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The hypergeometric distribution is a discrete probability distribution in probability theory and statistics that describes the probability of\({\rm{k}}\)successes (random draws for which the object drawn has a specified feature) in\({\rm{n}}\)draws, without replacement, from a finite population of size\({\rm{N}}\)that contains exactly\({\rm{K}}\)objects with that feature, where each draw is either successful or unsuccessful.

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

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

Then it has probability mass function 鈥�

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

For all integers \({\rm{x}}\) for which 鈥�

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

The probability distribution is called hypergeometric distribution.

There are total of \({\rm{N = 12}}\) jurors that can be interviewed. A success would represent a jury in favour of acquittal, therefore 鈥�

\({\rm{M = 4,}}\)

And since the first four are going to be interviewed, the sample size is \({\rm{n = 4}}\).

So, the given random variable \({\rm{X}}\) follows hypergeometric distribution with given parameters \({\rm{n,M,N}}\).

The \({\rm{pmf}}\) of the given random variable is 鈥�

\({\rm{h(x;4,4,12) = P(X = x) = }}\frac{{\left( {\begin{array}{*{20}{c}}{\rm{4}}\\{\rm{x}}\end{array}} \right)\left( {\begin{array}{*{20}{c}}{\rm{8}}\\{{\rm{4 - x}}}\end{array}} \right)}}{{\left( {\begin{array}{*{20}{c}}{{\rm{12}}}\\{\rm{4}}\end{array}} \right)}}{\rm{,}}\;\;\;{\rm{0}} \le {\rm{x}} \le {\rm{4}}\)

Proposition: The following is true for random variable \({\rm{X}}\) with hypergeometric distribution and \({\rm{pmf h(x;n,M,N)}}\) 鈥�

\(\begin{array}{c}{\rm{E(X) = n}} \cdot \frac{{\rm{M}}}{{\rm{N}}}\\{\rm{V(X) = }}\left( {\frac{{{\rm{N - n}}}}{{{\rm{N - 1}}}}} \right) \cdot {\rm{n}} \cdot \frac{{\rm{M}}}{{\rm{N}}} \cdot \left( {{\rm{1 - }}\frac{{\rm{M}}}{{\rm{N}}}} \right)\end{array}\)

Thus, the expected number of those favouring acquittal to be interviewed is 鈥�

\({\rm{E(X) = n}} \cdot \frac{{\rm{M}}}{{\rm{N}}}{\rm{ = 4}} \cdot \frac{{\rm{4}}}{{{\rm{12}}}}{\rm{ = 1}}{\rm{.33}}\)

Therefore, the value is obtained as \({\rm{1}}{\rm{.33}}\).