Студопедия
Случайная страница | ТОМ-1 | ТОМ-2 | ТОМ-3
АвтомобилиАстрономияБиологияГеографияДом и садДругие языкиДругоеИнформатика
ИсторияКультураЛитератураЛогикаМатематикаМедицинаМеталлургияМеханика
ОбразованиеОхрана трудаПедагогикаПолитикаПравоПсихологияРелигияРиторика
СоциологияСпортСтроительствоТехнологияТуризмФизикаФилософияФинансы
ХимияЧерчениеЭкологияЭкономикаЭлектроника

The hypergeometric probability distribution

Читайте также:
  1. Areas under continuous probability density functions
  2. Compute the required probability using the normal distribution.
  3. Conditional probability
  4. Distribution
  5. Distribution
  6. Formula for classical probability
  7. Frequency distribution. Grouped data and histograms

In previous section, we have learned that one of the conditions required to apply the binomial probability distribution is that the trials are independent so that the probabilities of the two outcomes (success and failure) remains constant. If the trials are not independent, we can not apply the binomial probability distribution to find probability of x successes in n trials. In such cases we replace the binomial distribution by the hypergeometric probability distribution. Such a case occurs when a sample is drawn without replacement from a finite population.

Definition:

Let

total number of elements in the population

number of successes in the population

number of failures in the population

number of trials (sample size)

number of successes in n trials

number of failures in n trials.

The probability of x successes in n trials is given by

Example:

A company has 12 employees who hold managerial positions. Of them,

7 are females and 5 are males. The company is planning to send 3 of these 12 managers to a conference. If 3 managers are randomly selected out

of 12,

a) find the probability that all 3 of them are female

b) find the probability that at most 1 of them is a female.

Solution:

Let the selection of a female be called a success and the selection of a male be called a failure.

a) From the given information,

N =total number of managers in the population=12

S = number of successes (females) in the population=7

N - S =number of failures (males) in the population=5

n = number of selections (sample size) =3

x = number of successes (females) in three selections =3 number of failures (males) in three selections =0.

Using the hypergeometric formula, we find

Thus, the probability that all three of the selected managers are female

is 0.1591.

b) The probability that at most one of them is a female is given by the sum of the probabilities that either none or one of the selected managers is a female.

To find the probability that none of the selected managers is a female:

N =12

S =7

N - S =5

n =3

x =0

n - x =3

To find the probability that one of the selected managers is a female:

N =12

S =7

N - S =5

n =3

x =1

n - x =2

In the end,

.

 


Дата добавления: 2015-08-05; просмотров: 151 | Нарушение авторских прав


Читайте в этой же книге: Variables | Exercises | Expected value | Variance and standard deviation of discrete random variable | Mean and variance of linear function of a random variable | Exercises | Jointly distributed discrete random variable | Covariance | Exercises | The binomial distribution |
<== предыдущая страница | следующая страница ==>
Exercises| Exercises

mybiblioteka.su - 2015-2024 год. (0.006 сек.)