Haiden Meyer

2022-10-08

Why is the sum of probability is not equal to 1 in a geometric distribution?

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.

The probability of each employee has at least one hospital admissions is $(0.2\cdot 1)/(1-2/3)=0.6$ instead of 0.2. Why is that?

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.

The probability of each employee has at least one hospital admissions is $(0.2\cdot 1)/(1-2/3)=0.6$ instead of 0.2. Why is that?

Joel Reese

Beginner2022-10-09Added 17 answers

Step 1

If the employee has a hospital admission, the number of admissions is geometrically distributed with mean $\frac{3}{2}$. That is the probability of exactly k admissions is $p(1-p{)}^{k-1},\text{}k=1,2,\dots $ and the mean is $\frac{1}{p}$ so $p=\frac{2}{3}.$.

Step 2

However, these probabilities are conditional probabilities, contingent on the employee's having had at least one admission (probability .2.) Therefore the probability ${p}_{k}$ that an employee has exactly k admissions is ${p}_{k}=\{\begin{array}{ll}.8,& {\textstyle k=0}\\ .4\cdot {3}^{-k},& {\textstyle k>0}\end{array}$

If the employee has a hospital admission, the number of admissions is geometrically distributed with mean $\frac{3}{2}$. That is the probability of exactly k admissions is $p(1-p{)}^{k-1},\text{}k=1,2,\dots $ and the mean is $\frac{1}{p}$ so $p=\frac{2}{3}.$.

Step 2

However, these probabilities are conditional probabilities, contingent on the employee's having had at least one admission (probability .2.) Therefore the probability ${p}_{k}$ that an employee has exactly k admissions is ${p}_{k}=\{\begin{array}{ll}.8,& {\textstyle k=0}\\ .4\cdot {3}^{-k},& {\textstyle k>0}\end{array}$

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