Chapter 13.5, Problem 29E

To determine

To calculate: The value of probability $P\left(\text{salmon, given bass}\right)$ when three fish are removed from an aquarium.

Answer to Problem 29E

The value of probability $P\left(\text{salmon, given bass}\right)$ when three fish are removed from an aquariumis $\frac{2}{5}$ .

Explanation of Solution

Given information:

There are 6 fish in an aquarium namely a trout, a bass, a perch, a catfish, a walleye and a salmon and three fish at random are drawn.

Formula used:

Probability of an event Eis $P\left(\text{E}\right)=\frac{\text{number of outcome}}{\text{total number of outcome}}$ .

Given an event Q , the conditional probability of event A is $P\left(A|Q\right)=\frac{P\left(A\text{ and Q}\right)}{P\left(Q\right)}$ where $P\left(Q\right)\ne 0$ .

Calculation:

Consider the provided information thatthere are 6 fish in an aquarium namely a trout, a bass, a perch, a catfish, a walleye and a salmon and three fish at random are drawn.

The probability of $P\left(\text{salmon, given bass}\right)$ .

Let A be an event that removed fish is a salmon and B be an event that removed fish is a bass.

Use combination to evaluate the probability, recall that if there are n objects taken r at a time then combination is defined as $C\left(n,r\right)=\frac{n!}{r!\left(n-r\right)!}$ .

Recall that probability of an event Eis $P\left(\text{E}\right)=\frac{\text{number of outcome}}{\text{total number of outcome}}$ .

Probability that removed fish is a bass,

  $\begin{array}{c}P\left(B\right)=\frac{C\left(1,1\right)\cdot C\left(5,2\right)}{C\left(6,3\right)}\\ =\frac{\frac{5!}{2!3!}}{\frac{6!}{3!3!}}\\ =\frac{10}{20}\end{array}$

Probability that removed fish is salmon and bass,

  $\begin{array}{c}P\left(A\text{ and B}\right)=\frac{C\left(1,1\right)\cdot C\left(1,1\right)\cdot C\left(4,1\right)}{C\left(6,3\right)}\\ =\frac{\frac{4!}{1!3!}}{\frac{6!}{3!3!}}\\ =\frac{4}{20}\end{array}$

Recall that given an event Q , the conditional probability of event A is $P\left(A|Q\right)=\frac{P\left(A\text{ and Q}\right)}{P\left(Q\right)}$ where $P\left(Q\right)\ne 0$ .

The probability of $P\left(\text{salmon, given bass}\right)$ .

  $\begin{array}{c}P\left(A|B\right)=\frac{\frac{4}{20}}{\frac{10}{20}}\\ =\frac{4}{10}\\ =\frac{2}{5}\end{array}$

Thus, the value of probability $P\left(\text{salmon, given bass}\right)$ when three fish are removed from an aquarium is $\frac{2}{5}$ .