91Ó°ÊÓ

the proportion of the total number of conceivable outcomes to the number of options in an exhaustive collection of equally likely outcomes that cause a given occurrence.

We have the following information:

\({\rm{10}}\)basaltic rock specimens;

\({\rm{10}}\)granite specimens

That is to say, there are

\({\rm{10 + 10 = 20}}\)

specimens in total \({\rm{15}}\)of the \({\rm{20}}\) specimens are chosen at random by the helper.

The probability mass function is then applied.

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

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

for every integer for where \({\rm{x}}\)

max{ 0,n - N + M} £x£ { n,M }.

Hypergeometric distribution is the name given to the probability distribution.

A success in our situation is granite, while a failure is basaltic rock. There are ten granite specimens, indicating that

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

\({\rm{N = 20}}\)specimens total, and

\({\rm{N - M = 20 - 10 = 10}}\)

basaltic rock specimens.

A random variable can take any number of values.

\({\rm{X = 5,6,7,8,9,10}}\)

Because if there are four granite specimens (any number less than five), there must be more than ten basaltic rock specimens, which is impossible because there are only ten overall. This is in accordance with the preceding proposition.

\(\begin{array}{*{20}{c}}{{\rm{max\{ 0,n - N + M\} }}}&{{\rm{£x£{ n,M\} }}}\\{{\rm{max\{ 0,15 - 20 + 10\} }}}&{{\rm{£x£\{ 15,10\} }}}\\{}&{{\rm{5 < r < 1\c{C}}}}\end{array}\)

So, the random variable \({\rm{X}}\) has a hypergeometric distribution with parameters \({\rm{n = 15}}\) (number of elements in the sample), \({\rm{N = 20}}\)and \({\rm{M = 10}}\), and \({\rm{X}}\)\({\rm{pmf}}\)is

\({\rm{h(x;15,10,20) = P(X = x) = }}\frac{{\left( {\begin{array}{*{20}{c}}{{\rm{10}}}\\{\rm{x}}\end{array}} \right)\left( {\begin{array}{*{20}{c}}{{\rm{20 - 10}}}\\{{\rm{15 - x}}}\end{array}} \right)}}{{\left( {\begin{array}{*{20}{c}}{{\rm{20}}}\\{{\rm{15}}}\end{array}} \right)}}\)

\({\rm{x^I \{ 5,6, \ldots ,10\} }}\)is zero if \({\rm{x^I \{ 5,6, \ldots ,10\} }}\)is zero. We get the following \({\rm{pmf}}\)by applying the algorithm above to all \({\rm{x^I \{ 5,6, \ldots ,10\} }}\)

To discover the probability we're looking for, we must first recognize that the only way for all 10 of one sort to be in a sample is if there are five of the other kind, so there are only two possibilities.

\(\begin{array}{l}{\rm{5 + 10 }}\\{\rm{and}}\\{\rm{10 + 5}}\end{array}\)

Because we calculated the \({\rm{pmf}}\)in (a), the probability that all specimens of one type are included in the sample is A.

\({\rm{P(A)}}\mathop {\rm{ = }}\limits^{{\rm{(1)}}} {\rm{P(X = 5) + P(X = 10) = 0}}{\rm{.0163 + 0}}{\rm{.0163 = 0}}{\rm{.0326}}\)

(1): If there are \({\rm{5}}\)of granite, there are \({\rm{10}}\)of basaltic rock (all), and if there are \({\rm{10}}\)of granite (all), there are \({\rm{5}}\)of basaltic rock on the left.

We must first calculate the mean and standard deviation of random \({\rm{X}}\).

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

\(\begin{array}{*{20}{c}}{{\rm{E(X) = n \times }}\frac{{\rm{M}}}{{\rm{N}}}}\\{{\rm{V(X) = }}\left( {\frac{{{\rm{N - n}}}}{{{\rm{N - 1}}}}} \right){\rm{ \times n \times }}\frac{{\rm{M}}}{{\rm{N}}}{\rm{ \times }}\left( {{\rm{1 - }}\frac{{\rm{M}}}{{\rm{N}}}} \right)}\end{array}\)

The following statement is correct:

\({\rm{E(X) = n \times }}\frac{{\rm{M}}}{{\rm{N}}}{\rm{ = 15 \times }}\frac{{{\rm{10}}}}{{{\rm{20}}}}{\rm{ = 7}}{\rm{.5}}\)

Furthermore, the variation is

\({\rm{V(X) = }}\left( {\frac{{{\rm{20 - 15}}}}{{{\rm{20 - 1}}}}} \right){\rm{ \times 15 \times }}\frac{{{\rm{10}}}}{{{\rm{20}}}}{\rm{ \times }}\left( {{\rm{1 - }}\frac{{{\rm{10}}}}{{{\rm{20}}}}} \right){\rm{ = 0}}{\rm{.9868}}\)

The standard deviation is finally

\({{\rm{\sigma }}_{\rm{X}}}{\rm{ = }}\sqrt {{\rm{V(X)}}} {\rm{ = }}\sqrt {{\rm{0}}{\rm{.9868}}} {\rm{ = 0}}{\rm{.9934}}\)

It is between the following two values if it is within one standard deviation of its mean values.

\(\begin{array}{*{20}{c}}{{\rm{E(X) - }}{{\rm{\sigma }}_{\rm{X}}}}&{{\rm{ < X < E(X) + }}{{\rm{\sigma }}_{\rm{X}}}}\\{{\rm{7}}{\rm{.5 - 0}}{\rm{.9934}}}&{{\rm{ < X < 7}}{\rm{.5 + 0}}{\rm{.9934}}}\\{{\rm{6}}{\rm{.5066}}}&{{\rm{ < X < 8}}{\rm{.4934}}}\end{array}\)

The following statement is correct:

\(\begin{array}{*{20}{c}}{{\rm{P(6}}{\rm{.5066 < X < 8}}{\rm{.4934)}}}&{\mathop {\rm{ = }}\limits^{{\rm{(1)}}} {\rm{P(7£ X£ 8)}}\mathop {\rm{ = }}\limits^{{\rm{(2)}}} {\rm{P(X = 7) + P(X = 8)}}}\\{}&{\mathop {\rm{ = }}\limits^{{\rm{(3)}}} {\rm{0}}{\rm{.3483 + 0}}{\rm{.3483 = 0}}{\rm{.6966}}}\end{array}\)

(1): The only possible values for \({\rm{X}}\) are \({\rm{5,6,7,8,9}}\)and\({\rm{10}}\)

(2): the event \({\rm{\{ 7£ X£ 8\} }}\) is the union of two disjoint events \({\rm{\{ X = 7\} and\{ X = 8\} }}\)

(3): the \({\rm{pmf}}\)has been calculated in (a).