Given information:

Given that a geologist has collected 15 specimens of basaltic rock and 15 specimens of granite.

It is also given that geologist instructs an assistant to randomly select 25 of the specimens for the analysis.

(a). Find the pmf of the number of granite specimens selected for the analysis:

The total number of specimens is N= 30.

The selected sample size of specimens is n=25.

The number of granite specimens is M-15.

The pmf of the number of granite specimens selected for the analysis is obtained as given below:

\[P(X=x)=\frac{(\begin{array}{c}M\\ x\end{array})(\begin{array}{c}N-M\\ n-x\end{array})}{(\begin{array}{c}N\\ n\end{array})};x=0,1,2,...,14,15\]

Discrete probabilities:

The probability that 10 of the selected specimens are granite specimens is obtained as given below:

\[P(X=10)=\frac{(\begin{array}{c}15\\ 10\end{array})(\begin{array}{c}30-15\\ 25-10\end{array})}{(\begin{array}{c}30\\ 25\end{array})}\]

\(\displaystyle={\frac{{{a}{3003}\times{1}}}{{{142506}}}}\)

=0.0211

The probability that 11 of the selected specimens are granite specimens is obtained as given below:

\[P(X=11)=\frac{(\begin{array}{c}15\\ 11\end{array})(\begin{array}{c}30-15\\ 25-11\end{array})}{(\begin{array}{c}30\\ 25\end{array})}\]

\(\displaystyle={\frac{{{1365}\times{15}}}{{{142506}}}}\)

=0.1437

The probability that 12 of the selected specimens are granite specimens is obtained as given below:

\(\displaystyle{P}{\left({X}={12}\right)}={\frac{{{\left({b}{e}{g}\in{\left\lbrace{a}{r}{r}{a}{y}\right\rbrace}{\left\lbrace{c}\right\rbrace}{15}\backslash{12}{e}{n}{d}{\left\lbrace{a}{r}{r}{a}{y}\right\rbrace}\right)}{\left({b}{e}{g}\in{\left\lbrace{a}{r}{r}{a}{y}\right\rbrace}{\left\lbrace{c}\right\rbrace}{30}-{15}\backslash{25}-{12}{e}{n}{d}{\left\lbrace{a}{r}{r}{a}{y}\right\rbrace}\right)}}}{{{\left({b}{e}{g}\in{\left\lbrace{a}{r}{r}{a}{y}\right\rbrace}{\left\lbrace{c}\right\rbrace}{30}\backslash{25}{e}{n}{d}{\left\lbrace{a}{r}{r}{a}{y}\right\rbrace}\right)}}}}\)

\(\displaystyle={\frac{{{455}\times{105}}}{{{142506}}}}\)

=0.3325

The probability that 13 of the selected specimens are granite specimens is obtained as given below:

\[P(X=13)=\frac{(\begin{array}{c}15\\ 13\end{array})(\begin{array}{c}30-15\\ 25-13\end{array})}{(\begin{array}{c}30\\ 25\end{array})}\]

\(\displaystyle={\frac{{{105}\times{455}}}{{{142506}}}}\)

=0.3325

The probability that 14 of the selected specimens are granite specimens is obtained as given below:

\[P(X=14)=\frac{(\begin{array}{c}15\\ 14\end{array})(\begin{array}{c}30-15\\ 25-14\end{array})}{(\begin{array}{c}30\\ 25\end{array})}\]

\(\displaystyle={\frac{{{15}\times{1365}}}{{{142506}}}}\)

=0.1437

The probability that 15 of the selected specimens are granite specimens is obtained as given below:

\[P(X=15)=\frac{(\begin{array}{c}15\\ 15\end{array})(\begin{array}{c}30-15\\ 25-15\end{array})}{(\begin{array}{c}30\\ 25\end{array})}\]

\(\displaystyle={\frac{{{1}\times{3003}}}{{{142506}}}}\)

=0.0211

Given that a geologist has collected 15 specimens of basaltic rock and 15 specimens of granite.

It is also given that geologist instructs an assistant to randomly select 25 of the specimens for the analysis.

(a). Find the pmf of the number of granite specimens selected for the analysis:

The total number of specimens is N= 30.

The selected sample size of specimens is n=25.

The number of granite specimens is M-15.

The pmf of the number of granite specimens selected for the analysis is obtained as given below:

\[P(X=x)=\frac{(\begin{array}{c}M\\ x\end{array})(\begin{array}{c}N-M\\ n-x\end{array})}{(\begin{array}{c}N\\ n\end{array})};x=0,1,2,...,14,15\]

Discrete probabilities:

The probability that 10 of the selected specimens are granite specimens is obtained as given below:

\[P(X=10)=\frac{(\begin{array}{c}15\\ 10\end{array})(\begin{array}{c}30-15\\ 25-10\end{array})}{(\begin{array}{c}30\\ 25\end{array})}\]

\(\displaystyle={\frac{{{a}{3003}\times{1}}}{{{142506}}}}\)

=0.0211

The probability that 11 of the selected specimens are granite specimens is obtained as given below:

\[P(X=11)=\frac{(\begin{array}{c}15\\ 11\end{array})(\begin{array}{c}30-15\\ 25-11\end{array})}{(\begin{array}{c}30\\ 25\end{array})}\]

\(\displaystyle={\frac{{{1365}\times{15}}}{{{142506}}}}\)

=0.1437

The probability that 12 of the selected specimens are granite specimens is obtained as given below:

\(\displaystyle{P}{\left({X}={12}\right)}={\frac{{{\left({b}{e}{g}\in{\left\lbrace{a}{r}{r}{a}{y}\right\rbrace}{\left\lbrace{c}\right\rbrace}{15}\backslash{12}{e}{n}{d}{\left\lbrace{a}{r}{r}{a}{y}\right\rbrace}\right)}{\left({b}{e}{g}\in{\left\lbrace{a}{r}{r}{a}{y}\right\rbrace}{\left\lbrace{c}\right\rbrace}{30}-{15}\backslash{25}-{12}{e}{n}{d}{\left\lbrace{a}{r}{r}{a}{y}\right\rbrace}\right)}}}{{{\left({b}{e}{g}\in{\left\lbrace{a}{r}{r}{a}{y}\right\rbrace}{\left\lbrace{c}\right\rbrace}{30}\backslash{25}{e}{n}{d}{\left\lbrace{a}{r}{r}{a}{y}\right\rbrace}\right)}}}}\)

\(\displaystyle={\frac{{{455}\times{105}}}{{{142506}}}}\)

=0.3325

The probability that 13 of the selected specimens are granite specimens is obtained as given below:

\[P(X=13)=\frac{(\begin{array}{c}15\\ 13\end{array})(\begin{array}{c}30-15\\ 25-13\end{array})}{(\begin{array}{c}30\\ 25\end{array})}\]

\(\displaystyle={\frac{{{105}\times{455}}}{{{142506}}}}\)

=0.3325

The probability that 14 of the selected specimens are granite specimens is obtained as given below:

\[P(X=14)=\frac{(\begin{array}{c}15\\ 14\end{array})(\begin{array}{c}30-15\\ 25-14\end{array})}{(\begin{array}{c}30\\ 25\end{array})}\]

\(\displaystyle={\frac{{{15}\times{1365}}}{{{142506}}}}\)

=0.1437

The probability that 15 of the selected specimens are granite specimens is obtained as given below:

\[P(X=15)=\frac{(\begin{array}{c}15\\ 15\end{array})(\begin{array}{c}30-15\\ 25-15\end{array})}{(\begin{array}{c}30\\ 25\end{array})}\]

\(\displaystyle={\frac{{{1}\times{3003}}}{{{142506}}}}\)

=0.0211