91影视

Proposition: Assume that population has\(M\)successes (S),\(N - M\)failures (F). If random variable\({\rm{X}}\)is

\(X = {\rm{ number of succeses in a random sample size n}},\)

then it has probability mass function

\(h(x;n,M,N) = P(X = x) = \frac{{\left( {\begin{aligned}{*{20}{c}}M\\x\end{aligned}} \right)\left( {\begin{aligned}{*{20}{c}}{N - M}\\{n - x}\end{aligned}} \right)}}{{\left( {\begin{aligned}{*{20}{c}}N\\n\end{aligned}} \right)}}\)

for all integers\(x\)for which

\(\max \{ 0,n - N + M\} \le x \le \min \{ n,M\} \)

The probability distribution is called hyper geometric distribution.

Theorem:

\(b(x;n,p) = \left\{ {\begin{aligned}{*{20}{l}}{\left( {\begin{aligned}{*{20}{l}}n\\x\end{aligned}} \right){p^x}{{(1 - p)}^{n - x}}}&{,x = 0,1,2, \ldots ,n}\\0&{,{\rm{ otherwise }}}\end{aligned}} \right.\)

Cumulative Density Function cdf of binomial random variable\(X\)with parameters\(n\)and\(p\)is

\(B(x;n,p) = P(X \le x) = \mathop \sum \limits^x b(y;n,p),\;\;\;x = 0,1, \ldots ,n.\)

For the binomial distribution, when\(n = 50\), the probabilities when\(c = 2\)can be computed using

\(\begin{aligned}{l}B(2;50,p) = P(X \le 2) = \sum\limits_{y = 0}^2 b (y;50,p)\\ = \left( {\begin{aligned}{*{20}{c}}{50}\\0\end{aligned}} \right){p^0}{(1 - p)^{50}} + \left( {\begin{aligned}{*{20}{c}}{50}\\1\end{aligned}} \right){p^1}{(1 - p)^{49}} + \left( {\begin{aligned}{*{20}{c}}{50}\\2\end{aligned}} \right){p^2}{(1 - p)^{48}}\end{aligned}\)

for\(p = 0.01,0.02, \ldots ,0.1\)

For the hyper geometric distribution, when\(n = 50\), the probabilities when\(c = 2\)can be computed using

\(\begin{aligned}{l}P(X \le 2) = \sum\limits_{y = 0}^2 h (y;50,M,500)\\ = \frac{{\left( {\begin{aligned}{*{20}{c}}M\\0\end{aligned}} \right)\left( {\begin{aligned}{*{20}{c}}{500 - M}\\{50}\end{aligned}} \right)}}{{\left( {\begin{aligned}{*{20}{c}}{500}\\{50}\end{aligned}} \right)}} + \frac{{\left( {\begin{aligned}{*{20}{c}}M\\1\end{aligned}} \right)\left( {\begin{aligned}{*{20}{c}}{500 - M}\\{49}\end{aligned}} \right)}}{{\left( {\begin{aligned}{*{20}{c}}{500}\\{50}\end{aligned}} \right)}} + \frac{{\left( {\begin{aligned}{*{20}{c}}M\\2\end{aligned}} \right)\left( {\begin{aligned}{*{20}{c}}{500 - M}\\{48}\end{aligned}} \right)}}{{\left( {\begin{aligned}{*{20}{c}}{500}\\{50}\end{aligned}} \right)}}\end{aligned}\)

for\(M = 5,10,15, \ldots ,50\).

\(\begin{aligned}{*{20}{c}}M&p&{{\rm{ Hypergeometric }}}&{{\rm{ Binomial }}}\\5&{0.01}&{0.9919}&{0.9862}\\{10}&{0.02}&{0.9317}&{0.9216}\\{15}&{0.03}&{0.8182}&{0.8108}\\{20}&{0.04}&{0.6775}&{0.6767}\\{25}&{0.05}&{0.5343}&{0.5405}\\{30}&{0.06}&{0.4041}&{0.4162}\\{35}&{0.07}&{0.2964}&{0.3108}\\{40}&{0.08}&{0.211}&{0.226}\\{45}&{0.09}&{0.1464}&{0.1605}\\{50}&{0.1}&{0.0994}&{0.1111}\\{}&{}&{}&{}\end{aligned}\)

Now, using corresponding\(p\)values and\(M\)values (depending on distribution), the following table represents the probabilities. One can notice that the binomial distribution gives

Satisfactory.