Answer:
0.1847367554
0.9450363585
0.75761
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