Question

Fish contaminated by a plant’s toxic discharge. Refer tothe U.S. Army Corps of Engineers’ study on the DDT contamination of fish in the Tennessee River (Alabama), Example 1.5 (p. 38). Part of the investigation focused on how far upstream the contaminated fish have migrated. (A fish is considered to be contaminated if its measured DDT concentration is greater than 5.0 parts per million.)

1. Considering only the contaminated fish captured from the Tennessee River, the data reveal thatof the fish are found between 275 and 300 miles upstream,are found 305 to 325 miles upstream, andare found 330 to 350 miles upstream. Use these percentages to determine the probabilities, P(275-300), P(305-325), and P(330-350).
2. Given that a contaminated fish is found a certain distance upstream, the probability that it is a channel catfish (CC) is determined from the data as$\mathbf{P}\left(\mathbf{CC}\mathbf{|}\mathbf{275}\mathbf{-}\mathbf{300}\right)\mathbf{=}\mathbf{.}\mathbf{775}$,$\mathbf{P}\left(\mathbf{CC}\mathbf{|}\mathbf{305}\mathbf{-}\mathbf{325}\right)\mathbf{=}\mathbf{.}\mathbf{77}$,and$\mathbf{P}\left(\mathbf{CC}\mathbf{|}\mathbf{330}\mathbf{-}\mathbf{350}\right)\mathbf{=}\mathbf{.}\mathbf{86}$. If a contaminated channel catfish is captured from the Tennessee River, what is the probability that it was captured 275–300 miles upstream?

Short answer

The probability that it was captured 275–300 miles upstream is 0.5162.

Step-by-step solution

Important formula

The Baye’s formula is

$\begin{array}{rcl}\mathbf{P}\left(\frac{{\mathbf{B}}_{\mathbf{i}}}{\mathbf{A}}\right)& \mathbf{=}& \frac{\mathbf{P}\mathbf{(}{\mathbf{B}}_{\mathbf{i}}\cap \mathbf{A}\mathbf{)}}{\mathbf{P}\mathbf{(}\mathbf{A})}\\ & \mathbf{=}& \frac{\mathbf{P}\mathbf{(}{\mathbf{B}}_{\mathbf{i}}\mathbf{)}\mathbf{P}\left(\frac{\mathbf{A}}{{\mathbf{B}}_{\mathbf{i}}}\right)}{\mathbf{P}\mathbf{(}{\mathbf{B}}_{\mathbf{1}}\mathbf{)}\mathbf{P}\left(\frac{\mathbf{A}}{{\mathbf{B}}_{\mathbf{1}}}\right)\mathbf{+}\mathbf{P}\mathbf{(}{\mathbf{B}}_{\mathbf{2}}\mathbf{)}\mathbf{P}\left(\frac{\mathbf{A}}{{\mathbf{B}}_{\mathbf{2}}}\right)\mathbf{+}...\mathbf{+}\mathbf{P}\mathbf{(}{\mathbf{B}}_{\mathbf{k}}\mathbf{)}\mathbf{P}\left(\frac{\mathbf{A}}{{\mathbf{B}}_{\mathbf{k}}}\right)}\\ & & \end{array}$

Determine the probabilities 

Now,

$\begin{array}{rcl}P\left(275-300\right)& =& 52\%\\ & =& 0.52\\ & & \end{array}$

$\begin{array}{rcl}P\left(305-325\right)& =& 39\\ & =& 0.39\\ & & \end{array}$

$\begin{array}{rcl}P\left(330-350\right)& =& 9\%\\ & =& 0.09\\ & & \end{array}$

The probability that it was captured 275–300 miles upstream.

Apply the Baye’s rule

$\begin{array}{rcl}P\left(275-300|CC\right)& =& \frac{\left(0.775\right)\left(0.52\right)}{\left(0.775\right)\left(0.52\right)+\left(0.77\right)\left(0.39\right)+\left(0.86\right)\left(0.09\right)}\\ & =& \frac{0.403}{0.7807}\\ & =& 0.5162\\ & =& 51.62\%\\ & & \end{array}$

Therefore, the probability is 0.5162.