# A geologist has collected 15 specimens of basaltic rock and 15 specime

A geologist has collected 15 specimens of basaltic rock and 15 specimens of granite. The geologist instructs a laboratory assistant to randomly select 25 of specimens for analysis.
a) What is the pmf of the number of granite specimens selected for analysis? (Round your probabilities to four decimal places.)
$\begin{array}{|c|c|}\hline x & & & & & & \\ \hline p(x) & & & & & & \\ \hline \end{array}$
b) What is the probability that all specimens of the two types of rock are selected for analysis? (Round your answer to four decimal places.)
c) What is the probability that the number of granite specimens selected for analysis is within 1 standard deviation of its mean value? (Round your answer to four decimal places.)

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Mason Hall
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
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amarantha41
(b). Find the probability that all specimens of one of the two types of rock are selected for analysis:
The probability that all specimens of one of the two types of rock are selected for analysis is obtained as given below:
$P(\begin{array}{c}Basaltic=15,\\ Granite=10\end{array})+P(\begin{array}{c}Basaltic=10,\\ Granite=15\end{array})=[\frac{(\begin{array}{c}15\\ 15\end{array})(\begin{array}{c}30-15\\ 25-15\end{array})}{(\begin{array}{c}30\\ 25\end{array})}+\frac{(\begin{array}{c}15\\ 10\end{array})(\begin{array}{c}30-15\\ 25-10\end{array})}{(\begin{array}{c}30\\ 25\end{array})}]$
$$\displaystyle={\frac{{{2}\times{\left({1}\times{3003}\right)}}}{{{142506}}}}$$
$$\displaystyle={2}\times{0.0211}$$
=0.0422
Thus, the probability that all specimens of one of the two types of rock are selected for analysis is 0.0422.