Hypergeometric Distribution

• There are two types of sampling methods from a finite population. If the population is infinite, two methods do not make any difference.
• Binomial distribution: the sampling with replacement ($p$ is constant)
• Hypergeometric distribution: the sampling without replacement ($p$ is not constant)
• Hypergeometric experiment:
  1. A random sample of size $n$ is selected without replacement from $N$ items.
  2. $k$ of the $N$ items may be classified as successes and $N-k$ as failures.
• Hypergeometric random variable: the number $X$ of successes of a hypergeometric experiment.

• The probability distribution of the hypergeometric variable $X$, the number of successes in a random sample of size $n$ selected from $N$ items of which $k$ are labeled success and $N-k$ labeled failure; $h(x;N,n,k)$.
  $$h(x;N,n,k)=$$
   
   
  
  $$\frac{ \left( \begin{array}{c} k\\ x\\ \end{array}\righ... ...) }{ \left( \begin{array}{c} N\\ n\\ \end{array}\right) }$$
  • the number of ways of selecting $x$ successes
    $$\left( \begin{array}{c} k\\ x\\ \end{array}\right)$$

  • the number of ways of selecting $n-x$ failures
    $$\left( \begin{array}{c} N-k\\ n-x\\ \end{array}\right)$$

  • the total number of samples of size $n$ chosen from $N$ items
    $$\left( \begin{array}{c} N\\ n\\ \end{array}\right)$$

• Example 5.12: Lots of 40 components each are called unacceptable if they contain as many as 3 defective or more.
• The procedure for sampling the lot is to select 5 components at random and to reject the lot if a defective is found.
• What is the probability that exactly 1 defective is found in the sample if there are 3 defectives in the entire lot?
• Solution: Using hypergeometric distribution with $n = 5, N = 40, k = 3$ and $x = 1$;
  $$h(x;N,n,k)=$$
  $$h(1;40,5,3)=$$
  $$\frac{ \left( \begin{array}{c} 3\\ 1\\ \end{array}\righ... ...ft( \begin{array}{c} 40\\ 5\\ \end{array}\right) }=0.3011$$

• So this plan is likely not desirable since it detects a bad lot (3 defectives) only about 30% of the time.

• Theorem 5.3:
  
• Example 5.14:Find the mean and variance of the random variable of Example 5.12 ( $n = 5, N = 40, ~and~ k = 3$) and then use Chebyshev's theorem to interpret the interval $\mu\pm 2\sigma$
• Solution:
  $$\mu=\frac{5*3}{40} and \sigma^2=\frac{40-5}{40-1}*5*\frac{3}{40}*\left( 1-\frac{3}{40} \right) \Rightarrow \sigma=0.558$$
  $$\mu\pm2\sigma=0.3775\pm2*0.558$$
  it has a probability of at least 3/ 4 of falling between -0.741 and 1.491.
• That is, at least three fourths of the time, the 5 components include less than 2 defectives.

• Relationship to the Binomial Distribution. If $n$ is small compared to $N$, the nature of the $N$ items changes very little in each draw. (when $\frac{n}{N}\leq 0.05$ )
• $\mu=np=\frac{nk}{N} and \sigma^2=npq=n*\frac{k}{N}*\left( 1-\frac{k}{N} \right)$, where $\frac{N-n}{n-1}$ is negligible when $n$ is smallrelative to $N$.
• The binomial distribution may be viewed as a large population edition of the hypergeometric distributions.
• Example 5.15: A manufacture of automobile tires reports that among a shipment of 5000 sent to a local distributor, 1000 are slightly faulty.
• If one purchases 10 of these tires at random from the distributor, what is the probability that exactly 3 are faulty?
• Solution: $h(x;N,n,k)=?$
  $$h(3;5000,10,1000)\approx$$
  $$\sum_{x=0}^3 b(x;10,0.2)-\sum_{x=0}^2 b(x;10,0.2)$$
  $$=0.8791-0.6778=0.2013$$

• Multivariate HypergeometricDistribution: If $N$ items can be partitioned into the $k$ cells $A_1,A_2,\ldots,A_k$ with $a_1,a_2,\ldots,a_k$ elements, respectively, then the probability distribution of the random variable $X_1,X_2, \ldots, X_k$, representing the number of elements selected from
  $A_1,A_2,\ldots,A_k$ in a random sample of size $n$, is
  $$f(x_1,x_2, \ldots, x_k; a_1,a_2, \ldots, a_k,N,n)= \frac{ \l... ...) }{ \left( \begin{array}{c} N\\ n\\ \end{array}\right) }$$

• with $\sum_{i=1}^kx_i=n$   and   $\sum_{i=1}^k a_i=N$

• Example 5.16: A group of 10 individuals are used for a biological case study.
• The group contains 3 people with blood type O, 4 with blood type A, and 3 with blood type B.
• What is the probability that a random sample of 5 will contain 1 person with blood type O, 2 with blood type A, and 2 with blood type B?
• Solution:
  $$f(1,2,2;3,4,3,10,5)= \frac{ \left( \begin{array}{c} 3\\ ... ...begin{array}{c} 10\\ 5\\ \end{array}\right) }=\frac{3}{14}$$