Chapter 2, Problem 106SE

One method used to distinguish between granitic (G) and basaltic (B) rocks is to examine a portion of the infrared spectrum of the sun’s energy reflected from the rock surface. Let R₁, R₂, and R₃ denote measured spectrum intensities at three different wavelengths; typically, for granite R₁ < R₂ < R₃, whereas for basalt R₃ < R₁ < R₂. When measurements are made remotely (using aircraft), various orderings of the R_is may arise whether the rock is basalt or granite. Flights over regions of known composition have yielded the following information:

Suppose that for a randomly selected rock in a certain region, P(granite) 5 .25 and P(basalt) 5 .75.

1. a. Show that P(granite | R₁ < R₂ < R₃) > P(basalt | R₁ < R₂ < R₃). If measurements yielded R₁ < R₂ < R₃, would you classify the rock as granite or basalt?
2. b. If measurements yielded R₁ < R₃ < R₂, how would you classify the rock? Answer the same question for R₃ < R₁ < R₂.
3. c. Using the classification rules indicated in parts (a) and (b), when selecting a rock from this region, what is the probability of an erroneous classification? [Hint: Either G could be classified as B or B as G, and P(B) and P(G) are known.]
4. d. If P(granite) = p rather than .25, are there values of p (other than 1) for which one would always classify a rock as granite?

To determine

Show that $P\left(\text{Granite|}{R}_1<R_2<R_3\right)>P\left(\text{basalt|}{R}_1<R_2<R_3\right)$.

Explain whether the granite or basalt is classified as rock.

Answer to Problem 106SE

It has been verified that $\underset{\_}{P\left(\text{Granite|}{R}_1<R_2<R_3\right)>P\left(\text{basalt|}{R}_1<R_2<R_3\right)}$.

The rock is classified as granite.

Explanation of Solution

Given info:

The information is based on the method that helps to differentiate between the granite and basaltic rocks to examine between the infrared spectrums of the sun’s energy from the rock surface. The events $R_1,R_2\text{ and }{R}_3$ represents wavelength that helps to measure the intensities of the spectrum. The granite has the wavelength of $R_1<R_2<R_3$ and the basalt has the wave length of $R_3<R_1<R_2$. Here, the orderings of $R_i\text{'}s$ denotes whether the rock is granite or basalt.

The composition of the salts is obtained in the table given below:

	Granite	Basalt
$R_1<R_2<R_3$	60%	10%
$R_1<R_3<R_2$	25%	20%
$R_3<R_1<R_2$	15%	70%

Calculation:

Bayles’ rule:

If $A_1,A_2\text{,}\mathrm{...}\text{,}{A}_k$ are k mutually exclusive and exhaustive events, such that the sum of the probability values of the events is 1 and there is an observed event B, then,

$\begin{array}{c}P\left(A_j|B\right)=\frac{P\left(A_j\cap B\right)}{P\left(B\right)}\\ =\frac{P\left(A_j\right)P\left(B|A_j\right)}{P\left(A_1\right)P\left(B|A_1\right)+P\left(A_2\right)P\left(B|A_2\right)+\mathrm{...}+P\left(A_k\right)P\left(B|A_k\right)}\end{array}$

Here, $j=1,\mathrm{...},k$.

Law of total probability:

Assume that the eventsA₁, A₂, A₃,…,A_k be mutually exclusive and exhaustive events.

For any event B,

The probability of B is obtained as:

$P\left(B\right)=P\left(B|A_1\right)P\left(A_1\right)+P\left(B|A_2\right)P\left(A_2\right)+\mathrm{...}+P\left(B|A_k\right)P\left(A_k\right)$

Define the events:

$\begin{array}{l}G:\text{Granite}\\ B:\text{Basalt}\\ R_1\text{:Wave length 1}\\ R_2\text{:Wave length 2}\\ R_3\text{:Wave length 3}\end{array}$

The probability of the corresponding event is given below:

$P\left(\text{Granite}\right)=0.25$ and $P\left(\text{Basalt}\right)=0.75$

The probability of the granite and $R_1<R_2<R_3$ is obtained as follows:

$\begin{array}{c}P\left(\text{Granite}\cap R_1<R_2<R_3\right)=P\left(\text{Granite}\right)\times P\left(R_1<R_2<R_3|\text{Granite}\right)\\ =0.25\times 0.6\\ =0.15\end{array}$

The probability of the granite and $R_1<R_3<R_2$ is obtained as follows:

$\begin{array}{c}P\left(\text{Granite}\cap R_1<R_3<R_2\right)=P\left(\text{Granite}\right)\times P\left(R_1<R_3<R_2\text{|Granite}\right)\\ =0.25\times 0.25\\ =0.625\end{array}$

The probability of the granite and $R_3<R_1<R_2$ is obtained as follows:

$\begin{array}{c}P\left(\text{Granite}\cap R_3<R_1<R_2\right)=P\left(\text{Granite}\right)\times P\left(R_3<R_1<R_2\text{|Granite}\right)\\ =0.25\times 0.15\\ =0.0375\end{array}$

The probability of the basalt and $R_1<R_2<R_3$ is obtained as follows:

$\begin{array}{c}P\left(\text{Basalt}\cap R_1<R_2<R_3\right)=P\left(\text{Basalt}\right)\times P\left(R_1<R_2<R_3\text{|Basalt}\right)\\ =0.75\times 0.1\\ =0.075\end{array}$

The probability of the basalt and $R_1<R_3<R_2$ is obtained as follows:

$\begin{array}{c}P\left(\text{basalt}\cap R_1<R_3<R_2\right)=P\left(\text{basalt}\right)\times P\left(R_1<R_3<R_2|\text{Basalt}\right)\\ =0.75\times 0.2\\ =0.15\end{array}$

The probability of the basalt and $R_3<R_1<R_2$ is obtained as follows:

$\begin{array}{c}P\left(\text{basalt}\cap R_3<R_1<R_2\right)=P\left(\text{basalt}\right)\times P\left(R_3<R_1<R_2\text{|Basalt}\right)\\ =0.75\times 0.7\\ =0.525\end{array}$

Here, the probabilities of the events are represented using tree diagram given below:

By using law of total probability, the probability of the $R_1<R_2<R_3$ is,

$\begin{array}{c}P\left(R_1<R_2<R_3\right)=\left\{\begin{array}{l}P\left(\text{Granite }\cap R_1<R_2<R_3\right)+\\ P\left(\text{Basalt }\cap R_1<R_2<R_3\right)\end{array}\right\}\\ =0.15+0.075\\ =0.225\end{array}$

Thus, the probability of $R_1<R_2<R_3$ is 0.225.

The probability of $\left(\text{Granite | }{R}_1<R_2<R_3\right)$ is obtained as given below:

$\begin{array}{c}P\left(\text{Granite | }{R}_1<R_2<R_3\right)=\frac{P\left(\text{Granite }\cap R_1<R_2<R_3\right)}{P\left(R_1<R_2<R_3\right)}\\ =\frac{0.15}{0.225}\\ =0.67\end{array}$

Thus the $P\left(\text{Granite | }{R}_1<R_2<R_3\right)$ is 0.67.

The probability of $\left(\text{Basalt | }{R}_1<R_2<R_3\right)$ is obtained as given below:

$\begin{array}{c}P\left(\text{Basalt | }{R}_1<R_2<R_3\right)=\frac{P\left(\text{Basalt }\cap R_1<R_2<R_3\right)}{P\left(R_1<R_2<R_3\right)}\\ =\frac{0.075}{0.225}\\ =0.33\end{array}$

Thus the $P\left(\text{Basalt | }{R}_1<R_2<R_3\right)$ is 0.33.

Hence, the probability of $\underset{\_}{\left(\text{Granite | }{R}_1<R_2<R_3\right)}$ is greater than $\underset{\_}{\left(\text{Basalt | }{R}_1<R_2<R_3\right)}$, that is $\underset{\_}{\left(0.67>0.33\right)}$. The $P\left(\text{Granite | }{R}_1<R_2<R_3\right)>\frac{1}{2}$, so, the rock is classified as granite.

To determine

Explain whether the rock is classified as granite or basalt.

Answer to Problem 106SE

For both $R_1<R_3<R_2$ and $R_1<R_3<R_2$ the rock is classified as basalt.

Explanation of Solution

Calculation:

By using law of total probability, the probability of the $R_1<R_3<R_2$ is,

$\begin{array}{c}P\left(R_1<R_3<R_2\right)=\left\{\begin{array}{l}P\left(\text{Granite }\cap R_1<R_3<R_2\right)+\\ P\left(\text{Basalt }\cap R_1<R_3<R_2\right)\end{array}\right\}\\ =0.0625+0.15\\ =0.2125\end{array}$

Thus, the probability of $R_1<R_3<R_2$ is 0.2125.

The probability of $\left(\text{Granite | }{R}_1<R_3<R_2\right)$ is obtained as given below:

$\begin{array}{c}P\left(\text{Granite | }{R}_1<R_3<R_2\right)=\frac{P\left(\text{Granite }\cap R_1<R_3<R_2\right)}{P\left(R_1<R_2<R_3\right)}\\ =\frac{0.0625}{0.2125}\\ =0.2941\end{array}$

Here, the probability is less than 0.5. It is classified as basalt.

The probability of $\left(\text{Granite | }{R}_1<R_3<R_2\right)$ is obtained as given below:

$\begin{array}{c}P\left(\text{Granite | }{R}_1<R_3<R_2\right)=\frac{P\left(\text{Granite }\cap R_1<R_3<R_2\right)}{P\left(R_1<R_3<R_2\right)}\\ =\frac{0.0375}{0.2125}\\ =0.0667\end{array}$

Here, the probability is less than 0.5. It is classified as basalt.

To determine

Obtain the probability of an erroneous classification.

Answer to Problem 106SE

The probability of an erroneous classification is 0.175.

Explanation of Solution

Here, G denotes granite and B denotes basalt.

The probability of the erroneous classification is obtained as shown below:

$\begin{array}{c}P\left(\text{Erroneous classification}\right)=P\left(\begin{array}{l}\text{Either G classified as B or}\\ \text{ B classified as G}\end{array}\right)\\ =P\left(\text{G classified as B}\right)+P\left(\text{B classified as G}\right)\\ =\left\{\begin{array}{l}P\left(B\right)P\left(R_1<R_3<R_2\text{ or }{R}_1<R_3<R_2|\text{G}\right)+\\ P\left(G\right)P\left(R_1<R_3<R_2|\text{B}\right)\end{array}\right\}\\ =\left\{\begin{array}{l}P\left(B\right)P\left(\left(R_1<R_3<R_2|\text{G}\right)\text{+}\left(R_1<R_3<R_2|\text{G}\right)\right)+\\ P\left(G\right)P\left(R_1<R_3<R_2|\text{B}\right)\end{array}\right\}\end{array}$

$\begin{array}{l}=\left\{\begin{array}{l}P\left(B\right)\left[P\left(R_1<R_3<R_2|\text{G}\right)\text{+}P\left(R_1<R_3<R_2|\text{G}\right)\right]+\\ P\left(G\right)P\left(R_1<R_3<R_2|\text{B}\right)\end{array}\right\}\\ =\left(0.25\right)\left(0.25+0.15\right)+\left(0.75\right)\left(0.10\right)\\ =\left(0.25\right)\left(0.35\right)+\left(0.075\right)\\ =0.0875+0.075\\ =0.1625\end{array}$

Thus, the probability of the erroneous classification is 0.1625.

To determine

Identify for which value of p the rock can be classified as granite.

Answer to Problem 106SE

The rock can be classified as granite if and only if $p>\frac{14}{17}$.

Explanation of Solution

Calculation:

The probability of $\left(\text{Granite | }{R}_1<R_2<R_3\right)$ is obtained as given below:

$\begin{array}{c}P\left(\text{Granite | }{R}_1<R_2<R_3\right)=\frac{P\left(\text{Granite }\cap R_1<R_2<R_3\right)}{P\left(R_1<R_2<R_3\right)}\\ =\frac{0.6p}{0.6p+0.1\left(1-p\right)}\\ =\frac{0.6p}{0.6p+0.1-0.1p}\\ \frac{0.6p}{0.1+0.5p}>0.5\end{array}$

$\begin{array}{c}\frac{0.6p}{0.1+0.5p}>0.5\\ 0.6p>0.05\left(0.1+0.5p\right)\\ 0.6p>0.05+0.25p\\ \left(0.6-0.25\right)p>0.05\end{array}$

$\begin{array}{c}p>\frac{0.05}{0.35}\\ p>\frac{1}{7}\end{array}$

Thus, the $P\left(\text{Granite | }{R}_1<R_2<R_3\right)>0.5$ if and only if the $p>\frac{1}{7}$

The probability of $\left(\text{Granite | }{R}_1<R_3<R_2\right)$ is obtained as given below:

$\begin{array}{c}P\left(\text{Granite | }{R}_1<R_3<R_2\right)=\frac{P\left(\text{Granite }\cap R_1<R_3<R_2\right)}{P\left(R_1<R_3<R_2\right)}\\ =\frac{0.25p}{0.25p+0.2\left(1-p\right)}\\ =\frac{0.25p}{0.25p+0.2-0.2p}\\ \frac{0.25p}{0.2+0.05p}>0.5\end{array}$

$\begin{array}{c}\frac{0.25p}{0.2+0.05p}>0.5\\ 0.25p>0.5\left(0.2+0.05p\right)\\ 0.25p>0.1+0.025p\\ \left(0.25-0.025\right)p>0.1\end{array}$

$\begin{array}{c}p>\frac{0.1}{0.225}\\ p>\frac{4}{9}\end{array}$

Thus, the $P\left(\text{Granite | }{R}_1<R_3<R_2\right)>0.5$ if and only if the $p>\frac{4}{9}$

The probability of $\left(\text{Granite | }{R}_3<R_1<R_2\right)$ is obtained as given below:

$\begin{array}{c}P\left(\text{Granite | }{R}_3<R_1<R_2\right)=\frac{P\left(\text{Granite }\cap R_3<R_1<R_2\right)}{P\left(R_3<R_1<R_2\right)}\\ =\frac{0.15p}{0.15p+0.7\left(1-p\right)}\\ =\frac{0.15p}{0.15p+0.7-0.7p}\\ \frac{0.15p}{0.7+0.55p}>0.5\end{array}$

$\begin{array}{c}\frac{0.15p}{0.7+0.55p}>0.5\\ 0.15p>0.5\left(0.7+0.55p\right)\\ 0.15p>0.35+0.0275p\\ \left(0.15-0.225\right)p>0.7\end{array}$

$\begin{array}{c}p>\frac{0.7}{0.0125}\\ p>\frac{14}{17}\end{array}$

Thus, the $P\left(\text{Granite | }{R}_3<R_1<R_2\right)>0.5$ if and if the $p>\frac{14}{17}$

Here, $\frac{14}{17}$ is greater than other two probability values. Hence, the rock can be classified as granite if and only if $p>\frac{14}{17}$.