The hypergeometric probability distribution

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,

.

 

Exercises

1. Let N=14, S=6, and n=5. Using the hypergeometric probability distribution formula, find

a) ;            b) ;             c)

 

2. Let N=16, S=10, and n=5. Using the hypergeometric probability distribution formula, find

a) ;             b) ;            c)

 

3. There are 25 goods, 5 of which are defective. We randomly select 4 goods. What is the probability that three of those four goods are nondefective and one is defective?

4. A committee of two members is to be formed from the list of 8 candidates. Of the 8 candidates, 5 are management and 3 are economics department students. Find the probability that

       a) both candidates are managers

       b) neither of the candidates are managers

       c) at most one of the candidates is a manager.

5. A company buys keyboards from another company. The keyboards are received in shipments of 100 boxes, each box containing 20 keyboards. The quality control department first randomly selects one box from each shipment, and then randomly selects five keyboards from that box. The shipment is accepted if not more than one of the five keyboards is defective. The quality control inspector selects a box from a recently received shipment of keyboards. Unknown to the inspector, this box contains six defective keyboards.

a) What is the probability that this shipment will be accepted?

b) What is the probability that this shipment will not be accepted?

                                   

 

Answers

1. a) 0.0599; b) 0.0030; c) 0.2378; 2. a) 0.0577; b) 0.0014; c) 0.0357;

3. 0.4506; 4. a) 0.3571; b) 0.1071; c) 0.6429; 5. a) 0.5165; b) 0.4835.         

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