Probability Explanation And Exercises Worksheet Page 26
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37Hypergeometric Distribution
by David M. Lane
Prerequisites
• Chapter 5: Binomial Distribution
• Chapter 5: Permutations and Combinations
The hypergeometric distribution is used to calculate probabilities when sampling
without replacement. For example, suppose you first randomly sample one card
from a deck of 52. Then, without putting the card back in the deck you sample a
second and then (again without replacing cards) a third. Given this sampling
procedure, what is the probability that exactly two of the sampled cards will be
aces (4 of the 52 cards in the deck are aces). You can calculate this probability
using the following formula based on the hypergeometric distribution:
C
C
k
x (N
k)
(n
x)
-
-
p
=
C
N
n
C
C
where
4
2 (52
4)
(3
2)
-
-
p
=
C
52
3
k is the number of “successes” in the
population
x is the number of “successes” in the sample
4!
48!
N is the size of the population
2!2!
47!1!
p
0.013
=
=
n is the number sampled
52!
p is the probability of obtaining exactly x
49!3!
successes
(n) (k)
C
is the number of combinations of k things
k
x
mean
=
taken x at a time
N
(n) (k) (N
k) (N
n)
-
-
In this example, k = 4 because there are four aces in the deck, x = 2 because the
sd
=
N (N
1)
2
-
problem asks about the probability of getting two aces, N = 52 because there are 52
cards in a deck, and n = 3 because 3 cards were sampled. Therefore,
210
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