Question

A company has five employees on its health insurance plan. Each year, each employee independently has an 80% probability of no hospital admissions. If an employee requires one or more hospital admissions, the number of admissions is modeled by a geometric distribution with a mean of 1.50. The numbers of hospital admissions of different employees are mutually independent. Each hospital admission costs 20,000.

Calculate the probability that the company's total hospital costs in a year are less than 50,000.

Answer 1

the probability that the company's total hospital costs in a year are less than 50,000 = 0.7828

Explanation:

From the given information:

the probability that the company's total hospital costs in a year are less than 50,000 will be the sum of the probability of the employees admitted.

If anyone is admitted to the hospital, they have
`(1)/(3)` probability of making at least one more visit, and a
`(2)/(3)` probability that this is their last visit.

If zero employee was admitted ;

Then:

Probability = (0.80)⁵

Probability = 0.3277

If one employee is admitted once;

Probability =
`(0.80)^4 * (0.20)^1 * (^5_1) * ((2)/(3))`

Probability =
`(0.80)^4 * (0.20)^1 * ((5!)/((5-1)!)) * ((2)/(3))`

Probability = 0.2731

If one employee is admitted twice

Probability =
`(0.80)^3 * (0.20)^2 * (^5_2) * ((2)/(3))^2`

Probability =
`(0.80)^3 * (0.20)^2 * ((5!)/((5-2)!)) * ((2)/(3))^2`

Probability = 0.1820

If two employees are admitted once

Probability =
`(0.80)^4* (0.20)^1 * (^5_1) * ((1)/(3)) * ((2)/(3))`

Probability =
`(0.80)^4 * (0.20)^1 * ((5!)/((5-1)!)) * ((1)/(3)) * ((2)/(3))`

Probability = 0.0910

∴

the probability that the company's total hospital costs in a year are less than 50,000 = 0.3277 + 0.2731 + 0.1820

the probability that the company's total hospital costs in a year are less than 50,000 = 0.7828