Consider a large insurance company with two types of policies: A and B. Suppose the number of claims the company sees in a given day has a Poissondistribution with a mean 12. Suppose further that a randomly selected claim isfrom a Type A policy is 1/4.

a. Find the probability that the company will receive at least 5 claims from A policies tomorrow.

b. Find the probability that the company will receive at least 5 claims from B policies tomorrow.

c. Find the probability that the company will receive at least 10 total claims tomorrow.

0.1847367554

0.9450363585

0.75761

Step-by-step explanation:

Given:

- The total number of claims n follows a possion distribution with mean λ = 12.

- The probability that claim is from Type A policy is p = 0.25

- The number of claims received from policy A = k.

Find:

a. Find the probability that the company will receive at least 5 claims from A policies tomorrow.

b. Find the probability that the company will receive at least 5 claims from B policies tomorrow.

c. Find the probability that the company will receive at least 10 total claims tomorrow.

Solution:

- We will use the derived results of multi-nomial theorem to calculate exactly k of n cases are of type A policy provided both distribution follows a Poisson distribution.

- The derived result is:

P (Type A policy = k) = (λ*p) ^k / k! e^ (λ*p)

Where λ*p = 12*0.25 = 3 for A

Where λ * (1-p) = 12*0.75 = 9 for B

Hence,

P (Type A policy = k) = 3^k / k! e^3

P (Type B policy = k) = 9^k / k! e^9

part a)

- The probability that company receives at-least 5 claims from Type A. Using the derived poisson distribution we have for Type A:

1 - [ P (k = 0) + P (k = 1) + P (k = 2) + P (k = 3) + P (k = 4) ]

1 - [ 1/e^3 + 3/e^3 + 9/2*e^3 + 27/6*e^3 + 81/24*e^3 + 243/120*e^3 ]

1 - [0.8152632446] = 0.1847367554

part b)

- The probability that company receives at-least 5 claims from Type B. Using the derived poisson distribution we have for Type B:

1 - [ P (k = 0) + P (k = 1) + P (k = 2) + P (k = 3) + P (k = 4) ]

1 - [ 1/e^9 + 9/e^9 + 81/2*e^9 + 729/6*e^9 + 6561/24*e^9 + 59049/120*e^9]

1 - [0.05496364145] = 0.9450363585

part c)

- The probability that company receives at-least 10 claims in total. The poisson distribution we have number of claims is:

P (X = k) = (λ) ^k / k! e^ (λ) = 12^k / k!*e^12

P (X > = 10) = 1 - P (X < 9)

1 - [ 1/e^12 + 12/e^12 + 12^2/2*e^12 + 12^3/6*e^12 + 12^4/24*e^12 + 12^5/120*e^12 + 12^6/720*e^12 + 12^7/5040*e^12 + 12^8/40320*e^12 + 12^9/362880*e^12]

P (X > = 10) = 1 - 0.2423921617 = 0.75761