(a)
Let $X$=number of granite specimens selected for analysis. It obeys the hypergeometric distribution.
- The population or set to be sampled consists of $N=10+10=20$ individuals.
- Each individual can be characterized as a success ($S$, here means it is a speciman of granite) or a failure ($F$, here means it is a speciman of basaltic rock), and there are $M=10$ successes in the population.
- A sample of $n=15$ individuals is selected without replacement in such a way that each subset of size $n$ is equally likely to be chosen.
Then the pmf of $X$ is
\[
\begin{split}
P(X=x)&=h(x;n,M,N)=h(x;15,10,20)=\dfrac{\binom{M}{x}\cdot\binom{N-M}{n-x}}{\binom{N}{n}}\\
&=\dfrac{\binom{10}{x}\cdot\binom{20-10}{15-x}}{\binom{20}{15}}\\
&=\dfrac{\binom{10}{x}\cdot\binom{10}{15-x}}{15504}\\
\end{split}
\]
where $x$ satisfies $\max\{0,n-N+M\}\leqslant x\leqslant\min\{n,M\}$, i.e., $\max\{0,15-20+10\}\leqslant x\leqslant\min\{15,10\}$. That is, $5\leqslant x\leqslant 10$.
(b)
The event of all specimens of one of the two types of rock are selected for analysis means that there are 10 specimens of basaltic rock or 10 specimens of granite in the sample.
Hence, the probability is equal to twice of $P(X=10)$.
\[
2\cdot P(X=10)=2\times\dfrac{\binom{10}{10}\cdot\binom{10}{15-10}}{15504}\approx 0.03250774
\]
(c)
By the formulas of mean value and variance of hypergeometric rv $X$,
\[
\mu=E(X)=n\cdot\frac{M}{N}=n\times\frac{10}{20}=7.5
\]
Then, the probability that the number of granite specimens selected for analysis is within 1 standard deviation of its mean value is
\[
\begin{split}
P(\mu-1\leqslant X\leqslant\mu+1)&=P(7.5-1\leqslant X\leqslant 7.5+1)=P(6.5\leqslant X\leqslant 8.5)\\
&=P(X=7\ \text{or}\ X=8)=P(X=7)+P(X=8)\\
&=h(7;15,10,20)+h(8;15,10,20)\\
&=\dfrac{\binom{10}{7}\cdot\binom{10}{15-7}}{15504}+\dfrac{\binom{10}{8}\cdot\binom{10}{15-8}}{15504}\\
&=\frac{120\times 45+45\times 120}{15504}\\
&\approx 0.69659443
\end{split}
\]
Also, we can calculate the variance
\[
\begin{split}
V(X)&=\frac{N-n}{N-1}\cdot n\cdot\frac{M}{N}\cdot\biggl(1-\frac{M}{N}\biggr)\\
&=\frac{20-15}{20-1}\cdot 15\cdot\frac{10}{20}\cdot\biggl(1-\frac{10}{20}\biggr)\\
&=\frac{5}{19}\cdot 15\cdot\frac{1}{2}\cdot\frac{1}{2}\\
&\approx 0.98684211
\end{split}
\]
