# Solved: A geologist has collected 10 specimens of basaltic rock and 10 specimens of granite. The geologist instructs a laboratory assistant

#### ByDr. Raju Chaudhari

Feb 25, 2021

A geologist has collected 10 specimens of basaltic rock and 10 specimens of granite. The geologist instructs a laboratory assistant to randomly select 15 of the specimens for analysis.

(a) What is the pmf of the number of granite specimens selected for analysis?
(b) What is the probability that all specimens of one of the two types of rock are selected for analysis?
(c) What is the probability that the number of granite specimens selected for analysis is within 1 standard deviation of its mean value?

#### Solution

The possible values of X are 5,6,7,8,9,10. (In order to have less than 5 of the granite, there would have to be more than 10 of the basaltic).

(a). The pmf of the number of granite specimens selected for analysis is

 \begin{aligned} P(X=x) & = \frac{\binom{10}{x}\binom{10}{15-x}}{\binom{20}{15}}\\ & \;\; x=5,6,7,8,9,10 \end{aligned}

 \begin{aligned} P(X=5) & = \frac{\binom{10}{5}\binom{10}{10}}{\binom{20}{15}}\\ &=0.0163 \end{aligned}

 \begin{aligned} P(X=6) & = \frac{\binom{10}{6}\binom{10}{9}}{\binom{20}{15}}\\ &=0.1354 \end{aligned}

 \begin{aligned} P(X=7) & = \frac{\binom{10}{7}\binom{10}{8}}{\binom{20}{15}}\\ &=0.3483 \end{aligned}

 \begin{aligned} P(X=8) & = \frac{\binom{10}{8}\binom{10}{7}}{\binom{20}{15}}\\ &=0.3483 \end{aligned}

 \begin{aligned} P(X=9) & = \frac{\binom{10}{9}\binom{10}{6}}{\binom{20}{15}}\\ &=0.1354 \end{aligned}

 \begin{aligned} P(X=10) & = \frac{\binom{10}{10}\binom{10}{5}}{\binom{20}{15}}\\ &=0.0163 \end{aligned}

The pmf is

x 5 6 7 8 9 10
P(x) .0163 .1354 .3483 .3483 .1354 .0163

(b) The probability that all specimens of one of the two types of rock are selected for analysis is

 \begin{aligned} P(\text{all 10 of one kind or the other}) & = P(X = 5) + P(X = 10)\\ & = .0163 + .0163 \\ & = .0326 \end{aligned}

(c)

 \begin{aligned} E(X) & = \frac{Mn}{N}\\ & = \frac{10*15}{20}\\ & = 7.5 \end{aligned}

 \begin{aligned} V(X) & = \dfrac{Mn(N-M)(N-n)}{N^2(N-1)}\\ & = .9869\\ \sigma_x &= \sqrt{V(X)}\\ & = 0.9934 \end{aligned}

$$\mu\pm \sigma = 7.5\pm .9934 = (6.5066, 8.4934)$$

The probability that the number of granite specimens selected for analysis is within 1 standard deviation of its mean value is

$$P( X = 7) + P( X = 8) = .3483 + .3483 = .6966$$