Chapter 12: Hypergeometric Distribution - School of Computing

Chapter 12:
Hypergeometric Distribution
CA266
School of Computing, DCU
Marija Bezbradica
mbezbradica@computing.dcu.ie
http://www.computing.dcu.ie/~mbezbradica/
(based on the book by Prof. Jane M. Horgan) 1

Example 1. Five cards are chosen from a well shuffled deck.
Lets denote:
X = number of diamonds selected
p – probability of selecting a diamond, i.e. the probability of “success”
n – number of selected cards
N – larger group, total size
p = 0.25, n = 5, N = 52
P(X=0) = P(0 diamond in the selected 5)
P(X=1) = P(1 diamond in the selected 5)
P(X=2) = P(2 diamonds in the selected 5)
Example 2. An audio amplifier contains six transistors. It has been ascertained that three of
the transistors are faulty but it is not known which three. Amy removes three transistors
at random, and inspects them.
X = number of defective transistors that Amy finds
p – probability of selecting a default, i.e. the probability of “success”
n – number of selected transistors
N – total size, total size
p = 0.05, n = 3, N = 6
P(X=0) = P(the sample will catch none of the defective transistors)
P(X=1) = P(the sample will catch one of the defective transistors)
P(X=2) = P(the sample will catch two of the defective transistors)
P(X=3) = P(the sample will catch three of the defective transistors)
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Example 3. A batch of 20 integrated circuit chips contains 20% defective chips. A sample
of 10 is drawn at random.
X = number of defective chips in the sample
p = 0.1, n = 10, N = 20
P(X=0) = P(0 defectives in the sample of size 10)
P(X=1) = P(1 defective in the sample of size 10)
P(X=2) = P(2 defectives in the sample of size 10)
Example 4. A batch of 100 printed circuit cards is populated with semiconductor chips.
20 of these are selected without replacement for function testing. If the original
batch contains 30 defective cards, how will these show up in the sample?
X = number of defective cards in the sample
p = 0.3, n = 20, N = 100
P(X=0) = P(0 defectives in the sample of size 20)
P(X=1) = P(1 defective in the sample of size 20)
P(X=2) = P(2 defectives in the sample of size 20)
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Hypergeometric Distribution
Recall:
• In Geometric – we investigated the number of trials necessary to arrive at the first
success
• In Binomial distribution – we investigated the number of successes in a fixed
number of n draws with replacement
In Hypergeometric distribution – probability of successes in n draws from a finite
population without replacement
Parameters are:
This gives:
n – number of chosen
N – total number
p – proportion of successes in the group to start with
M
Np
where M is number of successes in the group to start with
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Example 1.
p = 0.25, n = 5, N = 52
The deck can be visualised as:
Five cards are selected at random and without replacement from a deck.
P(X=0) = P(probability of getting 0 diamonds in the set)
P(X=1) = P(1 diamond in the selected 5) →
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Diamonds
P(X=2) = P(2 diamonds in the selected 5) →
similar for P(X=3), P(X=4), P(X=5) ...
39
Nondiamonds
5

Hypergeometric Distribution - PDF
Def
Conditions:
1. An experiment consists of n repeated trials
2. A finite population of size N consists of:
a) M elements called successes: M = Np
b) L elements called failures: L = N – M
X = number of successes
X is said to have a hypergeometric distribution
Np = M
Successes
N(1-p ) = N - M
Failures
A sample of n elements are selected at random without replacement
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Example 3. – cont.
What is probability that Amy finds three defective transistors when she selects three
of the six?
Solution 3:
Example 5.
Draw 6 cards from a deck without replacement. What is the probability of getting
two hearts?
Solution 5:
M = 13 number of hearts
L = 39 number of non-hearts
N = 52 total
Check in R:
> dhyper(2, 13, 39, 6)
[1] 0.3151299
> round(dhyper(2, 13, 39, 6), 5)
[1] 0.31513
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Calculating hypergeometric pdfs in R:
Use prefix “hyper” with “d” for pdf. Function is dhyper with arguments M and N - M.
Example 1: Five cards from a deck
x

Hypergeometric pdfs
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Example 6. Lotto a) with N = 36
The format of the game is such that you have to correctly chose n different numbers from N
Classic example of hypergeometric distribution:
n = 6
N = 36 (was in Ireland, now is 45, we will se why)
The population consists of:
for x = 0, 1, 2, 3, 4, 5, 6
Calculate in R:
x

Example 6. Lotto b) with N = 42
42 balls are numbered 1 - 42.
You select six numbers between 1 and 42. (The ones you write on your lotto card)
Number of possible ways to draw six numbers in the range [ 1, 42 ] =
What is the probability that they contain:
(i) match 6?
(ii) match 5?
(iii) match 4?
(iv) match 3?
Solution 6b:
Total, N = 42; Favourable, M = 6; Non-Favourable, L = 36
Sample size n = 6.
(iii) match 4:
x = 4
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Solution 6. Lotto – cont.
6 correct numbers, just one way of winning:
With 36 numbers
choose(36, 6)
Winning the Jackpot
[1] 1,947,792 # a chance of about 1 in 2 million!
With 39 numbers
choose(39, 6)
[1] 3,262,623 # a chance of less than 1 in 3 million!
With 42 numbers
choose(42, 6)
[1] 5,245,786 # less than 1 in 5 million chance!
The current situation of 45 numbers:
choose(45, 6)
[1] 8,145,060 # a chance of less than 1 in 8 million!
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Hypergeometric CDF
Cumulative Distribution Function (CDF) computes probabilities that the selected item will
contain up to and include x successes: F(x) = P(X ≤ x)
Example 1:
What is the probability of getting at most 2 diamonds in the 5 selected without
replacement from a well shuffled deck?
x = 2
n = 5
N = 52
M = 13
L = N – M = 39
Check in R:
> phyper(2, 13, 39, 5)
[1]0.9072329
Example 6:
What is the probability of getting something in lotto?
P(X ≥ 3) = 1 – P(X ≤ 2) = 1 – (P(match 0) + P(match 1) + P(match 2))
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Calculating hypergeometric cdfs in R:
par(mfrow = c(2,2))
x

Hypergeometric cdfs
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Binomial or Hypergeometric?
Boxes contain 20 items of which 10% are defective.
Find the probability that no more than 2 defectives will be obtained in a sample of size 10.
X = no of defectives
With Replacement Sampling
Without Replacement Sampling
P(X ≤ 2) = P(X = 0) + P(X = 1) + P(X = 2)
→ probabilities
are different for N = 20!
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Binomial or Hypergeometric? – cont.
Boxes contain 200 items of which 10% are defective.
Find the probability that no more than 2 defectives will be obtained in a sample of size 10.
X = no of defectives
With Replacement Sampling
Without Replacement Sampling
P(X ≤ 2) = P(X = 0) + P(X = 1) + P(X = 2)
→ near to probabilities
from binomial for N = 200!
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Binomial or Hypergeometric? – cont.
Boxes contain 2000 items of which 10% are defective.
Find the probability that no more than 2 defectives will be obtained in a sample of size 10.
X = no of defectives
With Replacement Sampling
Without Replacement Sampling
P(X ≤ 2) = P(X = 0) + P(X = 1) + P(X = 2)
→ very close to probabilities
from binomial for N = 2000!
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Binomial or Hypergeometric? – cont.
What is the probability of getting no more than 2 defectives in a random sample drawn
without replacement from a batch which has 10% defectives?
- we illustrated that when the batch size N is large the hypergeometric may be
approximated with binomial!
Batch size 20 200 2000 Binomial approx
P(X = 0) 0.2368 0.3398 0.3476 0.3487
P(X = 1) 0.5263 0.3974 0.3884 0.3874
P(X = 2) 0.2368 0.1975 0.1941 0.1937
P(X ≤ 2) 0.999 0.9347 0.9296 0.9298
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Hypergeometric and binomial pdfs with n = 10, p = 0.1
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Plotting hypergeometric and binomial
x

Hypergeometric vs. binomial pdfs with n = 10, p = 0.1 using matplot
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Matplot
Matplot - Plot the columns of one matrix against the
columns of another.
x

Let:
As N →∞, the hypergeometric distribution converges to the binomial!
Population Size = N
Proportion of successes = p
Number of successes in N = Np
Number of failures = N(1 − p)
X = number of successes in s sample of size n drawn without replacement
from N
Np = M
Successes
N(1-p ) = N - M
Failures
→ for N → ∞
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