Transcript Tree diagram and formula example

The Hypergeometric Probability Distribution



The hypergeometric distribution is closely related to
the binomial distribution.
With the hypergeometric distribution, the trials are
not independent, and the probability of success
changes from trial to trial.
You can use either the hypergeometric formula or the
tree diagram method (results should be the same).
Slide 1
The Hypergeometric Probability Distribution

Hypergeometric Probability Function
 r  N  r 
 

x  n  x 

f ( x) 
N
 
n
for 0 < x < r
where:
f(x) = probability of x successes in n trials
n = number of trials
N = number of elements in the population
r = number of elements in the population
labeled success
Slide 2
Example: Neveready

The Hypergeometric Probability Distribution
Bob Neveready has removed two dead batteries
from a flashlight and inadvertently mingled them
with the two good batteries he intended as
replacements. The four batteries look identical.
Bob now randomly selects two of the four
batteries. What is the probability he selects the two
good batteries?
Slide 3
Example: Neveready

The Hypergeometric Probability Distribution
 r  N  r   2  2   2!  2! 
 
      
x  n  x   2  0   2!0! 0!2! 1

f ( x) 


  .167
6
N
 4
 4! 
 
 


n
 2
 2!2!
where:
x = 2 = number of good batteries selected
n = 2 = number of batteries selected
N = 4 = number of batteries in total
r = 2 = number of good batteries in total
Slide 4
Bob Neveready’s dilemma
Let g = a good battery
Let b = a bad battery
P(g)=1/3
2/4*1/3=1/6
g
P(g)=2/4=1/2
g
2/4*2/3=1/3
b
b
2/4*2/3=1/3
g
P(b)=2/4=1/2
2/4*1/3=1/6
b
First selection
Second selection
The probability of selecting 1 good battery
out of 4 in the first selection is 2/4 or 1/2. (2
good batteries available out of 4 total). The
probability of selecting the 1 remaining good
battery out of the remaining 3 is 1/3. The
probability of selecting first a good battery,
then second a good battery is: (1/2)(1/3) =
1/6 (using the multiplication rule).
P(gg)=1/6
P(bb)=1/6
P(bg)=1/3
P(gb)=1/3
P of at least one bad is based on the
addition rule so:
P(bb)+P(bg)+P(gb)=1/6+1/3+1/3=5/6 the
complement of this is after two trials no bad
batteries are selected which would be 1-5/6
or 1/6. There are many ways to compute.
You could also use the addition rule which
would simply be P(gg…no bad)=1/6
All possibilities must equal to
one and this is a good way of
checking your numbers.
Slide 5