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

3/4 ÷ 3/4

step-by-step explanation:

basically divide any proper fraction with itself to fit the criteria.

so if we are using 3/4, then:

3/4 ÷ 3/4 = 1

so 1 is the quotient.

and the numerators for both these fractions is 3.   and if we divide the numerators:

3 ÷ 3 = 1

we see that quotient is equal to our original quotient as well.

welll its a very hard question but ill

step-by-step explanation:

first you take the 6 dollars and add 10 to it then add the 50 to that then multiply by 8 with xy and your answer will be

hope this