91Ó°ÊÓ

A discrete random variable is one that can only take on a finite number of different values

X=number of successes in an n-person random sample,

It has a probability mass function after that.

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

For all the integers of the value x we have:

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

The probability distribution is known as the hypergeometric distribution.

There are total of

\({\rm{N = 20}}\)

pairs. The number of successes then would be

\({\rm{M = 10}}\)

pairs, and then the sample size would have

\({\rm{n = 10}}\)

pairs.

As a result, the random variable will have a hypergeometric distribution, and the probability we must calculate will be:

\(\begin{array}{c}{\rm{h(x;10,10,20) = P(X = x)}}\\{\rm{ = }}\frac{{\left( {\begin{array}{*{20}{c}}{{\rm{10}}}\\{\rm{x}}\end{array}} \right)\left( {\begin{array}{*{20}{c}}{{\rm{20 - 10}}}\\{{\rm{10 - x}}}\end{array}} \right)}}{{\left( {\begin{array}{*{20}{c}}{{\rm{20}}}\\{{\rm{10}}}\end{array}} \right)}}\end{array}\)

For all other values of x, use \({\rm{x}} \in {\rm{\{ 0,1,}}.....{\rm{,10\} }}\), and zero.

The probability we required were as follows:

Therefore, the probability is:

\({\rm{M = 5}}\)

Instead than \({\rm{M = 10}}\). As a result, there's a good chance that all five of the top pairs will wind up playing in the same direction (denote the event as A)

\(\begin{array}{c}{\rm{P(A)}}\mathop {\rm{ = }}\limits^{{\rm{(1)}}} {\rm{P(X = 5) + P(X = 0)}}\\\mathop {\rm{ = }}\limits^{{\rm{(2)}}} {\rm{h(5;10,5,20) + h(0;10,5,20)}}\\{\rm{ = }}\frac{{\left( {\begin{array}{*{20}{l}}{\rm{5}}\\{\rm{5}}\end{array}} \right)\left( {\begin{array}{*{20}{l}}{{\rm{20 - 5}}}\\{{\rm{10 - 5}}}\end{array}} \right)}}{{\left( {\begin{array}{*{20}{l}}{{\rm{20}}}\\{{\rm{10}}}\end{array}} \right)}}{\rm{ + }}\frac{{\left( {\begin{array}{*{20}{l}}{\rm{5}}\\{\rm{0}}\end{array}} \right)\left( {\begin{array}{*{20}{l}}{{\rm{20 - 5}}}\\{{\rm{10 - 5}}}\end{array}} \right)}}{{\left( {\begin{array}{*{20}{l}}{{\rm{20}}}\\{{\rm{10}}}\end{array}} \right)}}\\{\rm{ = 0}}{\rm{.01625 + 0}}{\rm{.01625}}\\{\rm{ = 0}}{\rm{.0325}}\end{array}\)

(1):either all of the best players are playing east-west, or none of them are (resulting in five top players playing north-south); disjointed tournaments;

(2): hypergeometric distribution pmf.

Therefore, the probability is: \({\rm{P(A) = 0}}{\rm{.0325}}\).

\({\rm{M = n}}\)

Total of successes (top n pairings that end up playing east-west)

\({\rm{N = 2n}}\)

The sample size is n, and the number of pairings is n.

As a result, the random variable X has a hypergeometric distribution with the parameters stated previously. As stated in the proposition, the probability density function (pmf) is:

\(\begin{aligned}h(x;n,n,2n) &= P(X = x) \\&= \frac{{\left( {\begin{array}{*{20}{c}}{\rm{n}}\\{\rm{x}}\end{array}} \right)\left( {\begin{array}{*{20}{c}}{{\rm{2n - n}}}\\{{\rm{n - x}}}\end{array}} \right)}}{{\left( {\begin{array}{*{20}{c}}{{\rm{2n}}}\\{\rm{n}}\end{array}} \right)}}\\ &= \frac{{\left( {\begin{array}{*{20}{c}}{\rm{n}}\\{\rm{x}}\end{array}} \right)\left( {\begin{array}{*{20}{c}}{\rm{n}}\\{{\rm{n - x}}}\end{array}} \right)}}{{\left( {\begin{array}{*{20}{c}}{{\rm{2n}}}\\{\rm{n}}\end{array}} \right)}}\end{aligned}\)

It is then for\({\rm{x }} \in {\rm{ \{ 0,1,}}....{\rm{,n\} }}\).

The predicted values and variance must be determined in the second portion. Only n should determine both values.

For random variable X with hypergeometric distribution and pmf h(x; n, M, N), the following is true.

\(\begin{aligned} E(X) &= n \times \frac{{\rm{M}}}{{\rm{N}}}{\rm{;}}\\V(X) &= \left( {\frac{{{\rm{N - n}}}}{{{\rm{N - 1}}}}} \right){\rm{ \times n \times }}\frac{{\rm{M}}}{{\rm{N}}}{\rm{ \times }}\left( {{\rm{1 - }}\frac{{\rm{M}}}{{\rm{N}}}} \right)\end{aligned}\)

Then, the expected value is:

\(\begin{aligned} E(X) &= n \times \frac{{\rm{M}}}{{\rm{N}}}\\ &= n \times \frac{{\rm{n}}}{{{\rm{2n}}}}\\& = \frac{{\rm{n}}}{{\rm{2}}}\end{aligned}\)

The variance is evaluated as:

\(\begin{aligned}V(X) &= \left( {\frac{{{\rm{N - n}}}}{{{\rm{N - 1}}}}} \right){\rm{ \times n \times }}\frac{{\rm{M}}}{{\rm{N}}}{\rm{ \times }}\left( {{\rm{1 - }}\frac{{\rm{M}}}{{\rm{N}}}} \right)\\ &= \left( {\frac{{{\rm{2n - n}}}}{{{\rm{2n - 1}}}}} \right){\rm{ \times n \times }}\frac{{\rm{n}}}{{{\rm{2n}}}}{\rm{ \times }}\left( {{\rm{1 - }}\frac{{\rm{n}}}{{{\rm{2n}}}}} \right)\\ &= \frac{{\rm{n}}}{{{\rm{2n - 1}}}}{\rm{ \times }}\frac{{\rm{n}}}{{\rm{2}}}{\rm{ \times }}\frac{{\rm{1}}}{{\rm{2}}}\\ &= \frac{{{{\rm{n}}^{\rm{2}}}}}{{{\rm{4(2n - 1)}}}}\end{aligned}\)

Therefore, the value is: \(\begin{array}{c}{\rm{h(x;n,n,2n\} = }}\frac{{\left( {\begin{array}{*{20}{c}}{\rm{n}}\\{\rm{x}}\end{array}} \right)\left( {\begin{array}{*{20}{c}}{\rm{n}}\\{{\rm{n - x}}}\end{array}} \right)}}{{\left( {\begin{array}{*{20}{c}}{{\rm{2n}}}\\{\rm{n}}\end{array}} \right)}}\\{\rm{E(X) = }}\frac{{\rm{n}}}{{\rm{2}}}\\{\rm{V(X) = }}\frac{{{{\rm{n}}^{\rm{2}}}}}{{{\rm{4(2n - 1)}}}}\end{array}\).