Hypergeometric Random Variables

1
Hypergeometric Random Variables
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Sampling without replacement

• When sampling with replacement, each trial
  remains independent. For example,
• If balls are replaced, P(red ball on 2nd draw)
  P(red ball on 2nd draw first ball was red).

• If balls not replaced, then given the first ball
  is red, there is less chance of a red ball on the
  2nd draw.

Though for a large population of balls,
the effect may be minimal.
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n trials, y red balls

• Suppose there are r red balls, and N r other
  balls.
• Consider Y, the number of red balls in n
  selections,where now the trials may be
  dependent.(for sampling without replacement,
  when sample size is significant relative to the
  population)
• The probability y of the n selected balls are red
  is

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Hypergeometric R. V.

• A random variable has a hypergeometric
  distribution with parameters N, n, and r if its
  probability function is given by

where 0 lt y lt min( n, r ).
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Hypergeometric mean, variance

• If Y is a hypergeometric random variable with
  parameter p the expected value and variance for Y
  are given by

( Proof not as easy as previous distributions
and is not given at this time. )
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Sounds like

• If we let p r/N and q 1- p (N - r)/N, then
  the hypergeometric measures

Look quite similar to the expressions for the
binomial distribution, E(Y) np and V(Y) npq.
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Rule of Thumb

• For cases when n / N lt 0.05, it may be
  reasonable to approximate the hypergeometric
  probabilities using a binomial distribution.
• Suppose each hour, 1000 bottles are filled by a
  machine and on average 10 are underfilled.
• Each hour 20 of the bottles are randomly
  selected. Find probability at least 3 of the 20
  are underfilled.
• Since 20/1000 0.02, perhaps we could use the
  binomial distribution to approximate the answer.

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Easy binomial probability?

• Let p 0.10, the success of underfilling
• P( at least 3 underfilled )
• 1 P( 0, 1, or 2 underfilled) 1 P(Y
  0) P(Y 1) P(Y 2)
• Approximately equal to 1
  binomialcdf(20, 0.10, 2) 0.32307how close is
  this to actual hypergeometric?

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A hypergeometric probability

• P( at least 3 underfilled )
• 1 P( 0, 1, or 2 underfilled) 1 P(Y
  0) P(Y 1) P(Y 2)

As compared to 0.32307 using a binomial approx.
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The Binomial Approximation
The hypergeometric distribution
and a very similar binomial distribution
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As population increases

• Let N get large as n and pr/N remain constant,
  and we would see that

Hypergeometric probabilities converge to the
binomial probabilities, as the events become
almost independent.
Proof ?