Question

A trial has just resulted in a hung jury because eight members of the jury were in favour of a guilty verdict and the other four were for acquittal. If the jurors leave the jury room in random order and each of the first four leaving the room is accosted by a reporter in quest of an interview, what is the\({\rm{pmf}}\)of\({\rm{X = }}\)the number of jurors favouring acquittal among those interviewed? How many of those favouring acquittal do you expect to be interviewed?

Short answer

The \({\rm{pmf}}\) of \({\rm{X = }}\)the number of jurors favouring acquittal among those interviewed is \({\rm{h(x;4,4,12)}}\). The number of jurors those favouring acquittal are expected to be interviewed is \({\rm{E(X) = 1}}{\rm{.33}}\).

Step-by-step solution

Concept Introduction

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.

Hypergeometric Distribution

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

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

Then it has probability mass function –

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

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

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

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 –

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

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 \({\rm{n,M,N}}\).

Number of jurors favouring and interviewed

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

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

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

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

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

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

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