HYPERGEOMETRIC DISTRIBUTIONS
Some situations involve choosing items without replacement.
In these cases, combinations are used to calculate the probabilities.
The result is a hypergeometric distribution.
Example #1
Three marbles are drawn from a bag that
contains 5 blue and 4 yellow marbles.
Let the random variable, X, be the number of blue marbles drawn.
Complete the probability distribution for the experiment.
X
P(X)
0
1
2
3
CRITERIA of a HYPERGEOMETRIC DISTRIBUTION:
1.
2.
3.
Exactly two outcomes – success or failure.
Each trial is dependent (without replacement).
The probability changes with each trial.
The probability in a hypergeometric distribution:
P(x) =
 a  n  a
 

xr  x 
n
 
r 
where a is the number of successful outcomes
n is the total possible number of outcomes
r is the number of trials
The expectation (expected value) for a hypergeometric
distribution is
E(x) =
Example #2
ra
n
Returning to the blue marbles from example #1…
Determine the expected number of blue marbles drawn.
Example #3
A 12 member jury will be selected from a pool of
13 men and 10 women. Determine the
probability that a jury will have:
a)
5 women and 7 men;
b)
at least 2 women.
c)
Determine the expected number of men on the jury.