Welcome to a lesson
on the hypergeometric distribution.
There are five characteristics
of a hypergeometric experiment.
Number one, you take samples from two groups.
Number two, you are concerned with a group
of interest called the first group.
Number three, you sample without placement
from the combined groups for example,
you want to choose a softball team
from a combined group of 11 men and 13 women.
The team consists of 10 players.
Number four, each pick is not independent
since sampling is without replacement.
In the softball example, the probability of picking
a woman first is 13/24 because there are 13 women
out of a total of 24 possible players.
The probability of picking a man second is 11/23
if a woman is picked first.
Notice how the denominator drops from 24 to 23
because there is one less person to choose from.
And the numerator is 11 because there are 11 men.
The probability would be 10/23 if a man was picked first,
again, the denominator is 23 because one person
has already been picked and now the numerator
is 10 because the first person was a man this time
and therefore there are only 10 men to pick from.
The probability of the second pick depends
on what happened in the first pick.
And then finally number five,
you are not dealing with Bernoulli trials
or binomial trials.
The outcomes of a hypergeometric experiment fit
a hypergeometric probability distribution.
The random variable X is equal to the number of items
from the group of interests.
And we use the notation shown here
to represent a hypergeometric distribution.
We read this as X is a random variable
with a hyper geometric distribution.
The parameters are r, b and n where r equals the size
of the group of interest or the first group,
b is the size of the second group
and n is the size of the chosen sample.
Remember a random variable is capital X
and takes on values of small x.
The connection between values and events,
capital X is equal to small x
is the event that capital X takes
on the value small x and the probability X is equal to k
is equal to the product of r choose k
and B choose n minus k divided by r plus b choose n
and the mean is equal to the product
of r and n divided by the sum of r and b.
Let's take a look at an example.
The school site committee is chosen randomly
from six men and five women.
If the committee consists of four members chosen randomly,
what is the probability that two of them are men?
How many men do you expect to be on the committee?
To determine the probability that two of them are men,
we want to find the probability X is equal two
and to determine how many men we expect
to be on the committee, we need to define the mean.
So to set this up, the random variable X
is equal to the number of men on the committee of four.
The men are the group of interest or the first group
which is why r is equal to six.
The random variable X takes on the values
of zero one, two, three and four.
We stop at four because the committee size is four.
And then because there are five women B is equal to five
because we are choosing a committee of four
and is equal to four.
Now we have only information we need
to find the probability X equals two.
Notice because X equals two k is equal to two.
In the numerator we have r choose k,
which is six choose two, six choose two
is a number of ways we can choose two men from six men
times we choose and minus k will b is equal to five
the number of women and then n minus k
is equal to four minus two which is two.
If I choose two gives us number of ways
we can choose two women from five women.
In the denominator we have r plus b choose n
where r plus b is six plus five,
the number of men plus number of women
which gives us 11 and n is four
which gives us 11 choose four.
11 choose four is a number of ways
we can choose committees of four from 11 people.
And now to evaluate this on the TI-84.
we need the numerator and parentheses
so we enter open parenthesis
and then for six choose two enter six,
math right arrow to probability,
select option three for combination,
enter two right arrow times for five choose two
enter five math right arrow to probability,
select option three for combination,
enter two right arrow close parenthesis
for the numerator and then divided by 11 choose four,
enter 11 math probability number three
and then four and enter.
Let's first get the exact value
by pressing math enter, enter
which gives us 5/11 or to four decimal places
approximately 0.4545.
And now we need to find the mean
to find how many men we expect to have
on the committee of four in general.
The mean is equal to we have n times r
which is four times six divided by r plus b
which is six plus five.
Simplifying we have 24/11 converting to a decimal,
we have approximately 2.1818.
Let's take a look at a second example.
A bag of marbles contains 13 red marbles
and eight blue marbles,
notice how there are a total of 21 marbles.
If five marbles are drawn from the bag,
what is the probability that three of them are red?
How many red marbles would you expect in general
when drawing five marbles.
The random variable X is equal to the number
of red marbles of the five drawn.
The red marbles are the group of interest
or the first group which is why r equal to 13,
the random variable X takes on the values
of zero, one, two, three, four, five
because we are choosing five marbles.
B is a number of blue marbles which is eight
and n is the group size which is five.
To find that probability that three of the marbles are red,
we need to find the probability X is equal to three
because there are a total of 13 plus eight or 21 marbles
and we're selecting a group of five,
the denominator is 21 choose five and the numerator
we have r choose k times b choose n minus k.
R choose k is a number of ways
we can select three red marbles from 13 red marbles
which is 13 choose three times
if we're choosing three red marbles,
the remaining two marbles must be blue
and therefore the second combination is eight choose two.
The number of ways
we can choose two blue marbles from eight.
And now let's get back to the calculator
and evaluate this probability.
The exact probability is 1,144 over 2,907
that's an approximation if we're decimal places
is approximately 0.3935.
To find how many bed marbles we expect in general
when drawing five marbles under these conditions,
we defined the mean, the mean is equal to n times r
which is five times 13 divided by r plus b
which is 13 plus eight which gives us 65 divided by 21,
65 divided by 21 to four decimal places
is approximately 3.0952.
Before we go, I also want to show how
I can use an online tool
to determine hypergeometric distribution probabilities.
So following the link here shown on the screen,
if you want to verify this probability,
we enter the population size which is 13 plus eight or 21.
The number of successes in the population
would be the number of red marbles which is 13.
The sample size is five because we're drawing five marbles,
the number of successes in the sample
would be the number of red marbles we want which is three.
If we press Calculate, notice how it does verify
our probability here at the top, approximately 0.3935.
Also notice how this gives other possibilities
for X less than three, less than or equal to three,
X greater than three as well as X greater
than or equal to three.
I hope you found this helpful.