claim statuses of policies are mutually independent. Calculate the probability that Actuary Rahul examines fewer policies than Actuary Toby.

Answer 1

The probability that Actuary Rahul examines fewer policies than Actuary Toby is 0.2857.

Explanation:

It is provided that the automobile policies are separated into two groups: low-risk and high-risk. Actuary Rahul examines low-risk policies, continuing until a policy with a claim is found and then stopping. Actuary Toby follows the same procedure with high-risk policies.

The probability that a low-risk policy has a claim is, P (L) = 0.10.

The probability that a high-risk policy has a claim is, P (H) = 0.20.

For positive integer n, the probability that Actuary Rahul examines exactly n policies is:

P (Actuary Rahul examines exactly n policies) =
`(1-0.10)^(n-1) *0.10`

`=0.90^(n-1)* 0.10`

The probability that Actuary Toby examines more than n policies is:

P (Actuary Toby examines more than n policies) =
`(1-0.20)^(n)`

`=0.80^(n)`

It is provided that the claim statuses of policies are mutually independent.

Compute the probability that Actuary Toby examines more policies than Actuary Rahul as follows:

`P(\text{Toby}>\text{Rahul})=\sum\limits^(\infty)_(n=1){(0.90^(n-1)* 0.10)* 0.80^(n)}`

`=(0.10)/(0.90)\sum\limits^(\infty)_(n=1){0.90^(n)* 0.80^(n)}\\\\=0.1111* \sum\limits^(\infty)_(n=1){0.72^(n)}\\\\=0.1111* (0.72)/(1-0.72)\\\\=0.285686\\\\\approx 0.2857`

Thus, the probability that Actuary Rahul examines fewer policies than Actuary Toby is 0.2857.